Class 12 Medical Physics - Electric Charges And Fields

Find an expression for the oscillation frequency of an electric dipole of dipole moment $$\vec{p}$$ and rotational inertia $$I$$ for small amplitudes of oscillation about its equilibrium position in a unit form electric field of magnitude $$E$$.

We know that,

(\tau =-pE\sin \theta)$$ captures the sense

as well as the magnitude of the effect.

That is is a restoring torque to bring the tilted dipole back to its aligned equilibrium position.

If the amplitude of the motion is small,

we may replace $$\sin \theta$$ with $$\theta$$ in radians.

Thus $$\tau \approx -pE\theta$$.

Since this exhibits a simple negative proportionality to the angle of rotation the dipole oscillates in simple harmonic motion,

like a torsional pendulum with torsion constant

$$\kappa=pE$$. The angular frequency $$\omega$$ is given by
$$\omega^{2}=\dfrac{\kappa}{I}=\dfrac{pE}{I}$$
where $$I$$ is the rotational inertia of the dipole. The frequency of oscillation is
$$f=\dfrac{\omega}{2\pi}=\dfrac{1}{2\pi}\sqrt{\dfrac{pE}{I}}$$

A point charge Q is placed at point O as shown in the figure. The potential difference VA-VB is positive. Is the charge Q negative or positive

We know that, $$ V=\dfrac{1}{4\pi \epsilon 0}\dfrac{Q}{r} $$
$$ \Rightarrow V\propto \dfrac{1}{r} $$
The potential due to a point charge decreases with increase of distance.
$$ V_{A}- V_{B} > 0 $$ $$ \Rightarrow V_{A}> V_{B} $$
Hence, the charge Q is positive.

The figure shows the electric field lines around three point charges $$A$$, $$B$$ and $$C$$. Which charges are positive?

Electric lines of forces emerge out from positive charge. So, $$A$$ and $$C$$ are positive charges.

During dry weather, clothes made of synthetic fiber often stick to the skin. Which type of force is responsible for this phenomenon?

Due to electrostatic force between the clothes and the body, the clothes of synthetic fiber stick to the body.

Why does a plastic comb rubbed with dry hair attract bits of paper?

Plastic comb rubber with dry hair gets charged and attracts the bits of paper.

What are the two kinds of charges?

The 2 kinds of charges are 1.Positive and 2.Negative

Name the physical quantity of which the units are:
Coulomb

Coulomb - Charge

Describe an experiment to demonstrate that there are two $$ r $$ kinds of charges.

Aim-To Show that there are two kind of charges.
Appratus - gold leaf electroscope, glass rod rubbed with silk cloth.
any two non-identical bodies say $$ A $$ and $$ B $$.
Procedure -
(i) Charge the gold leaf electroscope positively by touching it with glass rod rubbed with silk cloth.
(ii) The gold leaf electroscope shows its charging by showing its leaves diverging.
(iii) Touch the brass cap of this electroscope with body A-note the divergence of the gold leaves.
(iv) Now touch the brass cap with body band note the divergence.
Observation - In case of body A - leaves further diverge.
In case of body $$ B $$ 0 Leaves converge.
Inference - Body $$ A $$ (positively charged)
Body $$ B $$ ( negatively charged )
So there are two types of charges

How will you show that like charges repel and unlike charges attract each other?

1.Rub two inflated balloons with a woolen cloth and hold the balloons together closely. They will to spring apart.

2.Rub two pieces of nylon on a sheet of paper and then hold them together closely. Again, they will try to spring apart.

3.Now, hold one balloon and one piece of nylon together closely. This time, the nylon clings to the balloon.

What is the condition of an ideal electric dipole?

We can think of an ideal dipole in which size $$2α → 0$$ and charge $$q → \infty$$ in such a way that the dipole moment, $$p = q \times 2\alpha$$ a has a finite value, such a dipole of negligibly small size is called an ideal or point dipole.

On what factors does the constant $$k =\dfrac{1}{4\pi \epsilon_0}$$ depends?

Constant $$k =\dfrac{1}{4\pi \epsilon_0}$$ depends on the medium and system of measurement.

State the Coulombs law for the force acting between two definite point charges.

According to Coulomb’s law, “the force of interaction between two point charges is directly proportional to the product of the magnitude of charges and inversely proportional to the square of the separation distance between them.”

For vacuum or air,$$F=\dfrac{1}{4\pi \epsilon_0}\,\dfrac{q_1q_2}{r^2}$$

Can electric potential be zero at a point in a vacuum, the electric field at that point is not zero. Give example.

Yes, it is possible.
Example:
1. Between line joining electric dipole
2. On equatorial line of an electric dipole.

To determine the electric field due to a point charge, the test charge should be small. Explain why?

The magnitude of the test charge must be small enough so that it does not disturb the distribution of the charges whose electric field we wish to measure otherwise the measured field will be different from the actual field.

An electric dipole is placed in a uniform electric field. Show that it will not accelerate.

In the uniform electric field on the electric dipole, two equal forces will act in opposite direction, which produces a torque. Therefore electric dipole will not perform translatory accelerating motion.

What is the electric field lines? Write any two of its properties.

An electric field line is an imaginary line or curve drawn from a point of an electric field such that tangent to it (at any point) gives the direction of the electric field at that point.

Properties :

1. Electric field lines starting from positive charge and terminating at a negative charge.

2. The number of field lines that start or end at a charge is directly proportional to the magnitude of the charge.

In given fig., an equilateral triangle ABC of side a has charges $$+2q, - q$$ and $$-q$$ at the vertices A, B, and C respectively. Calculate the dipole moment of this system.

Give arrangement is equivalent to two electric dipoles AB and AC, which are at 60° to each other.

The dipole moment of electric dipole AB

$$P_1 = qa$$ (along BA)

The dipole moment of electric dipole CA

$$p_2 = qa$$ (along CA)

Resultant of electric dipole moment of $$p_1$$ and $$p_2$$

$$P_R=\sqrt{{p_1}^2+{p_2}^2+2p_1p_1\cos \theta}$$

$$=\sqrt{(qa)^2+(qa)^2+2(qa)(qa)\cos \theta}$$

$$=\sqrt{2(qa)^2+2(qa)^2 \times \dfrac{1}{2}}=\sqrt{3}qa$$

The direction of electric dipole moment PR is along half of the angle between $$p_1$$ and $$p_2$$ due to the equal vectors.

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Explain why two field lines never cross each other at any point?

The tangents at a point on the line of force given the direction of electric field at that point. It two lines of force intersect each other at a point, then electric field at that point will have two directions. As the same cannot be true, tow lines of force will have two directions. As the same cannot be true, two lines of force will never intersect each other.

The electric field components due to a charge inside the cube of side 0.1 m are shown in figure.
where, $$E_x = \alpha$$, where $$\alpha$$ = 500 N/C-m, $$E_y$$ = 0, $$E_z = 0$$.
Calculate
(a) the flux through the cube and
(b) the charge inside the cube.

Given length of the cube=0.1 m

$$E_y=0,\space E_z=0,\space E_x=500\dfrac{N}{C-m}$$

the flux through the cube$$\psi=\int E.ds=\int (E_xi+E_yj+E_zk).ds=\int E_Xi.ds=E_xx=\alpha x=500x$$

$$\psi={(E_x.A)}_{t=0.1}+{(E_x.A)}_{x=0.2}=E_xAcos180+E_xcos0=500\times0.01\times (-1)+500\times 0.2\times 0.01=-0.5+1=0.5\dfrac{N}{Cm^2}$$

charge inside the cube$$\psi = \int E.ds=\dfrac{q}{E_0}\implies q=0.5\times E_0=0.5\times 8.85\times 10^{-12}=4.425\times 10^{-12}C$$

A hollow cylinder has a charge q coulomb within it. If $$\phi $$ is the electric flux in units of volt-meter associated with the curved surface B, the flux linked with the plane surface A in units of volt-meter will be?

Here, let the flux on A, B, C surfaces be $$\phi_A$$, $$\phi_B$$ and $$\phi_C$$

Then, $$\phi_n$$$$_e$$$$_t$$= $$\phi_A$$ + $$\phi_B$$ + $$\phi_C$$ = $$\dfrac{q}{\epsilon_0}$$

So, 2$$\phi_A$$ + $$\phi$$ = $$\dfrac{q}{\epsilon_0}$$

$$\phi_A$$ = $$\dfrac{1}{2}$$ ($$\dfrac{q}{\epsilon_0}$$ - $$\phi$$)

Hence the electric flux through the plane surface is $$\dfrac{1}{2}$$ ($$\dfrac{q}{\epsilon_0}$$ - $$\phi$$)

The figure shows the electric field lines around a conductor.
(a) At which of the four points is the field strongest?
(b) At which point is there a negative charge?

a) At the sharp turning, field strength will be high. At point D sharpness of curve is more in comparison with another point , So Point D will be the answer.

b) field lines converge to a negative charge and diverges to a positive charge.

Hence $$D$$ has (-)ve charge

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In figure, particle $$1$$ ( of charge $$+1.00\mu C$$), particle $$2$$ ( of charge $$+1.00\mu C$$), and particle $$3$$ ( of charge $$Q$$) form an equilateral triangle of edge length $$a$$. For what value of $$Q$$ ( both sign and magnitude) does the net electric field produced by the particles at the centre of the triangle vanish?

On the other hand, the conclusion ( that $$Q=+1.00\mu C)$$ is clear from symmetry. If a more in-depth justification is desired, for the electric field magnitudes of the three charges each at same distance

$$r=a/ \sqrt 3$$ from $$C$$ ) and then find field components along suitably chosen axes, requiring each component

-sum along that axis leads to $$2\ kq\sin 30^o /r^2=kQ/ r^2$$ where

$$q$$ refers to either of the charges at the bottom corners.

The yields $$Q=2q\sin 30^o =q$$ and thus to the conclusion mentioned above.

Given a uniform electric field $$\overrightarrow E = 2 \times 10^3 \widehat i$$N/C. Find the flux of this field through a square of side 20 cm, whose plane is parallel to the y-z plane. What would be the flux through the same square, if the plane makes an angle of $$30^o$$ with the x-axis?

$$\overrightarrow { E } =2\times { 10 }^{ 3 }\hat { i } \left( \frac { N }{ C } \right) \\ Flux=\overrightarrow { E } .\overrightarrow { S } \quad =EScos\theta $$

When plane is parallel to Y-Z plane then $$\theta =0$$

Flux =$$ES=2\times { 10 }^{ 3 }\times { (20\times { 10 }^{ -2 }) }^{ 2 }\\ \quad \quad =2\times { 10 }^{ 3 }\times 4\times { 10 }^{ -2 }=80$$

when plane makes angle 30 degree with X axis then perpendicular to plane makes angle 60 degree to X axis

hence

$$\theta ={ 60 }^{ 0 }\\ Flux=ES=2\times { 10 }^{ 3 }\times { (20\times { 10 }^{ -2 }) }^{ 2 }\times cos{ 60 }^{ 0 }\\ \quad \quad $$

=$$2\times { 10 }^{ 3 }\times 4\times { 10 }^{ -2 }\times \frac { \ { 1 } }{ 2 } =40\ $$

State Guass law in electrostatics. Derive an expression of electric field due to an infinitely long straight uniformly charged wire. Draw necessary diagram.

The Gauss law states that electric flux passing through any closed surface is equal to the charge enclosed by that surface divided by permittivity of vacuum.

By symmetry, the magnitude of the electric field will be the same at all points on the curved surface of the cylinder and directed radially outward. $$\vec E$$ and $$\vec {ds}$$ are along the same direction.

Now here we have the two surfaces, one curved and other the plane caps,

First, the flux through the curved surface,

$$\oint \vec E\cdot \vec {ds} = \dfrac {q_{in}}{\epsilon _0}$$

$$E(2\pi rl)=\lambda l/\epsilon _0$$

$$E= \dfrac{\lambda}{2\pi r\epsilon _0}$$

Now due to the plane caps,

The angle between $$\vec E$$ and $$\vec {ds}$$ is 90,

so the flux through that part is zero

so, Total flux through the closed surface is,

$$E= \dfrac{\lambda}{2\pi r\epsilon _0}$$

What is an electric dipole? Define electric dipole moment?

An electric dipole is a separation of positive and negative charges.
The electric dipole moment is the product of the magnitude of the charge and the distance between the centres of positive and negative charges.
$$\mu = Q \times r $$
Here, $$Q$$ is charge and $$r$$ is the distance.

Obtain the expression for the intensity of electric field near a uniformly charged straight wire of infinite length with the help of Gauss theorem.

Consider a long straight wire which carries the uniform charge per unit length $$\lambda$$ We expect the electric field generated by such a charge distribution to possess cylindrical symmetry. We also expect the field to point radially (in a cylindrical sense) away from the wire (assuming that the wire is positively charged).

Let us draw a cylindrical gaussian surface, co-axial with the wire, of radius $$R$$ and length $$L$$ see given fig. The above symmetry arguments imply that the electric field generated by the wire is everywhere perpendicular to the curved surface of the cylinder. Thus, according to Gauss' law,

$$E_R 2\pi RL=\dfrac{\lambda L}{\epsilon_o}$$,

where $$E_R$$ is the electric field-strength a perpendicular distance $$R$$ from the wire. Here, the left-hand side represents the electric flux through the gaussian surface. Note that there is no contribution from the two flat ends of the cylinder, since the field is parallel to the surface there. The right-hand side represents the total charge enclosed by the cylinder, divided by $$\epsilon_o$$. It follows that

$$\therefore E_R= \dfrac{\lambda}{2\pi \epsilon_oR}$$

The field points radially (in a cylindrical sense) away from the wire if $$\lambda>0$$ and radially towards the wire if $$\lambda<0$$.

Write Gauss's theorem of electrostatics. Obtain an expression for the intensity of electric field near a uniformly charged straight wire of infinite length with its help.

Explanation:

i) To write Gauss theorem of electrostatics.

According to Gauss theorem, the total electric flux ($$\phi$$) through any closed surface ($$S$$) in free space is equal to $$\dfrac{1}{\varepsilon _{0}}$$ times the total electric charge ($$q$$) enclosed by the surface, i.e, $$\phi=\oint_{S}^{ }\overrightarrow{E}.\overrightarrow{dS}=\dfrac{q}{\varepsilon _{0}}$$

ii) To find electric field intensity due to an infinitely long straight uniformly charged wire or line.

Explanation: Consider an infinite and very thin straight wire having uniform linear charge density $$\lambda$$. The wire is symmetrical about linear axis. To calculate the electric field $$E$$ at point $$P$$, distance $$r$$ from the line draw an imaginary cylinder (Gaussian surface) of radius $$r$$ and length $$l$$ around the charged line. The charge enclosed by the gaussian surface, $$q=\lambda l$$.

According to Gauss' Theorem,

$$\oint \overrightarrow{E}.\overrightarrow{dS}=\dfrac{q}{\varepsilon _{0}}=\dfrac{\lambda l}{\varepsilon _{0}}$$ ...(i)

The cylindrical gaussian surface is divided into three parts I, II, III.

Therefore, equation (i) can be written as,

$$\oint_{S}^{ }\overrightarrow{E}.\overrightarrow{dS}=\oint_{I}^{ }\overrightarrow{E}.\overrightarrow{dS}+\oint_{II}^{ }\overrightarrow{E}.\overrightarrow{dS}+\oint_{III}^{ }\overrightarrow{E}.\overrightarrow{dS}=\dfrac{\lambda l}{\varepsilon _{0}}$$ ...(ii)

For surfaces I and II, angle between $$\overrightarrow{E}$$ and $$\overrightarrow{dS}$$ is $$90^{\circ}$$, so $$\overrightarrow{E}.\overrightarrow{dS}= EdS\cos{\theta}=EdS\cos{90^{\circ}}=0$$ for those surfaces.

Equation (ii) becomes, $$\oint_{S}^{ }\overrightarrow{E}.\overrightarrow{dS}=\oint_{III}^{ }\overrightarrow{E}.\overrightarrow{dS}=\dfrac{\lambda l}{\varepsilon _{0}}$$

or,

$$\oint_{III}^{ }\overrightarrow{E}.\overrightarrow{dS}=\dfrac{\lambda l}{\varepsilon _{0}}$$ or $$\int E.dS\cos{0^{\circ}}=\dfrac{\lambda l}{\varepsilon _{0}}$$

Since $$E$$ is constant over the gaussian surface,

$$\therefore \: E\int_{III}^{ }dS=\dfrac{\lambda l}{\varepsilon _{0}}$$

$$\int_{III}^{ }dS=$$ area of curved surface of cylinder $$=2\pi rl$$

$$E\times 2\pi rl=\dfrac{\lambda l}{\varepsilon _{0}}\: \: \Rightarrow \:\: E=\dfrac{\lambda l}{2\pi\varepsilon _{0}r}\widehat{n}$$

where $$\widehat{n}$$ is a unit vector perpendicular to the current surface of the wire.

A charge $$Q$$ is uniformly distributed over a rod of length $$l$$. Consider a hypothetical cube of edge $$l$$ with the centre of the cube at one end of the rod. Find the minimum possible flux of the electric field through the entire surface of the cube.

Gauss law, $$\phi =\dfrac { { q }_{ enclosed } }{ { \varepsilon }_{ 0 } } $$

As half length is inside the cube charge enclosed by cube is $$Q/2$$

$$\phi=\dfrac{Q}{2\varepsilon_0}$$

Figure shows tracks of three charged particles in a uniform electrostatic field. Given the signs of the three charges. Which particle has the highest charge to mass ratio.

Look carefully at the path which the 3 particles have followed, before entering the field and after leaving it.

You can see that particle 3 has deviated the most.

By coulomb's law, the particle with highest charge will experience the greatest force, while passing through an electric filed.

As per Newton's laws of motion, lighter particle will deviate the most because it will undergo greater acceleration.

Thus we can say that $$deviation$$ proportional to $$\cfrac{charge}{mass}$$

Hence the particle with the highest $$\cfrac{charge}{mass}$$ ratio will undergo maximum deviation.

Hence $$3$$ has the maximum $$\cfrac{charge}{mass}$$ ratio.

State and explain coulomb's inverse square law

Coulomb's low of inverse square states that force between two charges proportionally depend on the product of magnitude of charges and inversely dependent on the square of distance between those charges.

i.e. $$F\alpha { q }_{ 1 }{ q }_{ 2 }\quad \& \quad F\alpha \dfrac { 1 }{ { r }^{ 2 } } $$

$$\Rightarrow \quad F\alpha \dfrac { { q }_{ 1 }{ q }_{ 2 } }{ { r }^{ 2 } } $$

Substituting constant,

$$\vec { F } =\dfrac { K{ q }_{ 1 }{ q }_{ 2 } }{ { r }^{ 2 } } $$

Now, Significance :- Product at both the charges implies that the force is dependent on both of charges not only one charge.

Also, $${ r }^{ 2 }$$ signifies about symmetry,i.e. in any direction, for ex . $$x=+S$$ & $$x=-S$$ will have some force. which is called Symmetry.

Valence band

The valence band is the band of electron orbitals that electrons can jump out of, moving into the conduction band when excited. The valence band is simply the outermost electron orbital of an atom of any specific material that electrons actually occupy. This is closely related to the idea of the valence electron.

The energy difference between the highest occupied energy state of the valence band and the lowest unoccupied state of the conduction band is called the band gap and is indicative of the electrical conductivity of a material A large band gap means that a lot of energy is required to excite valence electrons to the conduction band. Conversely, when the valence band and conduction band overlap as they do in metals, electrons can readily jump between the two bands) meaning the material is highly conductive.

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An infinite line charge produces an electric field of $$9.0\times 10^4$$N $$C^{-1}$$ at a distance of $$2.0$$cm. What is the linear charge density?

$$E=\dfrac { λ }{ (2πϵor) } =\dfrac { 2λ }{ (4πϵor) } $$

$$E=\dfrac { 9\times 104N }{ C } $$

$$\dfrac { 1 }{ (4πϵo) } =\dfrac { 9\times 109Nm^2 }{ C^2 } $$

Linear charge density$$λ=\dfrac { E.4πϵor }{ 2 } $$

$$=\dfrac { (9\times 10^4\dfrac { N }{ C } )(2\times 10^{ - }2) }{ \left( (9\times 109\dfrac { { Nm }^{ 2 } }{ { C }^{ 2 } } )\times 2 \right) } $$

$$=10-7\dfrac { C }{ m } $$

A small test charge is released at rest at a point in an electrostatic field configuration. Will it travel along the line of force?

If a small test charge is released at rest at a point in an electrostatic field configuaration, then it will travel along the field lines passing through the point, only if the field lines are straight. This is because the field lines give the direction of acceleration and not of velocity.

If an electric field is given by $$10\hat i + 3\hat j + 4\hat k,$$ calculate the electric flux through a surface of area $$10$$ units lying in $$yz$$ plane

$$\vec{E}= 10 \hat{i}+ 3 \hat{j}+ 4\hat {k}$$

Since, we have surface area of unit $$10$$ lying in $$Y_2$$ plane.\perp^r$$ to [lane $$Y_z$$]

Hence, the area vector $$

$$\vec{S}= 10 \hat{i}$$

We know that

flux $$(\phi )$$ is given as

$$\phi = \int \vec {E}\cdot \vec{ds}$$

$$\phi = \vec{E}\cdot \vec{S}$$

$$ = (10\hat{i}+3\hat{j}+4\hat{k})\cdot (10 \hat{i})$$

$$\phi = 100 \, N m^2c^{-1}$$

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Find the dimensions of permittivity of vacuum(\in_0).

Force $$F= \dfrac{1}{4\pi \epsilon_o }\dfrac{q_1q_2}{r^2}$$

So, $$\epsilon_o = \dfrac{q_1q_2}{4\pi r^2 F}$$

Dimensions of $$\epsilon_o = \dfrac{[AT][AT]}{[L^2][MLT^{-2}]} = [ML^{-3}A^{2}T^4]$$

In electric dipole what is the locus of Zero potential?

The locus of the point of zero potential of an electrical dipole is straight line drawn at the middle of the electric dipole$$.$$

NaCI molecule is bound due to the electric force between the sodium and the chlorine ions when one electron of sodium is transferred to chlorine. Taking the separation between the ions to be $$2.75\times10^{-8}$$cm, find the force of attraction between them. State the assumptions (if any) that you have made.

Let the given separation between the ions be $$2.75 \times {10^{ - 8}}\,cm$$

$${q_1} = 1.6 \times {10^{ - 19}}C$$

$${q_2} = 1.6 \times {10^{ - 19}}C$$

$$F = k{q_1}{q_2}/{r^2}$$

$$F = 1.6 \times {10^{ - 19}}C \times 1.6 \times {10^{ - 19}}C \times {\left( {2.75 \times {{10}^{ - 8}}} \right)^2}$$

$$F = 3.05 \times {10^{ - 9}}N$$

We get, $$F = 3.05 \times {10^{ - 9}}N.$$

What is the direction of electric field $$E$$ at point $$P$$ in the following diagram

Electric field moves towards the lower potential and perpendicular to the equi-potential surface.

So, at point P, electric field moves towards left perpendicular to the $$-10 \ V$$ surface.

Why do electric lines of force never intersect each other?

Electric lines of force never intersect each other because at the point of intersection$$,$$ two tangents can be drawn to the two lines of force$$.$$ This means two direction of electric field at the point of intersection$$,$$ which is not possible$$.$$

What is electric dipole moment state its SI unit and dimension.

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A person standing on an insulating stool touches a charge insult conductor . will the conductor get complete discharged ?

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State Coulomb's law of force in electrostatics.

Coulomb's law states that: The magnitude of the electrostatic force of attraction or repulsion between two point charges is directly proportional to the product of the magnitudes of charges and inversely proportional to the square of the distance between them.

Depict the orientation of the dipole moment. Is it a scalar or a vector quantity? What are its $$SI$$ unit?

The electric dipole moment is defined as the product of either charge and the distance between the two charges. Its direction is from negative to positive charge.
i.e., $$|p| = q(2l)$$
Electric dipole moment is a vector quantity.
Its $$SI$$ unit is coulomb-metre.
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Find the magnitude of the electric field at a point $$4 cm$$ away from a line charge of density $$2 \times 10^{-6} Cm^{-1}$$.

Drawing an electric field around the line charge we find a cylinder of radius $$4\times 10^{-2}m$$.

Given : $$\lambda = $$ linear charge density

Let the length be $$\ell = 2\times 10^{-6}c/m$$

We know $$\oint E \ dl = \dfrac{Q}{\varepsilon_0} = \dfrac{\lambda l}{\varepsilon_0}$$

$$\Rightarrow E \times 2\pi r \ell = \dfrac{\lambda \ell}{\varepsilon_0} $$

$$\Rightarrow E = \dfrac{\lambda}{\varepsilon_0 \times 2\pi r}$$

For, $$r = 2\times 10^{-2}m$$ & $$\lambda = 2\times 10^{-6}c/m$$

$$\Rightarrow E = \dfrac{2\times 10^{-6}}{8.85\times 10^{-12} \times 2\times 3.14 \times 2 \times 10^{-2}} = 8.99\times 10^5 N/C \approx 9\times 10^5N/C$$

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Calculate the number of coulombs of positive charge in $$250\ cm^{3}$$ of (neutral) water. (Hint: A hydrogen atom contains one pro eight protons.)

The volume of $$250\ cm^{3}$$ corresponds to a mass of $$250\ g$$ since the density of water is $$1.0\ g/cm^{3}$$. This mass corresponds to $$250/18=14$$ moles since the molar mass of water is $$18$$. There are ten proton (each with charge $$q=+e$$) in each molecule of $$H_{2}O$$, so
$$Q=14N_{A}q=14(6.02\times 10^{23})(10)(1.60\times 10^{-19}C)=1.3\times 10^{7}C$$.

We know that the negative charge on the electrons and the positive charge on the proton are equal. Suppose, however, hat these magnitudes differ from each other by $$0.00010 \%$$. With what force would two copper coins, placed $$1.0\ m$$ apart, repel reach other? Assume that each coin contains $$3 \times 10^{22}$$ copper atoms. (Hint: A neutral copper atom contains $$29$$ protons and $$29$$ electrons.) What do you conclude?

If the relative difference between the proton and electron charges (in absolute value)were$$\dfrac{q_p-|q_e|}{e}=0.0000010$$
then the actual difference would be $$q_p - |q_e|{e}=1.6 \times 10^{-25}\ C$$. Amplified by a factor of $$29 \times 3 \times 10^{22}$$ as indicated in the problem, this amounts to a deviation from perfect neutrality of$$\triangle q=(29 \times 3 \times 10^{22})(1.6 \times 10^{-25}\ C)=0.14\ C$$
in a copper penny. Two such pennies at $$r=1.0\ m$$, would therefore experience a very large force. Eq. gives$$F=k\dfrac{(\triangle q)^2}{r^2}=1.7 \times 10^{8}\ N$$.

Calculate the electric dipole moment of an electron and a proton $$4.30\ nm$$ apart.

The magnitude of the dipole moment is given by $$p=qd$$,

where $$q$$ is the positive charge in the dipole

and $$d$$ is the separation of the charges.

For the dipole described in the problem,
$$p=(1.60\times 10^{-19}C)(4.30\times 10^{-9}m)$$

$$=6.88\times 10^{-28}C.m$$.
The dipole moment is a vector that points from the negative toward the positive charge.

A certain electric dipole is placed in a uniform electric field $$\vec{E}$$ of magnitude $$20\ N/C$$. Figure $$22-62$$ gives the potential energy $$U$$ of the dipole versus the angle $$\theta$$ between $$\vec{E}$$ and the dipole moment $$\vec{p}$$. The vertical axis scale is set by $$U_{s}=100\times 10^{-28}J$$. What is the magnitude of $$\vec{p}$$?

Examining the lowest value on the graph,

we have,
$$U=-\vec{p}.\vec{E}$$

$$=-1.00\times 10^{-28}J$$
If $$E=20\ N/C$$,

we find $$p=5.0\times 10^{-28}C.m$$

A certain dipole is placed in a uniform electric field $$\vec{E}$$ of magnitude $$40\ N/C$$. Figure gives the magnitude $$\tau$$ of the torque on the dipole versus the angle $$\theta$$ between field $$\vec{E}$$ and the dipole moment $$\vec{p}$$. The vertical axis scale is set by $$\tau_{s}=100\times 10^{-28}N.m$$. What is the magnitude of $$\vec{p}$$?

We now, consider

$$\theta$$ as a variable,

we note that it reaches its maximum value when $$\theta=-90^{o}:\tau_{\max}=pE$$.

Thus with $$W=40\ N/C$$ and $$\tau_{\max}=100\times 10^{-28}N.m$$ (determined from the graph),

we obtain the dipole moment: $$p=2.5\times 10^{-28}C.m$$

If a cat repeatedly rubs against your cotton slacks on a dry day, the charge transfer between the cat hair and the cotton can leave you with an excess charge of $$-2.00 \mu C$$.
Just before the spark appears, do you induce positive or negative charge in the faucet?

Unlike charge attract, and the faucet (which is grounded and is able to gain or lose any number of electrons due to its contact with Earth's large reservoir of mobile charges)becomes positively charged, especially in the region closest to your (negatively charge)hand, just before the spark.

An electric dipole consisting of charges of magnitude $$1.50\ nC$$ separated by $$6.20\ \mu m$$ is in an electric field of strength $$1100\ N/C$$. What are
the magnitude of the electric dipole moment

The magnitude of the dipole moment is
$$p=qd$$

$$=(1.50\times 10^{-9}C)(6.20\times 10^{-6}m)$$

$$=9.30\times 10^{-15}C.m$$

Calculate the solid angle subtends by the periphery of an area $$1 cm^2$$ at a point situated symmetrically at a distance of $$5 cm$$ from the area.

$$solid\, angle\, \Omega = \dfrac{Area}{(dist)^2}$$

$$ = \dfrac{1 cm^2}{(5 cm)^2} = \dfrac{1}{25} = 4 \times 10^{-2} steradian.$$

Answer the following questions.
(i) Write $$SI$$ unit of electric field intensity.
(ii) Write $$SI$$ unit of electric dipole moment.

(i) $$SI$$ unit of electric field intensity is newton/ coulomb $$(NC^{-1})$$.

(ii) $$SI$$ unit of electric dipole moment is $$coulomb \times metre (C\ m)$$.

A hollow half cylinder surface of radius R and length l is placed in a uniform electric field $$\vec{E}$$. Electric field is acting perpendicular on the ABCD. If the flux through the curved surface of the hollow cylindrical surface is $$E\times X Rl$$. Find X?

Flux=$$\phi=\vec{E}.\vec{\Delta S}$$

Hence, flux =$$\phi=E\times$$ projection of area normal to the direction of field so that area vector can be along $$\vec{E}$$.

Here, projection of half cylinder perpendicular to field E gives a rectangular area ABCD of area $$AB\times AD$$

Hence, flux$$=\phi=E\times (2R\times l)$$

$$\implies \phi=E\times 2Rl$$

Hence, $$X=2$$

$$Match \: the \: statements \: in \: Column \: A, \: with \: those \: in \: Column \: B.$$

A body is neutral when the amount of positive and negative charge on it is equal.
When there is deficiency of electron, the overall resultant charge on the body is positive.
When there is excess of electron, the overall resultant charge on the body is negative.
Electrostatic force of attraction occurs between 2 oppositely charged bodies but it may also occur between a charged and an uncharged body due to induced charge. While electrostatic force of repulsion occur only between two like charges i.e. both bodies must be charged. Thus, repulsion is the surest test for electrification.
The electrons in the outermost shell are loosely bound and are called free electrons.
Silver is the best conductor of electricity.

If the electric field to the left of two sheets is $$K\sigma/\varepsilon_0$$. Find K?

to the left of infinite plane $$\sigma$$, $$E= \frac{ \sigma}{2 \epsilon}$$(towards left due to positive charge plane)$$-\frac{ \sigma}{2 \epsilon}$$(towards right due to negative charge plane) $$ =0 $$.Electric field of both the infinite planes cancel each other.

Figure shows some of the electric field lines due to three point charges $$q_1, q_2$$ and $$q_3$$ of equal magnitude. What are the signs of each of the three charges?

Electric field lines go away from charge $$ q_1 and q_3 $$, therefore these charges are positive.
field line come towards the charge $$ q_2$$ , therefore this charge is negative.

State and explain Coulomb's law in electrostatics.

Coulomb's law in electrostatic states that a charge at rest $$q_1$$ applies a force $$F$$ on the other charge $$q_2$$, also at rest, which are separated by a distance $$r$$ such that the force is directly proportional to the product of both charges and inversely proportional to the square of the distance between them.

$$\therefore$$ $$F \propto q_1 q_2$$ and $$F \propto \dfrac{1}{r^2}$$

Thus we get $$F \propto \dfrac{q_1q_2}{r^2}$$

$$\implies$$ $$F = \dfrac{kq_1 q_2}{r^2}$$ where $$k = \dfrac{1}{4\pi \epsilon_o}$$

Write any three properties of electric lines of force.

i) Lines of force start from positive charge and terminate at negative charge.

ii) Lines of force never intersect.

iii) The tangent to a line of force at any point gives the direction of the electric field $$\left( E \right) $$ at that point.

If $$\vec { E } =3\hat { i } +4\hat { j } -5\hat { k } $$, calculate the electric flux through a surface of area 50 units in z-x plane

$$\vec {E}=3\hat i+4\hat j-5\hat k$$

$$\vec{A}=50\hat j$$

$$\therefore$$ Electric flux $$Q=\vec{E}.\vec{A}$$

$$=(3\hat i+4\hat j-5\hat k).50\hat j$$

$$=200\ units$$.

1457456_1079103_ans_b67bb29ea5c546cfa6a628eb213ba513.png

A metallic sphere is placed in a uniform electric field as shown in the figure. Which path is followed by electric field line and why?

Path d is followed by electric field line as there are no electric field lines within the metallic sphere and field lines are normalat each point on the surface.

A,B and C are three charged bodies. If A and B repel each other and A attracts C, what is the nature of the force between B and C?

If $$A$$ & $$B$$ repel each other

then they are like charge (let them be $$+ve$$)

If $$A$$ & $$C$$ attracts then they are unlike

charge let then $$C$$ will be $$ -ve$$

$$\therefore \ $$ as $$A$$ is $$+ve, B$$ is $$+ve$$ & $$c$$ is $$-ve$$

$$B$$ and $$C$$ will attract each other and the nature of force will be attractive.

An electric field $$\vec E - \vec iAx$$ exists in the space, where $$A = V/{m^2}$$. take the potential at $$(10m, 20m)$$ to be zero. Find the potential at the origin.

We know electric field $$E=\dfrac{-dV}{dx}$$

$$-dV=Axdx$$

Integrating both sides we get

$$V=-\dfrac{A{x}^{2}}{2}+C$$------------(A)

Now at x=10,20 V=0

$$0=-\dfrac{400-100}{2}A+C$$

$$C=150A$$

Putting the above value in A also $$A=V/{m}^{2}$$we get,

$$V=150 V/{m}^{2}$$

An electric dipole is formed by charge $$\pm\ 4\mu C$$, separated by $$5\ mm$$. Calculate dipole moment and its direction.

1312938_1135745_ans_53c7ebac32ff4668bb463c947de1cfdd.jpg

Two point dipoles of dipole moment $$ \overline { { P }_{ 1 } } and\overline { { P }_{ 2 } } $$ are at a distance x from each other and $$ \overline { { P }_{ 1 } } ||\overline { { P }_{ 2 } } $$
The force between the dipoles is

$$\begin{array}{l} E\bot =-\frac { { \overline { { P_{ 1 } } } } }{ { A\pi \in { x^{ 3 } } } } \\ \Rightarrow Force\, \, on\, \, 2={ P_{ 2 } }.\Delta x\overline { \in } \bot =Force\, \, on\, \, 1 \\ F=\dfrac { { -6{ P_{ 1 } }{ P_{ 2 } } } }{ { 4\pi { \in _{ 0 } }{ x^{ 4 } } } } \end{array}$$

A system has two charges $$qA = 2.5\times 10^{-7}C$$ and $$qB = -2.5\times 10^{-7}C$$ located at point $$A(0, 0, -15) cm$$ and $$B(0, 0, 15) cm$$ respectively. What are the total charge and electric dipole moment vector of the system?

At A, amount of charge, $$q_A$$= $$2.5\times{10}^{-7}C$$ (Given)

At B, amount of charge, $$q_B= -2.5\times{10}^{-7}$$

Total charge of the system,

$$q=q_A+q_B= 2.5 \times {10}^{-7} C = 0$$

Distance between two charges at points A and B,

$$ d=15+15=30 cm = 0.3m$$

Electric dipole moment of the system is given by,

$$p= q_A \times d= 2.5 \times{10}^{-7}\times0.3=7.5\times{10}^{-8}cm$$ along positive z-axis.

Therefore, the electric dipole moment of the system is$$ 7.5\times {10}^{-8} $$ along the positive z-axis.

Complete the table:
Objects
Electron exchange
Positive
Negative
Glass, Silk
From glass rod to silk
....a....
Silk
Ebonite, Wool
Wool to ebonite
...b...
...c...
Rubber rod, Wool
...d...
Wool
...e...

The correct option for the blanks are:

a. Glass.
b. Wool.
c. Ebonite.
d. From wool to rubber rod.
e. Rubber rod.

From which one to which does electronic transfer occur when the pair of substances given below are rubbed against each other?
Glass rod Silk cloth.

When the glass rod and silk are rubbed, the electron transfer from glass rod to silk.

Why must hospital personnel wear special conducting shoes while working around oxygen in an operating room? What might happen if the personnel wore shoes with rubber soles?

Conducting shoes are worn to avoid the build up of a static charge on than as the wearer walks. Rubber-soled shoes acquire a charge by friction with the floor and could discharge with a spark, possibly causing an explosive burning situation, where the burning is enhanced by the oxygen.

From which one to which does electronic transfer occur when the pair of substances given below are rubbed against each other?
Ebonite Wool.

When the ebonite and wool are rubbed with each other, the transfer of electron is from wool to ebonite.

Conceptual question:
On the basis of the repulsive nature of the force between like charges and the freedom of motion of charge within a conductor, explain why excess charge on an isolated conductor must reside on its surface.

Inject some charge at arbitrary places within a conducting object. Every bit of the charge repels every other bit so each bit runs away as far as it can, stopping only when it reaches the outer surface of the conductor.

Answer to conceptual questions.
A student who grew up in a tropical country and is studying in the United States may have no experience with static electricity sparks and shocks until his or her first American winter. Explain.

All of the constituents of air are nonpolar except for water. The polar water molecule in the air quite readily "steal" charge from a charged object, as any physics teacher trying to perform electrostatics demonstrations in humid weather well knows. As a result-it is difficult to accumulate Iarge amounts of excess charge on an object in a humid dimate. During a North American winter, the cold, dry air allows accumulation of significant excess charge, giving the potential (pun intended) for a shocking (pun also intended) introduction to static electricity sparks.

Conceptual Questions :
A person is placed in a large, hollow, metallic sphere that is insulated from ground. If a large charge is placed on the sphere, will the person be harmed upon touching the inside of the sphere?

No. If the person is uncharged, the electric field inside the sphere is zero. The interior wail of the shell carries no charge The person is not harmed by touching this wall.

You might have seen iron chains, suspended from the body of lorries and trucks touching the earth. What is this for?

These lorries and trucks have to travel a long distance. Sometimes the friction between rod and tire or vehicle and oil in it produces static electricity. This discharge gathers at one end of the truck and it causes sparking and from it the petrol may catch fire. The iron chains suspended from the body of lorries and trucks provide earthing and help to avoid the hazard.

Application of Gausss Law to Various Charge Distributions
A large, flat, horizontal sheet of charge has a charge per unit area of $$9.00 \mu C/m^{2}$$. Find the electric field just above the middle of the sheet.

For a large uniformly charged sheet $$\overset{\rightarrow}{E}$$ will be perpendicular to sheet and wil have a magnitude of
$$E=\frac{\sigma}{2\in_{0}}=2\pi k_{e}\sigma$$
$$=(2 \pi)(8.99 \times 10^{9}N \cdot m^{2}/C^{2})(9.00 \times 10^{-6}C/m^{2})$$
$$\overset{\rightarrow}{E}=5.08 \times 10^{5}N/C \,\,j$$

Find the initial acceleration of the rod

Electric field between two large conducting plates, $$E=\dfrac{\sigma}{\epsilon_0}$$
Therefore, force on insulated rod, $$F=QE=\frac{Q\sigma}{\epsilon_0}$$
Note that gravity is ignored here.
Since the insulated rod is pivoted at point P it will experience torque which is given by
$$\tau=F \times \dfrac{l}{2}=\dfrac{Ql\sigma}{2\epsilon_0}$$

Now, $$\tau=I \alpha$$
For rod $$I=\dfrac{Ml^2}{3}$$
Therefore, angular accelration $$\alpha=\dfrac{3Ql\sigma}{2\epsilon_0Ml^2}$$.

A circle, having radius "r" has line charge distribution over its circumference having linear charge density $$\lambda =\lambda_0\cos^2\theta$$. Calculate the total electric charge residing on the circumference of the circle.
$$\left[\overset{2\pi}{\underset{0}{\displaystyle\int}}\cos^2\theta d\theta =\pi\right]$$

Let us take an element of angular width $$d\theta$$ at angle $$\theta$$.

Length of this element $$=Rd\theta$$

Charge on this element is $$dq=\lambda Rd\theta=\lambda_o \cos^2\theta Rd\theta$$

Total charge =$$Q=\int q=\lambda_o R\int_{0}^{2\pi}\cos^2\theta d\theta$$

$$\implies Q=\lambda_o \pi R$$

Define relative premittivity. Write its unit.

Relative permittivity is defined as the ratio of the actual or absolute permittivity of a medium to the absolute permittivity of vacuum. If the permittivity of the medium is $$\epsilon$$ then,

$$\epsilon_{r}=\dfrac{\epsilon}{\epsilon_{o}}$$

As it is the ratio of permittivities, hence, it has no unit.

What are electric lines of force? State the properties of lines of force.

Electric lines of force can be defined as a way or path, it may be straight or curved, so that the tangent at any point to it gives the direction of the electric field intensity at that point.

Properties of lines of force :-

1-Lines of force originate from a positive charge and terminate to a negative charge.

2-The tangent to theline of force indicates the direction of the electricfield and electric force.

3-Electric lines of force are always normal to the surface of charged body.

4-Two electric lines of force cannot intersect each other.

5-There are no lines of force inside the conductor.

Do we have to include the signs (positive or negative)of the charges while solving problems?

Yes, we have to include the signs($$+ve$$ or $$-ve$$)of the charges while solving problems.

for example we have to calculate force between two positive charges $${q}_{1}$$ and $${q}_{2}$$. then force between them is:

$$F=$$ constant$$\times\left(+{q}_{1}\right)\times\left(+{q}_{2}\right) =$$some positive quantity.

and remember positive force is always repulsive.

while in other case suppose we have to calculate force between one negative charge q1 and one positive charge q2 then force is :

$$F=$$ constant$$\times\left(-{q}_{1}\right)\times\left(+{q}_{2}\right) =$$ some negative quantity.

and remember negative force is always attractive.

so, we have to include the signs(+ve or -ve)of the charges while solving problems to determine numerically whether a force is attractve or repulsive.

No two electric lines of intersect each other. Why ?

The tangent at a point on the line of force gives the direction of electric field at the point.

If two lines of force intersect each other at a point , then electric field at that point will have two direction .This cannot be true.Hence two lines of force can never intersect each other .

Why do the electrostatic filed lines not from clossed loop ?

Electric field lines flow from positive to negative charge thus never come back to the positive charge. Therefore, they do not form a closed loop.

An electrostatic field line is a continuous curve. That is, a field line cannot have sudden breaks. Why not ?

An electrostatic field line is a continuous curve because a charge experiences a continuous force when placed in an electrostatic field. The field line cannot have sudden breaks because the charge moves continuously and does not jump fom one point to another.

Define one coulomb charge.

One coulomb is the amount of charge transferred in one second by a steady current of one ampere and equivalent to $$6.415\times { 10 }^{ 16 }$$ charges.

Define Permitivity of vacuum $$ (E_0) $$

The permittivity of free space (a vacuum) is a physical constant equal to approximately $$8.85\text{ }\times\text{ }{{10}^{-12}}~$$ farad per meter $$\left( F/m \right).$$ It is symbolized as $${{\varepsilon }_{o}}$$. In general, permittivity is symbolized $$\varepsilon $$ and is a constant of proportionality that exists between electric displacement and electric field intensity in a given medium. In many materials, particularly glass and certain plastics, $$\varepsilon $$ is substantially greater than $${{\varepsilon }_{o}}$$

A uniform electric field $$a\hat{i}+b\hat{j}$$ intersects a surface of area A. What is the flux through this area if the the surface lies (a) in the yz plane, (b) in the xz plane, (c) in the xy plane?

$$\vec{E}= a \hat i + b \hat j $$
(a) $$\vec A = A \hat i \Rightarrow \phi = \vec E . \vec A = Aa $$
(b) $$\vec A = A \hat j \Rightarrow \phi = \vec E . \vec A = Ab $$
(c) $$\vec A = A \hat k \Rightarrow \phi = \vec E . \vec A = 0 $$

If coulomb's law involved $$1/r^3$$ instead of $$(1/r^2)$$, would Gauss's law still be true?

In this case the the electric field due to point charge $$q$$ is $$E=\dfrac{kq}{r^2}$$

Now consider a gaussian surface as a sphere surface of radius $$r$$

The flux through the surface is $$\phi=\oint E.dS=(kq/r^3)\oint dS=kq/r^3 \times 4\pi r^2=\dfrac{4\pi kq}{r}$$

Thus this flux is dependent on $$r$$ but from Gauss's law the flux depends only on the charge inside the surface.

The branch of electricity which deals with the properties of electrified bodies due to a moving charge is called electrostatics. True/False?

Electrostatics deals with charges at rest.

An electric dipole is free to move in a uniform electric field. Explain its motion when it is placed (i) parallel to the field and (ii) perpendicular to the field:

(i) Parallel to the field: The net force and torque on the dipole is zero. Hence, dipole will not move.
(ii) Perpendicular to the field: The net force is zero but torque is maximum. Hence, dipole will rotate to align itself along the field.

Coulomb's law states that the electric force becomes weaker with increasing distance, Suppose that instead, the electric force between two charged particles were independent of distance. In this case, would a neutral insulator still be attracted towards the comb.

Attraction requires opposite charges. As the insulator is neutral and also charge can not be induced on insulators without contact, the insulator will not be attracted towards a charged comb.

What is the electric flux through a cube of side 1 cm whose encloses an electric dipole?

Since net charge enclosed in the surfaces bound by the cube is zero(dipole consists of equal and opposite charges), according to the Gauss Law, it can be claimed that the net flux through the cube is zero.

Why do the electric field lines never cross each other?

At any point, if electric field lines cross each other it means at that point there are two directions of electric field, which is impossible.

Write Coulomb's law in vector form. Also explain the terms.

$$\textbf{Step 1: Coulomb's law in vector Form:}$$ $$\textbf{[Refer Fig.]}$$

Consider two point charges $$q_1$$ and $$q_2$$ separated by a distance $$r$$.

Coulomb's Force acting on $$q_1$$ due to $$q_2$$ is given by:

$$(\vec{F_{12}})$$

Where $$\hat{r_{21}}$$ unit vector directed. From $$q_2$$ to $$q_1$$

$$\vec{F_{12}}=\dfrac{kq_1q_2}{r^2}\hat{r_{21}}$$

Similarly Force acting on que to on $$q_2$$ due to $$q_1$$:

$$\vec{F_{21}}=\dfrac{kq_1q_2\hat{r_{12}}}{r^2}$$

where $$\hat{r_{12}}$$ unit vector directed From $$q_1$$ to $$q_2$$

In general it can also be written as

$$\vec{F}=\dfrac{kq_1q_2\vec{r}}{r^3}$$ $$\left\{As\,\hat{r}=\dfrac{\vec{r}}{r}\right\}$$

2117690_623483_ans_ae07b4f416544fd9aa48261ff8a77016.png

Draw a diagram of equipotential surface for a single charge.

For a single, isolated point charge, potential is inversely dependent upon radial distance from the charge. Therefore, equipotential surface for a single point charge is spherical, with the point charge at the center.

State the $$SI$$ unit of magnetic dipole moment

The SI unit of magnetic dipole moment is $$Am^{-2}$$.

Write the definition of electric dipole moment.

The electric dipole moment for a pair of opposite charges of magnitude $$q$$ is defined as the magnitude of the charge times the distance between them and the defined direction is toward the positive charge.

What is an electric dipole?

Why do the electrostatic field lines not form closed loop?

The field lines starts (or diverges) from the positive charge and terminate (or converges) on negative charge. If the electric field lines form closed loop , these line must originate and terminate on the same charge which is not possible because electric field lines always moves from positive to negative.

What happens to the electrostatic force when both charges are doubled?

Initial case:

Charges are $$ q_1, q_2$$

DIstance =$$r$$

$$ \therefore F = K \cfrac{q_1q_2}{r^2}$$

Final case:

$$q_1' = 2q_1$$

$$q_2' = 2q_2$$

$$r'=r$$

$$ \therefore F' = K \cfrac{2q_1\times 2q_2}{r^2}$$

$$ F' = 4F$$

Hence electrostatic force increases $$4$$ times of its initial value.

State Coulomb's law.

Coulomb's law states that the force acting between the two charges $$q_1$$ and $$q_2$$ kept at a separation $$(r)$$ is directly proportional to the product of two charges and inversely proportional to $$r^2$$.
So, Coulombic force $$F \propto \dfrac{q_1q_2}{r^2}$$

Define an electric line of force.

An electric line of force is an imaginary continuous line or curve drawn in an electric field such that tangent to it at any point gives the direction of the electric force at that point. the direction of a line of force is the direction along which a small free positive charge will move along the line. It is always directed from positive charge to negative charge.

Some of its important characteristics are

1. Electric lines of force contract longitudinally.

2. They expand laterally.

3. Two electric lines of force cannot intersect each other.

4. Two electric lines of force proceeding in the same direction repel each other.

5. Two electric lines of force proceeding in the opposite direction attract each other.

6. The lines of force are imaginary but the field it represents as real.

7. There are no lines of force inside the conductor.

Tick the correct one:
Object with same charge
(attract / repel)

The objects with the same charge repel each other.

Tick the correct one:
Bodies with different charges
(attract / repel)

The two bodies with different type of charges will attract each other.

When a glass rod is rub with silk, electron loses from the glass rod. If we bring another glass rod near to it what can be observed? Give reason.

They will repel. Same charges repel each other.

Tick the correct one:
The charged body ............ the neutral body.
(attract / repel)

The charged body attracts the neutral body.

State and explain coulomb's law of electric charges in scalar form.

If a charged plastic straw is brought near another uncharged plastic straw, what will happen?

The charged plastic straw will attract another uncharged plastic straw, when brought near to it because charges of different kind attract each other. The aluminium strips in an electroscope as shown in figure are replaced by plastic strips and a charged body is brought in contact with the metal clip.

If the metal clip used in the electroscope is replaced by an ebonite rod and a charged body is brought in contact with it, will there be any effect on the aluminium strips? Explain.

The aluminium strips will not show any repulsion. The charged body will not transfer any charge to the ebonite rod as ebonite rod is an insulator. As a result there will be no charge on the aluminium strips and no repulsion will occur.

How many point charges of same magnitude are required to constitute an electric dipole?

Dipole is two point charges of opposite polarity and same magnitude separated from each other by some distance.

The aluminium strips in an electroscope as shown in fig. 15.1 are replaced by plastic strips and a charged body is brought in contact with the metal clip. What will happen?

If the metal clip used in the electroscope is replaced by an ebonite rod and a charged body is brought in contact with it, therefore, there will be no charge on the aluminium strips and no repulsion of the strips will occur.

The figure shows three infinite non-conducting plates of charge perpendicular to the plane of the paper with charge per unit area $$+ \sigma, \: +3 \sigma \: and \: - \sigma $$. The ratio of the net electric field at that point A to that at point B is $$1/x$$. Find $$x$$:

The net electric field at A is $$\displaystyle E_A=\dfrac{-\sigma}{2\epsilon_0}+\dfrac{3\sigma}{2\epsilon_0}-\dfrac{\sigma}{2\epsilon_0}=\dfrac{\sigma}{2\epsilon_0}$$

The net electric field at B is $$\displaystyle E_B=\dfrac{-\sigma}{2\epsilon_0}-\dfrac{3\sigma}{2\epsilon_0}-\dfrac{\sigma}{2\epsilon_0}=-\dfrac{5\sigma}{2\epsilon_0}$$

$$\displaystyle \therefore \left|\dfrac{E_A}{E_B}\right|=\dfrac{\sigma}{2\epsilon_0}\times \dfrac{2\epsilon_0}{5\sigma}=1/5$$

A point charge $$+q$$ & mass $$100 gm$$ experiences a force of $$100N$$ at a point at a distance $$20cm$$ from a long infinite uniformly charged wire. If it is released its speed is $$20\sqrt { l{ n }x } m/s$$ when it is at a distance $$40cm$$ from wire. Find $$x$$.

For the inifinite uniform charged wire, $$E=\lambda/2\pi\epsilon_o r$$

at $$r=20cm$$, Force $$F=\lambda q/2\pi\epsilon_o\times 0.2=100N$$

acceleration, $$a=\lambda q/2\pi\epsilon_o\times0.1=vdv/dr$$

so, $$v^2/2=\dfrac{\lambda}{2\pi\epsilon_o\times0.1}\int_{0.2}^{0.4}lnr=200ln2$$

$$\Rightarrow v=20\sqrt{ln 2}$$

Thus, $$x=2$$

If the total electric flux through the paraboloidal surface due to a uniform electric field of magnitude $$E_0$$ in the direction shown in fig is $$\phi=XE_0 \pi r^2/2$$. Find out the value of X?

total field lines entering the flat surface= total field lines leaving the curved surface
Therefore, flux = field perpendicular to flat surface $$ \times $$ Area of flat surface = $$ E_0 \pi r^2 $$

A ring of diameter d is rotated in a uniform electric field until be position of maximum electric flux is found. The flux is found to be $$\phi$$. If the electric field strength is $$\ E=\frac{X \phi}{\pi d^2}$$. Find out X ?

flux will be maximum when the plane of the ring is perpendicular to the uniform field.
Then,
$$ \phi=EA \Rightarrow \phi=E \frac{\pi d^2}{4} \Rightarrow E=\frac{4 \phi}{\pi d^2} $$

A point charge Q is located on the axis of a disc of radius R at a distance b from the plane of the disc (figure). Show that if one-fourth of the electric flux from the charge passes through the disc, then $$R = \sqrt 3$$b.

The total flux produced by the charge is, according to Gauss’s law, is $$\frac{Q_{enclosed}}{\epsilon_0}$$. Only one quarter of this flux passes through the disk. The flux through the disk is given by:
$$\phi_{disk}=\int E.dA$$
where the integration cover the entire area of the disk. Evaluating this integral and set it equal to $$\frac{Q}{4\epsilon_0}$$ relates b to R. As shown in figure, dA is the area of annular ring with radius s and width ds. The electric field at the ring make an angle $$\theta$$ with the normal to to the ring, the flux through the ring is:
$$d\phi_{ring}= E. dA = E dAcos\theta=E(2\pi sds)cos\theta$$
The magnitude of the electric field has the same value at all points on the ring:
$$E_{ring}=\frac{Q}{4\pi \epsilon_0 r^2}=\frac{Q}{4\pi \epsilon_0(s^2 + b^2)}$$
and, $$cos\theta=\frac{b}{\sqrt{s^2+b^2}}$$
Therefore, $$d\phi_{ring}=\frac{Qb}{2\epsilon_0}\frac{s}{(s^2 + b^2)^{3/2}}ds$$
To get the flux through the entire disk we integrate $$d\phi_{ring} $$ from s = 0 to s = R
$$\phi_{disk}=\int^R_0 d\phi_{ring}$$
After integration, $$\phi_{disk}=\frac{Q}{2\epsilon_0}[1-\frac{b}{\sqrt{R^2+b^2}}]$$
Since the flux through the disk is already given by $$\frac{Q}{4\epsilon_0}$$, then:
$$\frac{Q}{4\epsilon_0}=\frac{Q}{2\epsilon_0}[1-\frac{b}{\sqrt{R^2+b^2}}]$$
which will give
$$4b^2=R^2+b^2$$
Therefore, $$R=\sqrt3 b$$

What is the charge per unit area in $$C/m^2$$, of an infinite plane sheet of charge if the electric field produced by the sheet of charge has magnitude 3.0 N/C?

The electric field due to infinite charge sheet is $$E=\dfrac{\sigma}{2\epsilon_0}$$

So, the charge per unit area, $$\sigma=2\epsilon_0E=2\times (8.85\times 10^{-12})\times 3=5.31\times 10^{-11} C/m^2$$

Net charge within an imaginary cube drawn in a uniform electric field is always zero. Is this statement true or false.

Statement is true.
In a uniform electric field, the number of field lines entering the closed imaginary cube is equal to the number of field lines leaving the closed imaginary cube. Hence the total outwards electric flux through the closed imaginary cube is zero. By Gauss's Law , this means he enclosed charge within the closed imaginary cube is zero.

Find the net electric flux through the entire cube.

$$ \varphi _{s1} = -E_y l^2= -Cl^2$$,
$$ \varphi_{s2} = E_z l^2= -DL^2$$,
$$ \varphi_{s3} = E_y l^2= CL^2$$,
$$ \varphi_{s4} = -E_z l^2=DL^2$$,
$$ \varphi_{s5} = E_x l^2=-BL^2 $$,
$$ \varphi_{s6} = -E_x l^2 =BL^2$$
net flux = $$ \varphi_{s1}+\varphi_{s2}+\varphi_{s3}+\varphi_{s4}+\varphi_{s5}+\varphi_{s6} = 0$$

A cube has sides of length L = 0.2 m. It is placed with one corner at the origin as shown in figure.The electric field is uniform and given by $$\vec{E}=(2.5 N/C) \dot{I} - (4.2 N/C) \dot{J}$$. Find the electric flux through the entire cube.

For a cube, the leaving and entering flux through two opposite surfaces (like front and back) are same and opposite.

So the net flux through the entire cube will be zero.

Find the electric flux through each of the six cube faces $$S_1, S_2, S_3, S_4, S_5$$ and $$S_6$$

Given the following data, calculate the electrostatic force and gravitational force between two electrons. Show that the electrostatic force is much greater than the gravitational force for the electrons separated by a certain distance.

Charge on the electron $$= 1.6 \times 10^{-19} C$$
Mass of the electron $$= 9.1 \times 10^{-31} kg$$
Gravitational constant $$= 6.67 \times 10^{-11} Nm^{2} kg^{-2}$$

given,

charge on the electron$$=1.6\times10^{-19}C$$

mass of the electron$$=9.1\times10^{-31}kg$$

gravitational constant$$=6.67\times10^{-11}Nm^{2}kg^{-2}$$

i)to find the electrostatic force b/w 2 electron.

let 2 electrons are at a distance 'd' apart

the electrostatic force is $${ F }_{ e }=\cfrac { 1 }{ 4\pi { \varepsilon }_{ o } } \cfrac { { q }^{ 2 } }{ { d }^{ 2 } } \\ =\cfrac { 9\times { 10 }^{ 9 }\times { \left( 1.6\times { 10 }^{ -19 } \right) }^{ 2 } }{ { d }^{ 2 } } \\ =\cfrac { 23.04\times { 10 }^{ -29 } }{ { d }^{ 2 } } N$$

ii) To find the gravitational force,

$${ F }_{ g }=\cfrac { { Gm }^{ 2 } }{ { d }^{ 2 } } \\ =\cfrac { 6.67\times { 10 }^{ -11 }\times { \left( 9.1\times { 10 }^{ -31 } \right) }^{ 2 } }{ { d }^{ 2 } } \\ =\cfrac { 553.56\times { 10 }^{ -73 } }{ { d }^{ 2 } } N$$

iii) to show that electrostatic force is much grater then gravitational force.

for this we take ratio $$\cfrac { { F }_{ e } }{ { F }_{ g } } =\cfrac { 23.04\times { 10 }^{ -29 } }{ 553.56\times { 10 }^{ -73 } } \\ =0.0416\times { 10 }^{ 44 }\\ =0.416\times { 10 }^{ 43 }$$

we can see tham $$f_{e}>>f_{g}$$ ie. electrostatice force is much grater than the gravitational force for $$2$$ electrons separated by a certain distance

Calculate the electrostatic force of attraction between a proton and an electron in a hydrogen atom. The radius of the electron orbit is 0.05 nm and charge on the electron is $$1.6 \times 10^{-9} C.$$

From Coulomb's inverse square law: $$\displaystyle F= \frac {1}{4 \pi \varepsilon_{0}} \frac{q_1q_2}{r^2}$$

The magnitude of charge on electron and proton is the same $$= 1.6 \times 10^{-19} C$$

Therefore, electrostatic force is given by

$$\displaystyle F= \frac {9 \times 10^9 \times 1.6 \times 10^{-19} \times 1.6 \times 10^{-19}}{(0.05 \times 10^{-9})^2}$$

$$\displaystyle = \frac {9 \times 1.6 \times 1.6 \times 10^{-29}}{(5 \times 10^{-11})^2}$$

$$\displaystyle = \frac {9 \times 1.6 \times 1.6 \times 10^{-29}}{25 \times 10^{-22}}$$

$$\displaystyle = 0.9216 \times 10^{-29} \times 10^{+22}$$
$$\displaystyle = 0.9216 \times 10^{-7} N$$
$$\displaystyle = 9.216 \times 10^{-8} N$$

Answer the following questions.
(i) An electrostatic field line is a continuous curve. That is, a field line cannot have sudden breaks. Why is it so?
(ii) Explain why two field lines never cross each other at any point.

(a) An electrostatic field line is a continuous curve because a charge experiences a continuous force when traced in an electrostatic field. The field line cannot have sudden breaks because the charge moves continuously and does not jump from one point to the other.

(b) If two field lines cross each other at a point, then electric field intensity will show two directions at that point. This is not possible. Hence, two field lines never cross each other.

(a) Consider an arbitrary electrostatic field configuration. A small test charge is placed at a null point (i.e., where E = 0) of the configuration. Show that the equilibrium of the test charge is necessarily unstable.
(b) Verify this result for the simple configuration of two charges of the same magnitude and sign placed a certain distance apart.

(a)

Let the equilibrium of the test charge be stable. If a test charge is in equilibrium and displaced from its position in any direction, then it experiences a restoring force towards a null point, where the electric field is zero. All the field lines near the null point are directed inwards towards the null point. There is a net inward flux of electric field through a closed surface around the null point. According to Gauss's law, the flux of electric field through a surface, which is not enclosing any charge, is zero. Hence, the equilibrium of the test charge cannot be stable.

(b)

Two charges of same magnitude and same sign are placed at a certain distance. The mid-point of the joining line of the charges is the null point. When a test charged is displaced along the line, it experiences a restoring force. If it is displaced normal to the joining line, then the net force takes it away from the null point. Hence, the charge is unstable because stability of equilibrium requires restoring force in all directions.

Consider a uniform electric field $$E = 3\times 10^{3}\ \hat{i}\ NC^{-1}$$. (a) What is the flux of this field through a square of 10 cm on a side whose plane is parallel to the yz plane? (b) What is the flux through the same square if the normal to its plane makes a $$60^o$$ angle with the x-axis?

(a)

Electric field intensity, $$\overline{E}=3 \times 10^3$$N/C

Magnitude of electric field intensity, $$|E|=3 \times 10^3$$N/C

Side of the square, $$s = 10 cm = 0.1 m$$

Area of the square, $$A=s^2 = 0.01 m^2$$

The plane of the square is parallel to the y-z plane. Hence, angle between the unit vector normal to the plane and electric field, $$\theta = 0^o$$

Flux $$(\phi)$$ through the plane is given by the relation,

$$\phi=|\overline{E}|A\, cos\, \theta$$

$$=3\times 10^3\times 0.01\times cos\,0^o$$

$$=30 N m^2/C$$

(b)

Plane makes an angle of $$60^o$$ with the x-axis. Hence, $$\theta = 60^o$$

Flux, $$\phi = |\overline{E}|A\, cos\,\theta$$

$$=3\times 10^3\times 0.01\times cos\, 60^o$$

$$=30\times \frac{1}{2}=15N\,m^2/C$$

A system has two charges $$q_A = 2.5\times 10^{-7} C$$ and $$q_B = -2.5\times 10^{-7}C$$ located at points $$A: (0, 0, 15 cm)$$ and $$B: (0,0, +15 cm)$$, respectively. What are the total charge and electric dipole moment of the system?

According to the figure given below both the charges can be located in a coordinate frame of reference.

At A, amount of charge, $$q_A=2.5\times 10^{-7}C$$ (Given)

At B, amount of charge, $$q_B=-2.5\times 10^{-7}C$$ (Given)

Total charge of the system,

$$a=q_A+q_B=2.5\times 10^{-7}C-2.5\times 10^{-7}C=0$$

Distance between two charges at points A and B,

$$d=15+15=30cm=0.3m$$

Electric dipole moment of the system is given by,

$$p=q_A\times d=q_B\times d=2.5\times 10^{-7}\times 0.3=7.5\times 10^{-8}C\,m$$ along positive z-axis

Therefore, the electric dipole moment of the system is $$7.5\times 10^{-8}C\,m$$ along positive z−axis.

Consider a uniform electric field $$E = 3\times 10^{3}\ \hat{i}\ NC^{-1}$$. What is the net flux of the uniform electric field through a cube of side 20 cm oriented so that its faces are parallel to the coordinate planes?

Faces of cube are parallel to the coordinate axes. Therefore, the number of field lines entering the cube is equal to the number of field lines coming out of the cube,making the net flux zero through it.

Figure shows tracks of three charged particles in a uniform electrostatic field. Give the signs of the three charges. Which particle has the highest charge to mass ratio?

Opposite charges attract each other and same charges repel each other. It can be observed that particles 1 and 2 both move towards the positively charged plate and repel away from the negatively charged plate. Hence, these two particles are negatively charged. It can also be observed that particle 3 moves towards the negatively charged plate and repels away from the positively charged plate. Hence, particle 3 is positively charged.

The charge to mass ratio (emf) is directly proportional to the displacement or amount of deflection for a given velocity. Since the deflection of particle 3 is the maximum, it has the highest charge to mass ratio.

What is the electric flux through a cube of side $$1\ cm$$ which encloses an electric dipole?

As an electric dipole has equal and opposite charge, thus the net charge enclosed by the cube is zero i.e $$q_{enc} = 0$$

$$\therefore$$ Electric flux passing through the cube $$\phi = \dfrac{q_{enc}}{\epsilon_o} = 0$$

Why do the electrostatic field lines not form closed loops?

The electric line of force is the path occupied by a unit positive test charge when placed along the line. If electric field line is a closed loop it means that positive test charge would come back to the positive source charge i.e initially there was repulsion and now it has turned into the attraction, which is impossible .

Two charges of magnitudes $$-2Q$$ and $$+Q$$ are located at points $$(a, 0)$$ and $$(4a, 0)$$ respectively. What is the electric flux due to these charges through a sphere of radius $$'3a'$$ with its centre at the origin?

By Gauss's law, the flux $$\phi=\dfrac{Q_{en}}{\epsilon_0}$$

Here the charge enclosed by sphere of radius 3a is $$Q_{en}=-2Q$$

Thus, $$\phi=\dfrac{-2Q}{\epsilon_0}$$

Given a uniform electric field $$\vec {E} = 5\times 10^{3}\hat {i} N/C$$, find the flux of this field through a square of $$10\ cm$$ on a side whose plane is parallel to the $$Y-Z$$ plane. What would be the flux through the same square if the plane makes a $$30^{\circ}$$ angle with the $$X-axis?

$$\overrightarrow { E } =4\times { 10 }^{ 3 }\hat { i } \left( \frac { N }{ C } \right) \\ Flux=\overrightarrow { E } .\overrightarrow { S } \quad =EScos\theta $$ S is the area through which electric field lines crosses

When plane is parallel to Y-Z plane then $$\theta =0$$ here S = a X a = a^2 a= 5cm

Flux =$$ES=4\times { 10 }^{ 3 }\times { (5\times { 10 }^{ -2 }) }^{ 2 }\\ \quad \quad =4\times { 10 }^{ 3 }\times 25\times { 10 }^{ -4 }=10$$

when plane makes angle 30 degree with X axis then perpendicular to plane makes angle 60 degree to X axis

hence

$$\theta ={ 60 }^{ 0 }\\ Flux=ES=4\times { 10 }^{ 3 }\times { (5\times { 10 }^{ -2 }) }^{ 2 }\times cos{ 60 }^{ 0 }\\ \quad \quad $$

=$$4\times { 10 }^{ 3 }\times 25\times { 10 }^{ -4 }\times \frac { \ { 1 } }{ 2 } =5 $$

Answer the following questions:
(i) Define electric flux. Write its $$SI$$ units.
(ii) Using Gauss's law, prove that the electric field at a point due to a uniformly charged infinite plane sheet is independent of the distance from it.
(iii) How is the field directed if (i) the sheet is positively charged, (ii) negatively charged?

(a)

Electric flux is the measure of flow of the electric field through a given area. It is defined as the dot product of electric field and area vector through which the flux has to be computed. Hence,

$$\phi = \int \vec E. \vec {dS}$$

Its SI unit is Gauss ($$G$$) / $$Nm^2C^{-1}$$ / $$Vm$$.

(b)

Electric field near the surface of an infinite sheet of positive charge is directed away from the sheet of charge.

Consider a cylindrical gaussian surface with its axis perpendicular to the sheet of charge as shown in the figure.

The net flux through the curved surface is zero as the electric field and surface are parallel to each other.

Hence, $$\phi_{curved} = \int \vec E . \vec {dS} = 0$$

The net flux through the upper plane surface is given by:

$$\phi_{pu} = \int \vec E . \vec {dS} = EA$$

Similarly, The net flux through the lower plane surface is given by:

$$\phi_{lu} = \int \vec E . \vec {dS} = EA$$

Net flux through the closed cylindrical surface is:

$$\phi = \phi_{curved} + \phi_{pu} + \phi_{lu} = 2EA ..........(i)$$

By Gauss' Law,

$$\phi = \dfrac{Q_{enclosed}}{\varepsilon_o} = \dfrac{\sigma A}{\varepsilon_o} ..............(ii)$$

From (i) and (ii),

$$E = \dfrac{\sigma}{2 \varepsilon_o}$$

Hence, the electric field at a point due to a uniformly charged infinite plane sheet is independent of the distance from it.

(c)

(i) If the sheet is positively charged, field is directed away from the sheet of charge.

(ii) If the sheet is negatively charged, field is directed towards the sheet of charge.

Given a uniform electric field $$\overrightarrow E = 5 \times 10^3 \widehat i N/C$$, find the flux of this field through a square of 10 cm on a side whose plane is parallel to the y-z plane. What could be the flux through the same square if the plane makes $$30^o$$ angle with the x-axis?

Flux through a surface is given by:

$$\phi = \vec E. \vec {dS}$$

Area is $$S = EScos(90) = 0.1 \times 0.1 = 0.01\ m^2$$

$$\phi = 5 \times 10^3 \times 0.01 = 50 Nm^2C^{-1}$$

When the square makes an angle of $$30^o$$ with the x-axis, angle made by the normal with electric field is $$90^o -30^o = 60^o$$

Hence, $$\phi = ES cos (60^o) = 25 Nm^2C^{-1}$$

Symbol for steradian is ____

The symbol of steradian is $$sr$$. It is the SI unit of solid angle.
For example, the solid angle of a complete sphere is $$4\pi sr$$.

The lines of force exert a lateral pressure on each other. What does this property of lines of force leads to explain?

The lines of force exert lateral (sideways) pressure on one another explains that they tend to repel each other. This explains repulsion between like charges.

A large hollow metallic sphere has a positive charge of $$35.4\mu C$$ at its centre. Find how much electric flux will emanate from the sphere ?

Electric flux$$=\phi = \dfrac {q}{\epsilon_{0}} = \dfrac {35.4\times 10^{-6}}{8.85\times 10^{-12}} = 4\times 10^{6} N\ m^{2}/C$$

Define electric dipole moment. Give its unit.

Two equal and opposite charges separated by a very small distance constitute an electric dipole. Two point charges $$+q$$ and $$-q$$ are kept at a distance $$2d$$ apart. The magnitude of the dipole moment is given by the product of the magnitude of the one of the charges and the distance between them.

Electric dipole moment, $$p = 2qd$$.
It is a vector quantity and acts from $$-q$$ and $$+q$$. The unit of dipole moment is $$C m$$.

Write two properties of electric field lines.

The properties of electric field lines are: -

1. These are imaginary lines which shows pictorial influence of electric field intensity in space.

2. These are always emitted from positive charge and enter in negative charge.

3. They don't form closed loops.

4. They are closely packed in strong electric field region and farther apart in weak electric field region.

5. These lines are always perpendicular to surface.

6. These lines never intersect with each other.

Write the properties of electric lines of force.

The properties of electric lines of force are:

i) Lines of force start from positive charge and terminate at negative charge.

ii) Lines of force never intersect.

iii) The tangent to a line of force at any point gives the direction of the electric field at that point.

iv) The number of lines per unit area, through a plane at right angles to the lines is proportional to the magnitude of $$E$$. That is, when the lines of force are close together, $$E$$ is large and where they are far apart, $$E$$ is small.

v) Each unit positive charge gives rise to $$\dfrac { 1 }{ { \varepsilon }_{ 0 } } $$ lines of force in free space. Hence, number of lines of force originating from a point charge $$q$$ is $$N=\dfrac { q }{ { \varepsilon }_{ 0 } } $$ in free space.

Define one "Coulomb" on the basis of Coulomb's law.

One coulomb is defined as the quantity of charge, which when placed at a distance of $$1$$ metre in air or vacuum from an equal and similar charge, experiences a repulsive force of $$9\times { 10 }^{ 9 }N$$.

State Coulomb's law in electrostatics.

The force of attraction or repulsion between two point charges is.

i) directly proportional to the product of the charges and

ii) Inversely proportional to the square of the distance between them.

$${\displaystyle F=k_{e}{\frac {q_{1}q_{2}}{r^{2}}}}$$

where $$k_e$$ is Coulomb's constant ($$k_e = 9×10^9 N m^2 C^{−2})$$, $$q_1$$ and $$q_2$$ are the signed magnitudes of the charges, and the scalar $$r$$ is the distance between the charges.

Two infinitely large sheets having charge densities $$ { \sigma }_{ 1 }$$ and $$ { \sigma }_{ 2 }$$ respectively $$({ \sigma }_{ 1 } >{ \sigma }_{ 2 })$$ are placed near each other separated by distance $$d$$.A charge $$q$$ is placed in between two plates such that there is no effect on charge distribution on plates. Now this charge is moved at an angle of $${45}^{o}$$ with the horizontal towards plate having charge density $$ { \sigma }_{ 2 }$$ by distance $$a$$ $$(a<d)$$. Find work done by electric field in the process.

b'To find the work done on the electric field we need to find out the electric field of the given sheets.

thus the formula for finding the electric field can be directly given as:

$$E=\dfrac{\sigma_1-\sigma_2}{2\epsilon_0}$$

then applying the formula of the electric field in the formula of work we get:

$$W=qE.d$$

to find $$d$$ refer the figure,

$$\therefore d= \dfrac{a}{\sqrt2}$$

substituting the value of $$d$$ in the equation of work we get,

$$W=q\times E \times \dfrac{a}{\sqrt2}$$

then again substituting the value of $$E$$ we get,

$$W=\dfrac{q(\sigma_1-\sigma_2)a}{2\sqrt2}\epsilon_0$$

Find out the magnitude of electric field intensity and electric potential due to a dipole of dipole moment $$\bar P = \hat i + \sqrt 3 \hat j$$ kept at origin at following points.
(i) (2, 0, 0)
(ii) (-1, $$\sqrt 3 $$, 0)

$$E=\dfrac { k\left| \overrightarrow { p } \right| }{ { \left| z \right| }^{ 3 } } \\ \overrightarrow { z } =2\hat { i } \\ \overrightarrow { p } =\hat { i } +\sqrt { 3 } \hat { j } \\ k=9\times { 10 }^{ 9 }\\ E=\dfrac { 9\times { 10 }^{ 9 }\times \sqrt { 10 } }{ 8 } \\ E=35.575\times { 10 }^{ 8 }\\ V=\dfrac { \overrightarrow { p } .\hat { r } }{ 4\pi { \varepsilon }_{ 0 }{ r }^{ 2 } } \\ \overrightarrow { r } =-\hat { i } +\sqrt { 3 } \hat { j } \\ \left| \overrightarrow { r } \right| =\sqrt { { 1 }^{ 2 }+{ 3 }^{ 2 } } \\ \left| \overrightarrow { r } \right| =\sqrt { 10 } \\ V=\dfrac { (\hat { i } +\sqrt { 3 } \hat { j } ).(-\hat { i } +\sqrt { 3 } \hat { j } ) }{ 4\pi { \varepsilon }_{ 0 }\times 10 } \\ V=\dfrac { 9\times { 10 }^{ 9 } }{ 10 } \times (-1+3)\\ V=\dfrac { 9 }{ 5 } \times { 10 }^{ 9 }\\ V=1.8\times { 10 }^{ 9 }$$

Three identical metal plates with large surface areas are kept parallel to each other As shown in figure. The leftmost plate is given a charge $$Q$$, the rightmost a charge $$-2Q$$ and the middle one remain neutral. Find the charge appearing on the outer surface of the rightmost plate.

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The mutual force of repulsion between two point charges kept a fixed distance apart is $$9\times {10}^{-5}N$$ when in vacuum and $$4\times {10}^{-5}N$$ when placed in a dielectric medium. What is the value of dielectric constant of the medium?

Let the dielectric constant of the medium be $$K$$.

The force of interaction between the objects is $$\frac{F}{K}$$ when placed in

the medium of dielectric constant $$K$$ and $$F$$ is the force of interaction

between them in vacuum .

According to question ,

$$F = 9\times 10^{-5} N$$

and $$\frac{F}{K} = 4\times 10^{-5} N$$

Dividing both these equations ,

we get dielectric constant , $$K$$ = $$2.25$$

Define electric flux. Write its SI unit.

Electric flux is defined as the measure of count of number of electric field lines crossing an area.

Electric flux $$\phi = E A\cos\theta$$

SI unit of electric flux is $$Nm^2/C$$

Two field lines never intersect each other. Give reason.

Suppose two line crossing each other at a point, now draw tangent along both the lines on that particular point. You will get two direction of electric field or force at a single point which is not possible.

If flux in a coil changes by $$\Delta$$ $$\phi$$, and the resistance of the coil is R, prove that the charge flown in the coil during the flux change is $$\dfrac{{\Delta}\phi}{R}$$. (Note: It is independent of the time taken for the change in flux)

Flux in the coil changes by $$\triangle \phi $$

Resistance of coil $$=R$$

Now,

EMF $$=-\cfrac { d\phi }{ dt } $$ (from faradays law)

$$\therefore \quad V=-\cfrac { d\phi }{ dt } $$

Also,

$$I=\cfrac { dQ }{ dt } =\cfrac { V }{ R } $$

$$\Rightarrow \cfrac { dQ }{ dt } =\cfrac { V }{ R } $$

$$\left\{ dQ=-\cfrac { d\phi }{ R } =-\cfrac { \triangle \phi }{ R } \right\} $$

Hence proved

Two thin conducting plates (very large) parallel to each other carrying total charges $$\sigma A$$ and $$-2\sigma A$$ respectively (where A is the area of each plate), are placed in a uniform external electric field E as shown. Find the surface charge on each surface.

The charges on the plates will be distributed as

Charges on the outer surfaces of the plates is equal to the mean of total large of the capacitor

So, $$q_{1}=q_{4}=\dfrac{\sigma -2\sigma A}{2}=\dfrac{\sigma A}{2}$$

Charges on the inner surfaces are such that the total charges on rach plate os same as given.

So, $$q_{2}+q_{1}=\sigma A\Rightarrow q_{2}=\dfrac{3}{2}\sigma A$$

$$q_{3}+q_{4}=-2\sigma A\Rightarrow q_{3}=\dfrac{-3}{2}\sigma A$$

1427850_1044021_ans_3b1a46c9bd184c7c91bbec7f3a24cffb.png

When does an electric dipole placed in a uniform electric field experience
a. Maximum Torque
b. Minimum Torque

$$T=PE\sin \theta$$

$$(a)$$ When electric field is perpendicular to the axis

$$T=$$ maximum as $$\theta=90^o$$

$$(b)$$ When electric field is along the axis

$$T=0$$ minimum

If $$\vec { E } =3\widehat { i } +4\widehat { j } -5\widehat { k } $$ calculate the electric flux through the surface of area $$50$$ units in $$z-x$$ plane.

$$\vec { E } =3\widehat { i } +4\widehat { j } +5\widehat { k } $$

Area is $$50J$$

$$\phi=\vec{E}.\vec{A}$$

$$=(3\widehat { i } +4\widehat { j } -5\widehat { k }).50\widehat{j}$$

$$=100$$units.

Shown as electric dipole formed by two particle fixed at the ends of a light rod of length l. The mass of each particle is m and the charges are -q and +q. The system is placed in such a way that the dipole axis is parallel to a uniform electric field E that exists in the region. The dipole is slightly rotated about its centre and released. Show that for small angular displacement, the motion is angular simple harmonic and find its time period.

$$\begin{array}{l} We\, have \\ T=qEl\theta \\ I\alpha qEl\theta \\ \Rightarrow \alpha =\left( { \dfrac { { qEl } }{ I } } \right) \theta \, \, \, \, \, \, \therefore 54m \\ =\left\{ { \dfrac { { qEl } }{ { \left( { \dfrac { { m{ l^{ 2 } } } }{ 4 } \times 2 } \right) } } } \right\} \theta \\ \alpha =\left( { \dfrac { { 2qE } }{ { ml } } } \right) \theta \\ ={ w^{ 2 } }\theta \\ \Rightarrow w=\sqrt { \dfrac { { 2qE } }{ { ml } } } \\ T=\dfrac { { 2\pi } }{ w } =2\pi \sqrt { \dfrac { { ml } }{ { 2qE } } } \end{array}$$

Which is the required answer.

1221164_1498687_ans_1bbed4cfe91546d49203dec20db375d0.JPG

Write definition of electric field intensity.
Obtain an expression for electron force and electric pressure on the surcharge of a charged conductor.
Draw necessary diagram.

Electric field intensity : It is the strength of an electric field at any point. It is equal to the electric force per unit charge experienced by a test charge placed at that point. The unit of it i s volt/meter or Newton/ Coulomb..
Every element of a charged conductor experienced a normal mechanical force. This is the result of repulsive force from similar change present on the rest of the surface of the conductor.
Consider a small element ds on the surface of a charge conductor. if $$\sigma $$ is the surface density and the charge carried by the element ds be dq.
Consider a point P just outside the surface near the element ds. The electric intensity at a point P is given by $$E = \dfrac{\sigma}{\in_o}$$
The direction of the intensity is normally outwards.
The intensity made up of two components $$\overset{\rightarrow}{E_1}$$ due to the small change dq present on the element ds.
$$ \overset{\rightarrow}{E_2}$$ due to the remaining charge present on the rest of the surface of the conductor.
Hence at $$P,\overset{\rightarrow}{E_1} +\overset{\rightarrow}{E_2} =\dfrac{\sigma}{\in_o}$$
* Now consider a point Q inside the conductor the electric field {E_l} and {E_2} at this point are oppositely directed. However, electric intensity inside a changed conductor is zero
$$\overset{\rightarrow}{E_1} - \overset{\rightarrow}{E_2} = 0$$
$$\overset{\rightarrow}{E_1} = \overset{\rightarrow}{E_2}$$
Substituting for $$E_1$$ in eq.(3) we have
$$2E_2 = \dfrac{\sigma}{\in_o}; E_2 = \dfrac{\sigma}{2 \in_o}$$
Here $$E_2$$ is the electric field due to the charge on the rest of coductor. Hence repulsive force experienced by elements ds carrying dq to is given by
$$\overset{\rightarrow}{f} = \overset{\rightarrow}{E_2}. dq$$
$$ F = \dfrac{\sigma}{2 \in_o} dq$$
$$ f = \dfrac{\sigma}{2 \in_o}. \sigma. dq$$
$$F = \dfrac{sigma^2}{2 \in_o}. dq$$
The electric pressure on surface ds will be
$$pressure = \dfrac{force}{Area}$$
$$\dfrac { \sigma^ 2 }{ \dfrac { 2 \in_o }{ ds } } ds$$
$$P = \dfrac{\sigma^2}{2 \in_o} N/M^2$$

1205180_1514616_ans_3b9808e096c949cdb93863d3f5a44624.png

1205180_1514616_ans_3b9808e096c949cdb93863d3f5a44624.png

Answer the following Questions.
(i) Define electric flux. Write its $$SI$$ units.
(ii) A spherical rubber balloon carries a charge that is uniformly distributed over its surface. As the balloon is blown up and increases in size, how does the total electric flux coming out of the surface change? Give reason.

(i) Total number of electric field lines crossing a surface normally is called electric flux. Its $$SI$$ unit is $$Nm^{2}C^{-1}$$ or $$Vm$$.
(ii) Total electric flux through the surface $$= \dfrac {q}{\epsilon_{0}}$$
As charge remains unchanged when size of balloon increases, electric flux through the surface remains unchanged.

Who discovered the way of producing electricity by friction?

Philospher Thales discovered it but it was satisfactorily explained by Dr Gilbert.

Why two electric lines of force/field cannot intersect each other?

Tangent to any electric lines of field gives the direction of net electric field at that point. If the two lines intersect, there will be two tangents and thus two directions of net electric field at that point which is not possible. So, two electric field lines do not intersect each other.

What is the shape of equipotential surface for a given point charge $$q$$.

For a point charge, spherical equipotential surface are produced.

An infinite wire having charge density $$\lambda$$ passes through one of the edges of a cube having edge length $$l$$. Find the
a. total flux passing through the cube.
b. flux passing through the surfaces which are in contact with the wire.
c. flux passing through the surfaces which are not in contact with the wire.

a. If we construct three more cubes as shown in Fig. (by dotted line), then the net flux through all the four cubes is $$\phi=\dfrac {q_{in}}{\epsilon_{0}} = \dfrac {\lambda l}{\epsilon_{0}}$$
So, flux through anyone cube $$= \dfrac {\lambda l}{4\epsilon_{0}}$$
b. Field lines will be parallel to the surfaces in contact with the wire. Hence, the flux through these surfaces will be zero. There are four such surfaces.
c. There are two surfaces which are not in contact with the wire. So, the flux $$\lambda/4\epsilon_{0}$$ will be divided among these two surfaces. So, the flux through each surfaces is $$\lambda/ 8\epsilon_{0}$$.
1575299_1751102_ans_141a737519174d6baec05e4ee71474dd.jpg

1575299_1751102_ans_141a737519174d6baec05e4ee71474dd.jpg

Two particles A and B, having opposite charges $$2.0\times 10^{-6}$$C and $$-2.0\times 10^{-6}$$C, are placed at a separation of $$1.0$$cm. Write down the electric dipole moment of this pair.

The distance between the charges is $$1.0\ cm$$

The magnitude of two charges is $$2\times 10^{-6}\ C$$

The Dipole moment is given as:

$$=q\times l$$
(where $$q=$$ magnitude of charge $$l=$$separation between the charges)

Substitute the values in above expression:
$$=2\times 10^{-6}\times 10^{-2}cm$$

$$=2\times 10^{-8}cm$$.

Fig. shows two dipole moments parallel to each other and placed at a distance $$x$$ apart. What is the magnitude of force of interaction? What is the nature of force, attractive or repulsive?

Potential energy of dipole system
$$ U = - \overrightarrow {p_2}. \overrightarrow {F_{21} } = -p_2 \frac {1}{ 4 \pi \epsilon_0} \frac {p-1}{ x^3} cos \pi $$
$$ E_{21} $$ is the field due to $$ p_1 $$ at $$ p_2 $$ as shown .
$$ U = \frac {1}{ 4 \pi \epsilon_0} \frac {p_1 / p_2}{x^3} $$
$$ F = - \frac {\partial U }{ \partial x} = \frac {3}{ 4 \pi \epsilon_0} \frac {p_1p_2}{x^4} $$
1581213_1751658_ans_ec7a516cc8a44e9b955ef301512502fb.jpg

1581213_1751658_ans_ec7a516cc8a44e9b955ef301512502fb.jpg

Obtain the formula for rotating the electric dipole in a uniform electric field.

Work Done in Rotating of an Electric Dipole In a Uniform Electric Field
On placing an electric dipole in an uniform electric field, a torque is produced on the dipole which tries toalign it in the direction of the electric field. Ultimately the dipole gets aligned with the field and attain sequilibrium. Thus it is obvious that work will be done in rotating the dipole in uniform electric field.The torque acting on an electric dipole when it is placed in a uniform electric field $$ \vec{E} $$ making an angle $$ \theta $$ with the field,
$$ \tau=p E \sin \theta \ldots$$(1)
Where $$ \mathrm{p}(=\mathrm{q} \times 2 \text { I) is electric dipole moment and } \mathrm{E} $$, the intensity of electric field.
Therefore work, done in giving an angular displacement $$ d \theta $$
$$ d W=\tau d \theta $$
If the dipole is rotated from $$ \theta_{1} $$ orientation to $$ \theta_{2} $$
$$W=\int d W=\int_{\theta_{1}}^{\theta_{2}} \tau d \theta$$
or
$$\begin{aligned}W &=\int_{\theta_{1}}^{\theta_{2}} p E \sin \theta d \theta \\&=p E \int_{\theta_{1}}^{\theta_{2}} \sin \theta d \theta \\&=p E[-\cos \theta]_{\theta_{1}}^{\theta_{2}} \\&=p E\left[-\cos \theta_{2}-\left(-\cos \theta_{1}\right)\right] \\& W=p E\left(\cos \theta_{1}-\cos \theta_{2}\right)\ldots(2)\end{aligned}$$
1. If the dipole is rotated from position of equilibrium i.e., $$ \quad \theta_{1}=0^{\circ} $$ to $$ \theta $$ i.e., $$ \theta_{2}=\theta, $$ then
$$W=p E\left(\cos 0^{\circ}-\cos \theta\right)$$
or $$ W=p E(1-\cos \theta) \ldots$$(3)
2. If the dipole is rotated by $$ 90^{\circ} $$ from equilibrium then $$ \theta=90^{\circ} $$
$$\begin{array}{ll}\therefore & W=p E\left(1-\cos 90^{\circ}\right) \\\text { or } & W=p E \ldots(4)\end{array}$$
3. In rotating the dipole by $$ 180^{\circ} $$ from equilibrium
$$\begin{array}{l}W=p E\left(1-\cos 180^{\circ}\right) \\\mathrm{or \quad} W=p E[1-(-1)]=p E(1+1)\end{array}$$
Or
$$W=2 p E$$
1760333_1835910_ans_15f14dab770347acb9b18e3aea28c460.png

1760333_1835910_ans_15f14dab770347acb9b18e3aea28c460.png

Why does the ebonite rod becomes negatively charged when it is rubbed against fur?

Because electrons in fur are loosely bound as compared to ebonite.

Two large thin metal plates are parallel and close to each other. On their inner faces, the plates have surface charge densities of opposite signs and of magnitude $$17\times 10^{-22}C/m^2$$. What is $$\vec E$$.
Between the plates.

Here, $$\sigma =17\times 10^{-22}Cm^{-2}$$

Between two plates

The electric field at this region is given by

$$E=\dfrac {\sigma}{\in_0}=\dfrac {17\times 10^{-22}}{8.854\times 10^{-2}}=1.92\times 10^{-10}NC^{-1}$$

1766435_1842595_ans_9f0bb3be0cde4cd582d7c05c83b30f93.jpg

A system has two charges $$q_A=2.5\times 10^{-7}C$$ and $$q_B=2.5\times 10^{-7}C$$ located at points
$$A:(0,015\ cm)$$, respectively. What are the total charge and electric dipole moment of the system?

The charges $$q_A$$ and $$q_B$$ are located at the points $$A(0,0,-15cm)$$ and $$B(0,0,+15cm)$$. $$B$$ The points $$A$$ and $$B$$ lie on $$z-$$axis as shown.

If follows that,

$$AB=OA+OB=30ci =0.3m$$

Total charge, $$q=q_A+q_B$$

$$=2.5\times 10^{-7}+(-2.5\times 10^{-7})$$

$$=0$$

Electric dipole moment, $$p=$$ either charge $$\times AB$$

$$=2.5\times 10^{-7}\times 0.3$$

$$=7.5\times 10^{-8}m$$

The electric dipole moment is directed from $$B$$ to $$A$$ along $$z-$$axis.

1766277_1842527_ans_b94730d5afad4aeeb34b4a2fd652150a.bmp

Two large thin metal plates are parallel and close to each other. On their inner faces, the plates have surface charge densities of opposite signs and of magnitude $$17\times 10^{-22}C/m^2$$. What is $$\vec E$$.
In the outer region of the first plate.

Here, $$\sigma =17\times 10^{-22}Cm^{-2}$$

To the left of plates.

The electric field at this region is zero.

1766431_1842592_ans_5463ff3fafd9404cb5bf2aaaea706164.jpg

Does the curve shown in figure represent electrostatic field lines?

It is not because, field lines cannot from closed loops without having touching the source o any charge.

Does the curve shown in figure represent electrostatic field lines?

It is not because, field lines cannot intersect each other.

Two large thin metal plates are parallel and close to each other. On their inner faces, the plates have surface charge densities of opposite signs and of magnitude $$17\times 10^{-22}C/m^2$$. What is $$\vec E$$.
In the outer region of the second plate.

Here, $$\sigma =17\times 10^{-22}Cm^{-2}$$

To the right of plates.

The electric field at this region is zero.

1766433_1842594_ans_4349f86b3c364af5af6e4dd76d474979.jpg

Does the curves shown in figure represent electrostatic field lines?

It do not represent electrostatic field lines because the field lines cannot emerge from a negative charge and cannot terminate at a positive charge.

Is it representing the electrostatic field lines?

It is not because field lines are not normal to the surfaces of the conductors.

Show that the tangential component of electrostatics field is continuous from one side of a charged surface to another [Hint : For $$(a)$$, use Gauss law, For $$(b)$$, use the fact that work done by electrostatic field on a closed loop is zero.]

Consider a closed path $$ABCD$$. If $$E_1^{''}=E_2^{''}$$, showing that tangential components then work done is $$E_I^{''} \Delta 1+E_2^{''}(- \Delta I)=0$$ (property of electric field) $$\therefore E_1^{''}-E_2{''}=0,\ E_1{''}-E_2^{''},$$ showing that tangential components ate continuous.

Does the curve shown in figure represent electrostatic field lines?

It is correct representation of electric field lines because electric field line emerge from the positive charge and repel each other.

A point dipole with an electric moment $$p$$ is located at a distance $$l$$ from an infinite conducting plane. Find the modulus of the vector of the force acting on the dipole if the vector $$p$$ is perpendicular to the plane.

Using the concept of electrical image, force on the dipole $$\vec {p}$$,
$$\vec {F} = p \dfrac {\partial \vec {E}}{\partial l}$$, where $$\vec {E}$$ is field at the location of $$\vec {p}$$ due to $$(-\vec {p})$$
or, $$|\vec {F}| = \left |\dfrac {\partial \vec {E}}{\partial l}\right | p = \dfrac {3p^{2}}{32\pi \epsilon_{0} l^{4}}$$
as, $$|\vec {E}| = \dfrac {p}{4\pi \epsilon_{0}(2l)^{3}}$$.
1784336_1851215_ans_59fcd357cf5949359001e87ae7f0a42b.jpg

1784336_1851215_ans_59fcd357cf5949359001e87ae7f0a42b.jpg

Electric Flux
A nonuniform electric field is given by the expression
$$\overset{\rightarrow}{E}=ay\,\overset{\rightarrow}{i} + bz \,\overset{\rightarrow}{j}+cx \,\overset{\rightarrow}{k}$$
where $$a, b,$$ and $$c$$ are constants. Determine the electric flux through a rectangular surface in the xy plane,extending from $$x = 0$$ to $$x =w$$ and from $$y=0$$ to $$y=h$$

We are given an electric field in the general form
$$\overset{\rightarrow}{E}=ay\hat{i}+bz\hat{j}+cx\hat{k}$$
In the $$xy$$ plane $$z=0$$ so that the electric field reduces to
$$\overset{\rightarrow}{E}=ay\hat{i}+cx\hat{k}$$
To obtain the flux, we integrate (see ANS Fig P24.6 for the definition of dA):
$$\phi_{E}=\int{\overset{\rightarrow}{E}\cdot d\overset{\rightarrow}{A}}=\int(ay\hat{i}+cx\hat{k})\cdot \hat{k}dA$$
$$\phi_{E}=ch \int _{x=0}^{w}{xdx}=\dfrac{ch\omega^{2}}{2}$$
where the $$\hat{k}$$ term was eliminated since $$\hat{k}\cdot \hat{k}=0$$
1807796_1860830_ans_e0f0262c9d27494ca9e583ebc62922c5.png

1807796_1860830_ans_e0f0262c9d27494ca9e583ebc62922c5.png

A line of charge starts at $$x = +x_0$$ and extends to positive infinity. The linear charge density is $$\lambda=\lambda_0 x_0/x$$,where $$\lambda_0$$ is a constant. Determine the electric field at the origin.

We integrate the expression for the incremental electric field to obtain
$$\overrightarrow{E}=\int d\overrightarrow{E}=\overset{\infty}{\underset{x_0}{\int}}\left[\dfrac{k_e \lambda_0 x_0 dx(-\hat{i})}{x^3}\right]=-k_e \lambda_0 x_0 \hat{i}\overset{\infty}{\underset{x_0}{\int}}x^{-3} dx$$
$$=-k_e \lambda_0 x_0\hat{i}\left(-\dfrac{1}{2x^2}|^{\infty}_{x_0}\right)$$
$$=\boxed{\dfrac{k_e \lambda_0}{2x_0}(-\hat{i})}$$

Electric Flux
An electric field of magnitude $$3.50 kN/C$$ is applied along the x axis. Calculate the electric flux through a rectangular plane $$0.35 m$$ wide and $$0.700 m$$ long
(a) if the plane is parallel to the yz plane,
(b) if the plane is parallel to the xy plane, and
(c) if the plane contains the y axis and its normal makes an angle of $$40.0^{o}$$ with the x axis.

For a uniform electric field passing through a plane surface,
$$\phi_{E}=\overset{\rightarrow}{E}\cdot \overset{\rightarrow}{A}=EA\cos \theta$$, where $$\theta$$ is the angle between the electric field and the normal to the surface,
(a) The electric field is perpendicular to the surface so $$\theta=0^{o}$$
$$\phi_{E}=(3,50\times 10^{3}N/C)[(0.350m)(0.700m)]\cos 0^{o}$$
$$=858N\cdot m^{2}/C$$

(b) The electric field is parallel to the surface $$\theta=90^{o}$$, so $$\cos \theta=0$$ and the flux is zero.

Two long parallel threads carry charge per unit length. The threads are separated by a distance $$l$$. The maximum field strength in the symmetry plane between them is given as $$\displaystyle \frac{\lambda}{x\pi \varepsilon_0 l}$$. Find $$x$$:

The electric field at a distance r due to long thread is $$\displaystyle E_1=\dfrac{\lambda}{2\pi \epsilon_0 r}$$

The field is symmetry due to both threads at middle of them. so $$\displaystyle r=\sqrt{z^2+(l/2)^2}$$

The net field at middle of them is $$E=2E_1\sin\theta$$

or $$E=2(\dfrac{\lambda}{2\pi \epsilon_0 r}) \sin\theta=\dfrac{\lambda}{\pi \epsilon_0 r} \dfrac{z}{r}$$

or $$E=\dfrac{\lambda}{\pi \epsilon_0} \dfrac{z}{z^2+\dfrac{l^2}{4}}$$

For max E, $$\dfrac{dE}{dz}=0$$

$$\displaystyle \Rightarrow [\dfrac{1}{z^2+\dfrac{l^2}{4}}-\dfrac{2z^2}{(z^2+\dfrac{l^2}{4})^2}]=0$$

or $$\displaystyle 2z^2=z^2+\dfrac{l^2}{4}$$

or $$\displaystyle z^2=\dfrac{l^2}{4} \Rightarrow z=\dfrac{l}{2}$$

thus, $$\displaystyle E_{max}=\dfrac{\lambda}{\pi \epsilon_0} \dfrac{(l/2)}{(l/2)^2+\dfrac{l^2}{4}}=\dfrac{\lambda}{\pi \epsilon_0 l}$$

so $$x=1$$

Two point charges $$q$$ and $$-q$$ are separated by a distance $$2a$$ (Fig.). Evaluate the flux of electric field strength vector across a circle of radius $$R$$.

Electric field due to a dipole at equatorial line

$$E = \dfrac{2ql}{4\pi \varepsilon_0(l^2+x^2)^{3/2}}$$
Let elemental surface area be $$xd\theta dx$$
$$\displaystyle \phi = \int{Ex d\theta dx} = \dfrac{2ql}{4\pi\varepsilon_0}\int_{0}^{R} \dfrac{xdx}{(l^2+l^2)^{3/2}}\int_{0}^{2\pi}d\theta$$
$$=\displaystyle \dfrac{ql}{\varepsilon_0}\int_{0}^{R}\dfrac{xdx}{(l^2+x^2)^{3/2}}=\dfrac{q}{\varepsilon_0}\left[1-\dfrac{l}{\sqrt{l^2+R^2}}\right]$$

A charged particle q of mass m is in equilibrium at a height h from a horizontal infinite line charge with uniform linear charge density $$\lambda$$. The charge lies in the vertical plane containing the line charge. If the particle is displaced slightly (vertically), prove that the motion of the charged particle will be simple harmonic. Also find its time period

At equilibrium $$mg = qE$$

$$\displaystyle mg = \frac{q 2 k \lambda}{h}$$ (i)

When particle is displaced to a distance x in upward direction, then net force in upward direction is

$$\displaystyle F = q \left ( \frac{2k \lambda}{h + x} \right ) - mg$$

$$= \displaystyle \frac{2 k q \lambda}{h} \left ( 1 + \frac{x}{h} \right )^{-1} - mg = \frac{2 k q \lambda}{h} \left ( 1 - \frac{x}{h} \right ) - mg$$

$$= \displaystyle - \left ( \frac{2 K q \lambda}{h^2} \right ) x = - K_0 x$$

$$\therefore F \propto - x $$ (Motion is SHM)

Time period of motion is

$$\displaystyle T = 2 \pi \sqrt{\frac{m}{K_0}} = 2 \pi \sqrt{\frac{mh^2}{2 kq \lambda}}$$

Two mutually perpendicular long straight conductors carrying uniformly distributed charges of linear charge densitites $$\lambda_1$$ and $$\lambda_2$$ are positioned at a distance a from each other. How does the interaction between the rods depend on a?

Let us make two-dimensional view of the situation. Because of symmetry we can say the force on wire 2 ($$\lambda_2$$) will be along the negative y-direction.

$$dF = E_x dq cos \theta$$

$$\displaystyle F = \int dF = \int \frac{\lambda_1}{2 \pi \varepsilon_0 \sqrt{a^2 + x^2}} (\lambda_2 dx) cos \theta$$

where $$cos \theta = \displaystyle \frac{a}{\sqrt{a^2 + x^2}}$$

$$\displaystyle F = \frac{\lambda_1 \lambda_2}{2 \pi \varepsilon_0} a \int_{- \infty}^{\infty} \frac{dx}{a^2 + x^2} = \frac{\lambda_1 \lambda_2 a}{2 \pi \varepsilon_0} \times \frac{1}{a} \left [ tan^{-1} \frac{x}{a} \right ]$$

$$F = \displaystyle \frac{\lambda_1 \lambda_2}{2 \pi \varepsilon_0}$$.

It is independent of a.

Four charge particles each having charge $$Q = 1$$ C are fixed at the corners of the base (A, B, C and D) of a square pyramid with slant length $$a(AP = BP = BP = PC = a = \sqrt{2}m),$$ a charge -Q is fixed at point P. A dipole with dipole moment $$p = 1$$ Cm is placed at the center of the base and perpendicular to its plane as shown in figure. Force on the dipole due to the charge particles is $$\displaystyle \frac{x}{4 \pi \varepsilon_0} N$$. Find $$x$$.

Here, $$\displaystyle OA=OB=OC=OD=r=\sqrt{(a/2)^2+(a/2)^2}=a/\sqrt{2}=1 m$$ and $$\displaystyle OP=\sqrt{a^2-(a/\sqrt{2})^2}=a/\sqrt{2}=r$$
The electric field at the each vertex of the square ABCD due to dipole is $$\displaystyle E=\frac{p}{4\pi \epsilon_0 r^3}$$
Hence, force on each of them due to dipole is $$\displaystyle F_1=QE=\frac{Qp}{4\pi \epsilon_0 r^3}$$
This force is downward on charges. Hence, force due to these charges on dipole is $$4F_1$$ upward.
Now the force on dipole due to charge at P is $$\displaystyle F_2=\frac{Qp}{4\pi \epsilon_0 (OP)^3}=\frac{Qp}{4\pi \epsilon_0 r^3}$$ upward.
Net force on dipole is $$\displaystyle F=4F_1+F_2=\frac{4Qp}{4\pi \epsilon_0 r^3}+\frac{Qp}{4\pi \epsilon_0 r^3}=\frac{5}{4\pi \epsilon_0 }$$ as $$Q=1, p=1, r=1$$
thus, $$x=5$$

A short dipole is placed along the x-axis at x =x.
a. Find the force acting on the dipole due to a point charge q placed at the origin.
b. Find the force on the dipole if the dipole is rotated by $$180^{\circ}$$ about the z-axis.
c. Find the force on dipole if the dipole is rotated by $$90^{\circ}$$ anticlockwise about z-axis, i.e., it becomes parallel to the y-axis.

$$\displaystyle E = \frac{1}{4 \pi \varepsilon_0} \frac{q}{x^2}$$

$$\displaystyle U = -pE cos 0^{\circ} = - \frac{qp}{4 \pi \varepsilon_0 x^2}$$

$$\displaystyle = - \frac{\partial U}{\partial x} = \frac{-pq}{2 \pi \varepsilon_0 x^3}$$

Negative sign indicates that force on dipole is toward the positive x-direction or the force is attractive.

b. $$\displaystyle U = -pE cos 180^{\circ} = \frac{qp}{4 \pi \varepsilon_0 x^2}$$

$$\displaystyle = - \frac{\partial U}{\partial x} = \frac{-pq}{2 \pi \varepsilon_0 x^3}$$

Positive sign indicates that force on dipole is toward the positive x-direction or the force is repulsive.

C. $$\displaystyle E=\frac{1}{4\pi \varepsilon_0} \frac{p}{x^3}$$

Let us first find force on q due to p.

$$\displaystyle F = qE = \frac{qp}{4\pi \varepsilon_0 x^3}$$

Charge q will also apply the same force on dipole but in an opposite direction, so the force on dipole is

$$\displaystyle F = \frac{qp}{4\pi \varepsilon_0 x^3}$$ along $$\vec{p}$$ or parallel to y-axis.

The electric field in a region is given by $$\displaystyle \vec{E} = \frac{3}{5} E_o \widehat{i} + \frac{4}{5} E_o \widehat{j}$$ with $$E_o = 2.0 \times 10^3$$ N/C. Find the flux of this field (in $$Nm^2C^{-1}$$) through a rectangular surface of area 0.2 m$$^2$$ parallel to the Y-Z plane.

Figure 3.118 shows two dipole moments parallel to each other and placed at a distance x apart. What is the magnitude of force of interaction? What is the nature of force, attractive or repulsive?

Potential energy of dipole system

$$\displaystyle U = -\bar{p_2}. \vec{E_{21}} = - p_2 \frac{1}{4 \pi \varepsilon_0} \frac{p_1}{x^3} cos \pi$$

$$E_{21}$$ is the field due to $$p_1 at p_2$$ as shown.

$$\displaystyle U = \frac{1}{4 \pi \varepsilon_0}\frac{p_1 p_2}{x^3}$$

$$\displaystyle F = -\frac{\partial U}{\partial x} = \frac{3}{4 \pi \varepsilon_0} \frac{p_1 p_2}{x^4}$$

F comes out to be positive, so it is a repulsive force

Two point charges q and -q are separated by a distance 2l, Find the flux of electric field strength vector across the circle of radius R placed with its centre coinciding with the midpoint of line joining the two charges in the perpendicular plane.

$$E=\dfrac{kq}{x^2+y^2+l^2}\dfrac{x \hat i+y \hat j-l \hat k}{\sqrt{x^2+y^2+z^2}} - \dfrac{kq}{x^2 + y^2 + l^2} \dfrac{x \hat i +y \hat j +l \hat z}{\sqrt{x^2 + y^2 + z^2}}$$
$$=-\dfrac{2}{4 \pi \epsilon_0} \dfrac{ql \hat k}{(x^2 + y^2 + z^2)^{3/2}}$$
$$\phi= \int \overrightarrow E. d \overrightarrow A= \int E dx dy \hat k$$
$$=\dfrac{-1}{2 \pi \epsilon_0} ql \int \dfrac{dx dy}{(x^2 + y^2+z^2)^{3/2}}$$
By doing change of coordinates,
$$E=\dfrac{-1}{2 \pi \epsilon_0} ql \int \dfrac{r d \theta dr}{(r^2 + l^2)^{3/2}}$$
$$=\dfrac{-1}{2 \pi \epsilon_0} ql 2 \pi \int^R _0 \dfrac{r dr}{(r^2 + l^2)^{3/2}}$$
By doing change of coordinates, $$u=r^2 + l^2$$
$$E=\dfrac{-ql}{\epsilon_0} \int \dfrac{du}{2 u^{3/2}}$$
After solving this integral finally,
$$E = \dfrac{q}{\epsilon_0}(1 - \dfrac{1}{\sqrt{1+ (R/l)^2}})$$

(a) Show that the normal component of electrostatic field has a discontinuity from one side of a charged surface to another given by
$$(E_2\,-\, E_1)\, \dot\, \hat{n}\,\displaystyle \frac{\sigma }{\varepsilon _0}$$
where $$\hat{n}$$ is a unit vector normal to the surface at a point and $$\sigma$$ is the surface charge density at that point. (The direction of $$\hat{n}$$ is from side 1 to side 2.) Hence show that just outside a conductor, the electric field is $$\sigma \hat{n} /\epsilon _0$$
(b) Show that the tangential component of electrostatic field is continuous from one side of a charged surface to another. [Hint: For (a), use Gauss's law. For, (b) use the fact that work done by electrostatic field on a closed loop is zero.]

(a)

Electric field on one side of a charged body is $$E_1$$ and electric field on the other side of the same body is $$E_2$$. If infinite plane charged body has a uniform thickness, then electric field due to one surface of the charged body is given by,

$$\displaystyle \overline{E_1}=-\frac{\sigma}{2\in_0}\hat{n}$$ ...(i)

Where,

$$\hat{n}=$$ Unit vector normal to the surface at a point

$$\sigma =$$Surface charge density at that point

Electric field due to the other surface of the charged body,

$$\displaystyle \overline{E_2}=\frac{\sigma}{2\in_0}\hat{n}$$ ...(ii)

Electric field at any point due to the two surfaces,

$$\displaystyle \overline{E_2}-\overline{E_1} = \frac{\sigma}{2\in_0}\hat{n}+\frac{\sigma}{2\in_0}\hat{n}=\frac{\sigma}{\in_0}\hat{n}$$

$$\displaystyle (\overline{E_2}-\overline{E_1}).\hat{n}=\frac{\sigma}{\in_0}$$ ...(iii)

Since inside a closed conductor, $$\overline{E_1}=0$$,

$$\displaystyle \therefore \overline{E}=\overline{E_2}=\frac{\sigma}{\in_0}\hat{n}$$

Therefore, the electric field just outside the conductor is $$\displaystyle \frac{\sigma}{\in_0}\hat{n}$$.

(b)

When a charged particle is moved from one point to the other on a closed loop, the work done by the electrostatic field is zero. Hence, the tangential component of electrostatic field is continuous from one side of a charged surface to the other.

A long charged cylinder of linear charged density $$\lambda $$ is surrounded by a hollow co-axial conducting cylinder. What is the electric field in the space between the two cylinders?

Charge density of the long charged cylinder of length L and radius r is $$\lambda$$.

Another cylinder of same length surrounds the pervious cylinder. The radius of this cylinder is R.

Let E be the electric field produced in the space between the two cylinders.

Electric flux through the Gaussian surface is given by Gauss’s theorem as,

$$\phi=E(2\pi d)L$$

Where, $$d=$$ Distance of a point from the common axis of the cylinders

Let q be the total charge on the cylinder.

It can be written as

$$\therefore \phi = E(2\pi dL)=\frac{q}{\in_0}$$

Where,

$$q=$$ Charge on the inner sphere of the outer cylinder

$$\in_0=$$ Permittivity of free space

$$E(2\pi dL)=\frac{\lambda L}{\in_0}$$

$$\displaystyle E=\frac{\lambda}{2\pi \in_0d}$$

Therefore, the electric field in the space between the two cylinders is

$$\displaystyle \frac{\lambda}{2\pi \in_0d}$$.

In a certain region of space, electric field is along the z-direction throughout. The magnitude of electric field is, however, not constant but increases uniformly along the positive z-direction, at the rate of $$10^5\, NC^{-1}\ m^{-1}$$ per metre. What are the force and torque experienced by a system having a total dipole moment equal to $$10^{-7}$$ Cm in the negative z-direction ?

Dipole moment of the system, $$p=q\times dl=-10^{-7}Cm$$

Rate of increase of electric field per unit length,

$$\displaystyle \frac{dE}{dl}=10^{+5}NC^{-1}$$

Force (F) experienced by the system is given by the relation,

$$F=qE$$

$$F=q\frac{dE}{dl}\times dl$$

$$=p\times \frac{dE}{dl}$$

$$=10^{-7}\times 10^{-5}$$

$$=-10^{-2}N$$

The force is $$-10^{-2}N$$ in the negative z-direction i.e., opposite to the direction of electric field. Hence, the angle between electric field and dipole moment is $$180^o$$.

Torque (τ) is given by the relation,

$$T=pE\, sin\, 180^o=0$$

Therefore, the torque experienced by the system is zero.

Two large, thin metal plates are parallel and close to each other. On their inner faces, the plates have surface charge densities of opposite signs and of magnitude $$17.0\times10^{-22}\,C/m^{2}$$. What is E: (a) in the outer region of the first plate. (b) in the outer region of the second plate, and (c) between the plates?

The situation is represented in the following figure.

A and B are two parallel plates close to each other. Outer region of plate A is labelled as I, outer region of plate B is labelled as III, and the region between the plates, A and B, is labelled as II.

Charge density of plate A, $$\sigma = 17.0\times 10^{-22}C/m^2$$

Charge density of plate B, $$\sigma = -17.0\times 10^{-22}C/m^2$$

Electric field in regions can be found with the help of Gauss Law. In the regions, I and III, electric field E is zero. This is because charge is not enclosed by the Gaussian surfaces of the plates.

Electric field E in region II is given by the relation,

$$\displaystyle E=\frac{\sigma}{\in_0}$$

Where,

$$\in_0=$$ Permittivity of free space $$=8.854\times 10^{-12}N^{-1}C^2m^{-2}$$

$$\therefore E=\displaystyle \frac{17.0\times 10^{-22}}{8.854\times 10^{-12}}$$

$$=1.92\times 10^{-10}N/C$$

Therefore, electric field between the plates is $$1.92\times 10^{-10}N/C$$.

State Gauss' law. Using this find an expression for electric field due to an infinitely long straight charged wire uniform charge density.

Gauss's law :
The law relates the flux through any closed surface and the net charge enclosed with in the surface. The law states that the total flux of the electric field $$E$$ over any closed surface is equal to $$\dfrac { 1 }{ { \varepsilon }_{ 0 } }$$ times the net charge enclosed by the surface.
$$\phi =\dfrac { q }{ { \varepsilon }_{ 0 } }$$
This closed imaginary surface is called Gaussian surface. Gauss's law tells us that the flux of $$E$$ through a closed surface $$S$$ depends only on the value of net charge inside the surface and not on the location of the charges. Charges outside the surface will not contribute to flux.

Electric field due to an infinitely long straight uniformly charged wire :
Consider an uniformly charged wire of infinite length having a constant linear charge density $$\lambda$$ (Charge per unit length). Let $$P$$ be a point at a distance $$r$$ from the wire and $$E$$ be the electric field at the point $$P$$. A cylinder of length $$l$$, radius $$r$$, closed at each end by plane caps normal to the axis is chosen as Gaussian surface. Consider a very small area $$ds$$ on the Gaussian surface. By symmetry, the magnitude of the electric field will be the same at all points on the curved surface of the cylinder and directed radially outward. $$\vec { E }$$ and $$\vec { ds }$$ are along the same direction.
The electric flux $$\left( \phi \right)$$ through curved surface $$=\displaystyle\oint { Eds\cos { \theta } }$$
$$\phi =\displaystyle\oint { Eds }$$ $$ \left[ \because \theta =0;\cos { \theta } =1 \right]$$
$$=E\left( 2r\pi l \right)$$ [The surface area of the curved part is ] since $$\vec { E }$$ and $$\vec { ds }$$ are right angles $$2\pi rl$$ to each other, the electric flux through the plane caps $$=0$$.
$$\therefore$$ Total flux through the Gaussian surface, $$\phi =E\left( 2\pi rl \right)$$. The net charge enclosed by Gaussian surface is, $$q=\lambda l$$
$$\therefore$$ By Gauss's law,
$$ =E\left( 2\pi rl \right) \dfrac { \lambda l }{ { \varepsilon }_{ 0 } }$$ or $$ E=\dfrac { \lambda }{ 2\pi { \varepsilon }_{ 0 }r }$$
The direction of electric field $$E$$ is radially outward, if line charge is positive and inward, if the line charge is negative.

Two plates A and B are placed one above the other in the gravitational field and a block of mass m is connected to the upper plate by a spring of spring constant k. Its time period is found to be T. Now the space between the plates is made gravity free and a charge $$+q$$ is given to the block of mass m and an electric field E is produced in the direction shown. The new time period is?

Two plates A and B are placed one above the other in the gravitation field.

Spring constant$$=K$$

Time period$$=T$$

Charge$$=+q$$

Mass$$=m$$

Electric field $$=E$$

New time period $$=?$$

Actually in equilibrium

$$Kx=mg\\x=\cfrac{mg}{K}\\x=\cfrac{2mg}{K}$$

When if it will dropped the block when the spring is in its natural length then block will go upto.

$$T=2\pi\sqrt{\cfrac{x}{m}}\\ \quad=2\pi\sqrt{\cfrac{\cfrac{mg}{K}}{m}}\\ \quad=2\pi\sqrt{\cfrac{g}{K}}$$

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Find the electric field intensity at a point $$P$$ which is at a distance $$R$$ (point lying on the perpendicular drawn to the wire at one of its end) from a semi-infinite uniformly charged wire. (Linear charge density $$= \lambda$$.)

Field at point $$P$$ is due to an elemental charge $$dq (= \lambda d\ell)$$
$$dE = k\dfrac{dq}{(R^2 + \ell^2)}$$

The component along x-axis

$$dE_x = \dfrac{kdq \cos \theta}{(R^2 + \ell^2)} = \dfrac{k(dq)R}{(R^2 + \ell^2)^{3/2}}$$

$$\therefore E_x = \underset{0}{\overset{\infty}{\int}}\dfrac{1}{4\pi \in_0}\dfrac{\lambda d\ell R}{(R^2 + \ell^2)^{3/2}} \Longrightarrow E_x = \dfrac{\lambda}{4\pi \in_0 R}$$

Similarly, $$E_y = \dfrac{1}{4\pi \in_0}\dfrac{\lambda (d\ell) \ell}{(R^2 + \ell^2)^{3/2}}$$

$$\therefore E_y = \dfrac{\lambda}{4\pi \in_0 R}$$

$$\therefore E = \sqrt{E_x^2 + E_y^2} = \dfrac{\lambda}{2 \sqrt{2}\pi \in_0 R}, \tan \theta = \dfrac{E_y}{E_x}$$

$$\Longrightarrow \theta = 45^o$$
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Consider a uniform electric field $$E=3\times 10^3$$ $$\hat{i}$$ N/C. (a) What is the flux of this field through a square of $$10$$ cm on a side whose plane is parallel to the yz plane? (b) What is the flux through the same square if the normal to its plane makes a $$60^o$$ angle with the x-axis?

Area $$A=(0.1)^2=0.01m^2$$

Flux $$\phi=EA\cos\theta$$

$$(i)\theta=0^o$$

$$\phi=3\times 10^3\times 0.01\times \cos0^o$$

$$\implies \phi=30Nm^2/C$$

$$(i)\theta=60^o$$

$$\phi=3\times 10^3\times 0.01\times\dfrac{1}{2}$$

$$\implies \phi=15Nm^2/C$$

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A system has two charges $${q}_{A}=2.5\times {10}^{-7}$$ and $${q}_{B}=2.5\times {10}^{-7}$$, located at point $$A$$: $$(0, 0, -15\ cm)$$, $$B$$: $$(0, 0, 15\ cm)$$respectively. What are the total charge and electric dipole moment of the system?

Charge on A, $$q_A=2.5 \times 10 ^{-7} C$$

Charge on B, $$q_B=2.5 \times 10 ^{-7} C$$

Therefore, total charge q will be

$$q=q_A+q_B$$

$$=2.5\times10^{-7}+2.5\times10^{-7}$$

$$=5\times10^{-7}C$$

Distance between two charges, $$a=15-(-15)=30cm=0.3m$$

Electron dipole moment,

$$p=q.a$$

$$=2.5\times10^{-7}\times0.3$$

$$=7.5\times 10^{-7}Cm$$ (along z-axis)

Because the direction of dipole is always negative to positive.

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Two identical helium filled balloon $$A$$ and $$B$$m fastened to a weight of $$5g$$ by threads floats in equilibrium as shown in fig. Calculate the charge on each balloons, assuming that they carry equal charges

$$ OC = \sqrt {OA^2 - AC^2} $$

= $$\sqrt {(1)^2 - (0.3)^2} $$ = 0.954

As the balloons float in equilibrium.

Using triangle law of vectors, We get -

$$\frac{F}{AC} = \frac{w}{OC}$$

where, w = upthrust acting on each ballon.

As W = $$w + w $$

$$w = \frac{W}{2}$$

F = $$\frac{AC}{OC} \times \frac {W}{2}$$

$$\frac{q \times q }{4 \pi \epsilon_0 (AB)^2} $$ = $$\frac{AC}{OC} \times \frac{mg}{2}$$

$$q^2 = \frac{(0.6)^2 \times (0.3) \times (5 \times 10^{-3} \times 9.8 )}{9 \times 10^9 \times 0.954 \times 2}$$

$$q^2 = \frac{0.36 \times 0.30 \times 49}{9 \times 0.954 \times 2 } \times 10^{-7} C$$

q = $$ ^+_- 5.6 \times 10^{-7} C $$

Two ball $$A$$ and $$B$$ of same mass $$(M)$$ and charges $$Q, -Q$$ are suspended by two strings of same length from two different suspension points $${S}_{1}$$ and $${S}_{2}$$ if $${S}_{1}{S}_{2}=3x$$ and $$AB=x$$ then show that-
tension in the string is $$T=\dfrac { { Q }^{ 2 }L }{ 4\pi { \in }_{ 0 }{ x }^{ 3 } }$$
$$\tan { \theta } =\dfrac { { Q }^{ 2 } }{ 4\pi { \in }_{ 0 }{ x }^{ 2 }Mg }$$

Angle $$\theta$$ is not defined from where it is measure in complete question .

let angle made by eachh string is $$\theta$$ with vertical

$$sin \theta = \frac{x}{L} $$from the forces .

force by electrostat F = $$\frac{KQ^2}{x^2}$$

F = t sin $$\theta $$

$$\frac{KQ^2}{sin \theta x^2}$$ = T

T = $$ \frac{KQ^2}{(\frac{x}{L})x^2}$$ tension in the string

T cos$$\theta = Mg$$

divide both the eqn

$$tan \theta = \frac{F}{Mg}$$

= $$\frac{\frac{KQ^2}{x^2}}{Mg}$$

= $$\frac{KQ^2}{Mg x^2}$$

A dipole is formed by taking two charges $$+8\ \mu C$$ and $$-8\mu C$$ and keeping them $$5\ mm$$ apart. Find the electric field at an axial point $$P$$ situated at $$20\ cm$$ from the center of a dipole. What would be the electric intensity if the point $$P$$ were on the equator of the dipole?

Electric field due to dipole on axial position , $$E_a=\dfrac{2kp}{r^3}=\dfrac { 2\times9\times { 10 }^{ 9 }\times 5\times { 10 }^{ -3 }\times 8\times { 10 }^{ -6 } }{ { 0.2 }^{ 3 } } =9\times10^{3}{ N/C}$$

electric field on equtorial point , $$E_e=\dfrac{kp}{r^3}=\dfrac { 9\times { 10 }^{ 9 }\times 5\times { 10 }^{ -3 }\times 8\times { 10 }^{ -6 } }{ { 0.2 }^{ 3 } } =4.5\times10^3N/C$$

A particle, having a charge $$+5\ \mu C$$, is initially at rest at the point $$x=30\ cm$$ on the $$x-$$axis. The particle begins to move due to the presence of a charge $$Q$$ that is kept fixed at the origin. Find the kinetic energy of the particle at the instant it has moved $$15\ cm$$ from its initial position if
(a) $$Q=+5\mu C$$ and (b) $$Q=-15\mu C$$

Gain in K.E$$=$$Difference in P.E

Given:$${q}_{1}=5\mu$$C, $${q}_{2}=\pm 15\mu$$C and $${r}_{12}=30$$

K.E at the instant $${q}_{1}$$ moved $$15$$cm from the initial position.

Initial P.E, $$E=\dfrac{9\times{10}^{9}\times 5\times{10}^{-6}\times 15\times{10}^{-6}}{3\times{10}^{-2}}$$

$$(a)$$Since both charges are positive, the charges repel.Hence, the charge $${q}_{1}$$ moves away by $$15$$cm.

Hence, $${r}_{12}=30+15=45$$cm

$${E}^{\prime}=\dfrac{9\times{10}^{9}\times 5\times{10}^{-6}\times 15\times {10}^{-6}}{45\times{10}^{-2}}=1.5$$

K.E$$=E-{E}^{\prime}=2.25-1.5=0.75$$

$$(b)$$Since both charges are opposite, the charges attract.Hence, the charge $${q}_{1}$$ moves closer by $$15$$cm.

Hence, $${r}_{12}=30-15=15$$cm

$${E}^{\prime}=\dfrac{9\times{10}^{9}\times 5\times{10}^{-6}\times 15\times {10}^{-6}}{15\times{10}^{-2}}=4.5$$

K.E$$={E}^{\prime}-E=4.5-2.25=2.25$$

An electric dipole of momentum $$\vec{p}$$ is placed in a uniform electric field. The dipole is rotated through a very small angle $$\theta$$ from equilibrium and is released, find the time period of simple harmonic motion.

Whendisplacedatanangletheta,fromitsmeanpositionthemagnitudeofrestoringtorqueis

$$τ=−Psinθ$$

For small angular displacements in $$θ≈θ$$

$$τ=−PEθ$$

$$α=\frac { τ }{ I } =−\frac { PE }{ I } θ=−{ w }^{ 2 }θ$$

$${ w }^{ 2 }=\frac { PE }{ I } $$

$$T=2π\sqrt { \frac { IP }{ E } } $$

Write the expression for electric intensity at a point due to point charge and explain the terms.

Electric Intensity due to a point charge:-

$$F=Eq_0$$ [force experienced by the charge $$q_0$$]

$$F=\dfrac {k q q_0}{r^2}$$ [according to collomb's law]

$$r=$$ distance between the $$q$$ and $$q_0$$ source charge

we know, $$E=\dfrac {F}{q_0}$$

$$E=\dfrac {1}{q_0}\times \dfrac {kqq_0}{r^2}$$

$$E=\dfrac {kq}{r^2}$$ we know $$k=\dfrac {1}{4\pi \in_0}$$

Final Expression $$\rightarrow E=\dfrac {1}{4\pi \in_0}\times \dfrac {q}{r^2}$$

$$\in _0=$$ dielectric constant of letan $$k$$ space

$$q=$$ point charge

$$r=distance between two point charge

If there is a medium of dielectric constant $$\in_{r}$$ then,

$$E=\dfrac {1}{4\pi \in_0\in_r}.\dfrac {q}{r^2}$$

Vector from $$\rightarrow \ \vec E=\dfrac {1}{4\pi \in_0 \in_r}.\dfrac {q}{r^2}.\vec r$$

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An electric field is uniform, and in the positive $$x$$ direction for positive $$x$$, and uniform with the same magnitude but in the negative $$x$$ direction for negative $$x$$. It is given that $$E = 200 \,\hat{i} \,N/C$$ for $$x > 0$$ and $$E = -200 \,\hat{i} \,N/C$$ for $$x < 0$$. A right circular cylinder of length $$20 \,cm$$ and radius $$5 \,cm$$ has its centre at the origin and its axis along the x-xis so that one face is at $$x = +10 \,cm$$ and the other is at $$x = -10 \,cm$$.
(a) What is the net outward flux through each flat face?
(b) What is the flux through the side of the cylinder?
(c) What is the net outward flux through the side of the cylinder?

(a) Net $$\phi$$ through flat
surface $$= \overrightarrow{E} \cdot \overrightarrow{A} [\theta = 0^o]$$
$$= EA \cos \theta$$
$$= (200) (\pi 0.05^2)$$

$$= \dfrac{\pi}{2} \dfrac{Nm^2}{C^2}$$

$$= 1.5707 \dfrac{Nm^2}{C^2}$$

(b) Net $$\phi$$ through curved
surface $$\overrightarrow{E} \cdot \overrightarrow{A}$$
$$= 0 [\because \overrightarrow{E} \bot \overrightarrow{A}]$$

(c) Net outward flux $$= 2 \times \phi$$ flat surface

$$= \pi \dfrac{Nm^2}{C^2}$$

$$= 3.14 \dfrac{Nm^2}{C^2}$$

Two equal and opposite charges each of $$2\mu C$$ are placed at a distance of 3.0 cm. Dipole moment of the system will be

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A charged particle of mass 1.0 g is suspended through a silk thread of length 40cm in a horizontal electric field of $$4\times 10^{ 4 }\quad N/C$$. If the particle stays at a distance of 24 cm from the vertical line passing through the suspension points in equilibrium , The charges on the particle is:

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An arbitrary surface encloses a dipole. What is the electric flux through this surface ?

Net charge on a dipole $$= -q + q = 0$$.

According to Gauss's theorem, electric flux through the surface,
$$= \dfrac {q}{\epsilon_{0}} = \dfrac {0}{\epsilon_{0}} = 0$$.

State Coulomb's law, explain its vector form and define S.I. unit of electric charge. State two limitations of Coulomb's law.

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(i) Using Gauss Theorem show mathematically that for any point outside the shell, the field due to a uniformly charged spherical shell is same as the entire charge on the shell, is concentrated at the centre.
(ii) Why do you expect the electric field inside the shell to be zero according to this theorem?

(i) Electric field intensity at a point outside a uniformly charged thin spherical shell: Consider a uniformly charged than spherical shell of radius $$R$$ carrying charge $$Q$$. To find the electric field outside the shell. We consider a spherical Gaussian surface of radius $$r(> R)$$, concentric with given shell. If $$\vec {E}$$ is electric field outside the shell, then by symmetry electric field strength has same magnitude $$E_{0}$$ on the Gaussian surface and is directed radially outward. Also the directions of normal at each point radially outward, so angle between $$\vec {E}_{i}$$ and $$\vec {dS}$$ is zero at each point
Hence, electric flux through Gaussian surface.
$$\oint S = \vec {E}_{0}\cdot \vec {dS}$$
$$\oint = E_{0} dS\cos 0 = E_{0}\cdot 4\pi r^{2}$$
Now, Gaussian surface is outside the given charged shell, so charge enclosed by Gaussian surface is $$Q$$.
Hence, by Gauss's theorem
$$\oint s = \vec {E}_{0} \cdot \vec {dS} = \dfrac {1}{\epsilon_{0}} \times charged\ enclosed$$
$$\Rightarrow E_{0} \cdot 4\pi r^{2} = \dfrac {1}{\epsilon_{0}} \times Q \Rightarrow E_{0} = \dfrac {1}{4\pi \epsilon_{0}} \dfrac {Q}{r^{2}}$$
Thus, electric field outside a charged thin spherical shell is the same as if the whole charge $$Q$$ is concentrated at the centre.
If $$\sigma$$ is the surface charge density of the spherical shell, then
$$Q = 4\pi R^{2}\sigma$$ coulomb
$$\therefore E_{0} = \dfrac {1}{4\pi \epsilon_{0}} \dfrac {4\pi R^{2}\sigma}{r^{2}} = \dfrac {R^{2}\sigma}{\epsilon_{0}r^{2}}$$
(ii) Electric field inside the shell (hollow charged conducting sphere) : The charge resides on the surface of a conductor. Thus a hollow charged conductor is equivalent to a charged spherical shell. To find the electric field inside the shell, we consider a spherical Gaussian surface of radius $$r(< R)$$ concentric with the given shell. If $$\vec {E}$$ is the electric field inside the shell, then by symmetry electric field strength has the same magnitude $$E_{i}$$ on the Gaussian surface and is directed radially outward. Also the directions of normal at each point is radially outward, so angle between $$\vec {E_{i}}$$ and $$\vec {dS}$$ is zero at each point.
Hence, electric flux through Gaussian surface
$$= \int_{S} \vec {E}_{i}\cdot \vec {dS} = \int E_{i} dS \cos 0 = E_{i} \cdot 4\pi r^{2}$$
Now, Gaussian surface is inside the given charged shell, so charge enclosed by Gaussian surface is zero.
Hence, by Gauss's theorem
$$\int_{S} \vec {E}_{i}\cdot \vec {dS} = \dfrac {1}{\epsilon_{0}}\times charge\ enclosed$$
$$\Rightarrow E_{i} 4\pi r^{2} = \dfrac {1}{\epsilon_{0}} \times 0 \Rightarrow E_{i} = 0$$
Thus, electric field at each point inside a charged thin spherical shell is zero. The graph is shown in fig.
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The figure shows a section of a long, thin-walled metal tube of radius $$R = 3.00$$ cm, with a charge per unit length of $$\lambda = 2.00 \times10^{-8} C/m$$. What is the magnitude E of the electric field at a radial distance (a) $$r= R/2.00$$ and (b) $$r = 2.00R$$? (c) Graph $$E$$ versus $$r$$ for the range $$r =0\ to\ 2.00R$$

We imagine a cylindrical Gaussian surface A of radius r and unit length concentric with the metal tube. Then by symmetry $$\oint_A\vec E.d\vec A=2\pi rE=\dfrac{q_{enc}}{\epsilon_0}$$
(a) For $$r < R$$, $$q_{enc} = 0$$, so $$E = 0$$.
(b) For $$r > R$$, $$q_{enc} = \lambda$$, so $$E(r)= \lambda/2\pi r \epsilon_0$$ .With $$\lambda=2.00\times10^{-8}C/m$$ and $$r = 2.00R =
0.0600$$ m, we obtain
$$E=\dfrac{(2.0 \times 10^8 \,C/m)}{2\pi (0.0600 m)(8.85 \times 10^{-12}\, C^2/N.m^2)}$$
(c) The plot of $$E$$ vs. $$r$$ is shown below.
Here, the maximum value is :
$$E_{max}=\dfrac{\lambda}{2\pi r\epsilon_0}=\dfrac{(2.0 \times 10^8 \,C/m)}{2\pi (0.030 m)(8.85 \times 10^{-12}\, C^2/N.m^2)}=1.2\times10^{4}\,N/C$$

Two identical conducting sphere, fixed in place, attract each other with an electrostatic force of $$0.108\ N$$ when their center-to-center separation is $$50.0\ cm$$. The sphere are then connected by a thin conducing wire. When the wire is removed, the sphere repel each other with an electrostatic force of $$0.0360\ N$$. Of the initial charges on the sphere, with a positive net charge, what was the positive charge on the other?

psilon }_{ 0 }}\dfrac {\left(\dfrac {q_1+q_2}{2}\right)\left(\dfrac {q_1 +q_2}{2}\right)}{r^2}=k\dfrac {(q_1 +q_2)^2}{4r^2}$$
We solve the two force equations simultaneously for $$q_1$$ and $$q_2$$. The first gives the product
$$F_b =\dfrac {1}{4\pi { \varepsilon }_{ 0 }}\dfrac {\left(\dfrac {q_1+q_2}{2}\right)\left(\dfrac {q_1 +q_2}{2}\right)}{r^2}=k\dfrac {(q_1 +q_2)^2}{4r^2}$$
We solve the two force equations simultaneously for $$q_1$$ and $$q_2$$. The first gives the product
$$q_1q_2=-\dfrac {r^2 F_q}{k}=-\dfrac {(0.500\ m)^2 (0.108\ N)}{8.99\times 10^{9}N.m^2 /C^2}=-3.00\times 10^{-12} C^2$$
and the second gives the sum
$$q_1+q_2=2r\sqrt {\dfrac {F_b}{k}}=2(0.500\ m)\sqrt {\dfrac {0.360\ N}{8.99\times 10^9 N. m^2 /C^2}}=2.00\times 10^{-6}C$$
where we have taken the positive root. Thus, the product results provides the relation
$$q_2=\dfrac {-(3.00\times 10^{-12}C^2)}{q_1} $$
which we substitute into the sum result, producing
$$q_1-\dfrac {3.00\times 10^{-12} C^2}{q_1}=2.00\times 10^{-6}C$$.
Multiplication by $$q_1$$ and rearranging, we obtain a quadratic equation
$$q_1^2 -(2.00\times 10^{-6}C)q_1-3.00\times 10^{-12}C^2=0$$
The solutions are
$$q_1=\dfrac {2.00\times 10^{-6}C\pm \sqrt {(-2.00\times 10^{-6}C)^2}-4(3.00\times 10^{-12}C^2)}{2}$$
If the positive sign is used, $$q_1=3.00\times 10^{-6}C$$, and if the negative is used, $$q_1=-1.00\times 10^{-6}C$$.
If we insted work with the $$q_1=-1.00\times 10^{-6}C$$, then we find $$q_2=3.00\times 10^{-6}C$$.
Note that since spheres are identical, the solutions are essentially the same: one sphere originally had charge $$-1.00\times 10^{-6}C$$ and the other had charge $$+3.00\times 10^{-6}C$$.
What if we had not made the assumption, above, that $$q_1+q_2 \ge 0$$ ? If the signs of the charges were reserved (so $$q_1+q_2 < 0$$), then the forces remain the same, so a charge of $$+1.00\times 10^{-6}C$$ on one sphere and a charge of $$-3.00\times 10^{-6}C$$ on the other also satisfies the condition of the problem.

Two identical conducting sphere, fixed in place, attract each other with an electrostatic force of $$0.108\ N$$ when their center-to-center separation is $$50.0\ cm$$. The sphere are then connected by a thin conducing wire. When the wire is removed, the sphere repel each other with an electrostatic force of $$0.0360\ N$$. Of the initial charges on the sphere, with a positive net charge, what was the negative charge on one of them?

We assume the spheres are far apart. Then the charge distribution on each of them is spherically symmetric and Coulomb's law can be used. Let $$q_1$$ and $$q_2$$ be the original charges. We choose the coordinate system so the force on $$q_2$$ is positive if it repelled by $$q_1$$. Then, the force on $$q_2$$ is
$$F_a=-\dfrac {1}{4\pi { \varepsilon }_{ 0 }}\dfrac {q_1 q_2}{r^2}=-k\dfrac {q_1 q_2}{r^2}$$
where $$r=0.0500\ m$$. The negative sign indicates that the spheres attract each other. After the wire is connected, the spheres, being identical, acquire the same charge. Since charge is conserved, the total charge is the same it was originally. This means the charge on each sphere is $$(q_1 +q_2)/2$$. The force is now one of repulsion and is given by
$$F_b =\dfrac {1}{4\pi { \varepsilon }_{ 0 }}\dfrac {\left(\dfrac {q_1+q_2}{2}\right)\left(\dfrac {q_1 +q_2}{2}\right)}{r^2}=k\dfrac {(q_1 +q_2)^2}{4r^2}$$
We solve the two force equations simultaneously for $$q_1$$ and $$q_2$$. The first gives the product
$$F_b =\dfrac {1}{4\pi { \varepsilon }_{ 0 }}\dfrac {\left(\dfrac {q_1+q_2}{2}\right)\left(\dfrac {q_1 +q_2}{2}\right)}{r^2}=k\dfrac {(q_1 +q_2)^2}{4r^2}$$
We solve the two force equations simultaneously for $$q_1$$ and $$q_2$$. The first gives the product
$$q_1q_2=-\dfrac {r^2 F_q}{k}=-\dfrac {(0.500\ m)^2 (0.108\ N)}{8.99\times 10^{9}N.m^2 /C^2}=-3.00\times 10^{-12} C^2$$
and the second gives the sum
$$q_1+q_2=2r\sqrt {\dfrac {F_b}{k}}=2(0.500\ m)\sqrt {\dfrac {0.360\ N}{8.99\times 10^9 N. m^2 /C^2}}=2.00\times 10^{-6}C$$
where we have taken the positive root. Thus, the product results provides the relation
$$q_2=\dfrac {-(3.00\times 10^{-12}C^2)}{q_1} $$
which we substitute into the sum result, producing
$$q_1-\dfrac {3.00\times 10^{-12} C^2}{q_1}=2.00\times 10^{-6}C$$.
Multiplication by $$q_1$$ and rearranging, we obtain a quadratic equation
$$q_1^2 -(2.00\times 10^{-6}C)q_1-3.00\times 10^{-12}C^2=0$$
The solutions are
$$q_1=\dfrac {2.00\times 10^{-6}C\pm \sqrt {(-2.00\times 10^{-6}C)^2}-4(3.00\times 10^{-12}C^2)}{2}$$
If the positive sign is used, $$q_1=3.00\times 10^{-6}C$$, and if the negative is used, $$q_1=-1.00\times 10^{-6}C$$.
Using $$q_2=(-3.00\times 10^{-12})/q_1=3.00\times 10^{-6}C$$, we get $$q_2=-1.00\times 10^{-6}C$$.

The figure shows a closed Gaussian surface in the shape of a cube of edge length $$2.00$$ m, with one corner at $$x_1=5.00\, m$$,$$y_1 = 4.00$$ m.The cube lies in a region where the electric field vector is given by $$\vec E =-3.00i-4.00y^2\,j +3.00k\, N/C$$ with y in meters. What is the net charge contained by the cube?

None of the constant terms will result in a nonzero contribution to the flux, so we focus on the x dependent term only:
$$E_{nonconstant}=(-4.00y^2)i$$ (in SI units)
The face of the cube located at $$y = 4.00$$ has area $$A = 4.00\, m^2$$ (and it “faces” the $$+j$$ direction) and has a “contribution” to the flux equal to
$$E_{nonconstant}=(−4)(4^2)(4) = –256 \,N·m/C^2$$ .
The face of the cube located at $$y = 2.00$$ m has the same area A (however, this one “faces” the $$–j$$ direction) and a contribution to the flux:
$$-E_{nonconstant}=(−4)(2^2 )(4) = 64 \,N·m/C^2$$ .
Thus, the net flux is $$Φ = (−256 + 64) \,N·m/C^2
= −192 \,N·m/C^2$$ .

According to Gauss’s law, we therefore have :
$$q_{enc}=\epsilon_0\phi=(8.85 \times10^{-12} \,C^2 /N m^2 )( 192 \,N.m^2/C)=1.70 \times10^{-9}$$ C

An electron on the axis of an electric dipole is $$25\ nm$$ from the centre of the dipole. What is the magnitude of the electrostatic force on the electron if the dipole moment is $$3.6\times 10^{-29}\ C.m$$?
Assume that $$25\ nm$$ is much larger than the separation of the charged particles that form the dipole.

We know that,
$$F=|q|E=|q| \left( \dfrac{p}{2\pi \varepsilon_0z^3}\right)$$

$$=\dfrac{2\ kep}{z^3}$$
where the constant $$k$$ in the last step.

Thus, we obtain
$$F=\dfrac{2(8.99\times 10^9\ N.m^2/C^2)(1.60\times 10^{-19}\ C)(3.6\times 10^{-29}\ C.m)}{(25\times 10^{-9}\ m)^3}$$

$$=6.6\times 10^{-15}\ N$$.
If the dipole is oriented such that $$\vec p$$ is in the $$+z$$ direction, then $$\vec F$$ points in the $$-z$$ direction.

A very long straight uniformly charged thread carries a charge $$\lambda$$ per unit length. Find the magnitude and direction of the electric field strength at a point which is at a distance $$y$$ from the thread and lies on the perpendicular passing through one of the thread's ends.

The problem is reduced to finding $$E_{x}$$ and $$E_{y}$$ viz. the projections of $$\vec {E}$$ in Fig, where it is assumed that $$\lambda > 0$$.
Let us start with $$E_{x}$$. The contribution to $$E_{x}$$ from the charge element of the segment $$dx$$ is
$$dE_{x} = \dfrac {1}{4\pi \epsilon_{0}} \dfrac {\lambda dx}{r^{2}}\sin \alpha ....(1)$$
Let us reduce this expression to the form convenient for integration. In our case, $$dx = r d\alpha/ \cos \alpha, r = y/\cos \alpha$$. Then
$$dE_{x} = \dfrac {\lambda}{4\pi \epsilon_{0}y} \sin \alpha d\alpha$$.
Integrating this expression over $$\alpha$$ between $$0$$ and $$\pi/2$$, we find
$$E_{x} = \lambda/4 \pi \epsilon_{0} y$$.
In order to find the projection $$E_{y}$$ it is sufficient to recall that $$dE_{y}$$ differs from $$dE_{x}$$ in that $$\sin \alpha$$ in (1) is simply replaced by $$\cos \alpha$$.
This gives
$$dE_{y} = (\lambda \cos \alpha d\alpha)/ 4\pi \epsilon_{0}y$$ and $$E_{y} = \lambda /4\pi \epsilon_{0}y$$.
We have obtained an interesting result:
$$E_{x} = E_{y}$$ independently of $$y$$,
i.e. $$\vec {E}$$ is oriented at the angle of $$45^{\circ}$$ to the rod. The modulus of $$\vec {E}$$ is
$$E = \sqrt {E_{x}^{2} + E_{y}^{2}} = \lambda \sqrt {2}/ 4\pi \epsilon_{0}y$$.
1784310_1850974_ans_a1803b37a69d4df88932be26291591c8.jpg

1784310_1850974_ans_a1803b37a69d4df88932be26291591c8.jpg

Consider a uniform electric. Held $$\vec E=3\times 10^3 \hat i N/C$$
A point charge $$+10\ p\ C$$ is a distance $$5\ cm$$ directly above the centre of a square of side $$10cm$$, as shown in figure. What is the magnitude of the electric flux through the square of $$10cm$$ on a side whose plane is parallel to the plane?

Here, angle between normal to the square,

ie.e, arae vector and the electric field is $$60^o$$

$$\phi =\vec E. \Delta \vec s$$

$$=E\ \Delta S\ \cos 60^o$$

$$=3\times 10^3 \times 10^{-2}\times 0.5$$

$$=15\ Nm^2 C^{-1}$$

1766329_1842557_ans_c40413b48edf43da84d003d9790fc1da.jpg

Show that the normal component of electrostatic field has a discontinuity from one side of a charge surface to another given by $$(E_2-E_1),\ \cap n=?$$ Where is a unit vector normal to the surface at a point and $$a$$ is the surface charge density at that point (The direction of $$ft$$ from side $$1$$ to $$2$$.) Hence show that just outside a conductor, the electric field is

In figure, a Gaussian surface, in the form of a pill box is shown. It is of area $$\Delta A$$ (on end faces) and of negligible thickness. Let $$\bar E_1$$ be the field below and $$\bar E_2$$, the field above.

The flux through upper surface lower face

$$=\bar E_1. \Delta \bar A =\dfrac{q}{q_0}=\dfrac{\sigma \Delta A}{\varepsilon_0}$$

$$\therefore (\bar E_2-\bar E_1). \cap n=\varepsilon_0$$

Inside a conducting surface, $$E=0$$

$$\therefore \bar E_2=\bar E=\dfrac{\sigma \cap n}{\varepsilon_0}$$

1769198_1843057_ans_e34a9dbdd17f4a828965cb58b7a095c5.png

Two charge $$q_A = 2.5 \times 10^{-7}C$$ and $$q_B = -2.5 \times 10^{-7}C$$ are at point A$$(0, 0, - 15\,cm)$$ and B $$(0, 0, +15\,cm)$$ respectively form a system. Determine the electric dipole moment of the system.

$$\vec{r_A}=-15\hat{k}\,cm$$

and position vector of point $$B$$

$$\vec{r_B}=+15\hat{k}cm$$

$$\vec{r_{AB}}=\vec{r_{B}}-\vec{r_A}=\vec{15\hat{k}}-(-15\vec{\hat{k}})$$

$$=\vec{15\hat{k}}+(15\vec{\hat{k}})=30\hat{k}$$

$$\vec{r_{AB}}=30\hat{k}$$

Magnitude of distance between A and B

$$r_{AB}=|\vec{r_{AB}}|=\sqrt{(30)^2)}=30cm$$

$$r_{AB}=0.30m$$

$$\therefore$$ Effective lenght of electric dipole

$$2l=r_{AB}=0.30m$$

$$\therefore$$ Electric dipole moment

$$p=q\times 2l=2.5\times 10^{-7}\times 0.30$$

$$=0.75\times 10^{-7}=7.5\times 10^{-8}C.m$$

Position vector of point A
1937307_1834909_ans_363e9de13de64a7ea1cbc478ccc5d7e0.png

A nonconducting solid sphere has a uniform volume charge density $$\rho$$. Let be the vector from the center of the sphere to a general point P within the sphere. (a) Show that the electric field at P is given by $$\vec E=\rho \vec r/3\epsilon_0$$ (Note that the result is independent of the radius of the sphere.) (b) A spherical cavity is hollowed out of the sphere, as shown in the figure. Using superposition concepts, show that the electric field at all points within the cavity is uniform and equal to $$\vec E=\rho \vec a/3\epsilon_0$$ where is the position vector from the center of the sphere to the center of the cavity.

(a) From Gauss’ law, we get
$$\vec E(\vec r)=\dfrac{1}{4\pi \epsilon_0}\dfrac{q_{enc}}{r^3}\vec r=\dfrac{1}{4\pi \epsilon_0}\dfrac{(4\pi\rho r^3/3)}{r^3}\vec r=\dfrac{\rho \vec r}{3\epsilon_0}$$
(b) The charge distribution in this case is equivalent to that of a whole sphere of charge density ρ plus a smaller sphere of charge density –$$ρ$$ that fills the void. By superposition :

$$\vec E(\vec r)=\dfrac{\rho \vec r}{3\epsilon_0}=\dfrac{-\rho (\vec r-\vec a)}{3\epsilon_0}=\dfrac{\rho \vec a}{3\epsilon_0}$$

What must be the magnitude of a uniform electric field if it is to have the same energy density as that possessed by a 0.50 T magnetic field?

We set $$u_{E}=\dfrac{1}{2}
\varepsilon_{0} E^{2}=u_{B}=\dfrac{1}{2} B^{2} / \mu_{0}$$ and solve for the
magnitude of the electric field:

$$E=\dfrac{B}{\sqrt{\varepsilon_{0}
\mu_{0}}}=\dfrac{0.50 \mathrm{T}}{\sqrt{\left(8.85 \times 10^{-12} \mathrm{F} /
\mathrm{m}\right)\left(4 \pi \times 10^{-7} \mathrm{H} /
\mathrm{m}\right)}}=1.5 \times 10^{8} \mathrm{V} / \mathrm{m}$$

In Figure, short sections of two very long parallel lines of charge are shown, fixed in place, separated by $$L = 8.0$$ cm. The uniform linear charge densities are $$6.0 \mu C/m$$ for line 1 and $$2.0 \mu C/m$$ for lineWhere along the $$x$$ axis shown is the net electric field from the two lines zero?

We reason that point P (the point on the x-axis where the net electric field is zero) cannot be between the lines of charge (since their charges have opposite signs). We reason further that P is not to the left of “line 1” since its magnitude of charge (per unit length) exceeds that of “line 2”; thus, we look in the region to the right of “line 2” for P.
Using Eq. , we have :

$$E_{net}=E_1+E_2=\dfrac{2\lambda_1}{4\pi \epsilon_0(x+\dfrac{L}{2})}+\dfrac{2\lambda_2}{4\pi \epsilon_0(x-\dfrac{L}{2})}$$

Setting this equal to zero and solving for $$x$$ we find :
$$ x=\Bigg(\dfrac{\lambda_1-\lambda_2}{\lambda_1+\lambda_2}\Bigg)\dfrac{L}{2}=\Bigg(\dfrac{6.0\mu C/m-( -2.0 \mu C/m)}{6.0\mu C/m+( -2.0 \mu C/m)}\Bigg)\dfrac{8.0\,cm}{2}=8.0$$ cm

If the total charge enclosed by a surface is zero, does it imply that the electric field everywhere on the surface is zero? Conversely, if the electric field everywhere on a surface is zero, does it imply that the net charge inside is zero.

By Gauss's law of electro statics,

$$\oint E.ds =\dfrac{q}{\epsilon_{0}}$$

Consider a Gaussian surface enclosing charges $$q_{1}$$, $$q_{2}$$, $$q_{3}$$ and $$q_{4}$$ as shown in figure.

$$q_{1}+q_{2}+q_{3}=0$$

But, electric field on $$R.H.S.$$ due to $$q_{4}$$ will not be zero. But, on the $$L.H.S.$$, it will be zero. So the statement is true.

If electric field on Gaussian surface is zero then, in the above case, it is possible only when $$q_{4}=0$$. i.e. if everywhere, on Gaussian surface, electric field is zero then net charge will be zero.

1662951_1795423_ans_7cd56f3621dd4cfea14cff1f00b83c33.png

The given figure shows the electric field lines around three point charges $$A, B$$ and $$C$$.
(i) Which charges are positive?
(ii) Which charge has the largest magnitude? Why?
(iii) In which region or regions of the picture could the electric field be zero?
Justify your answer.
(i) near $$A$$ (ii) near $$B$$ (iii) near $$C$$ (iv) nowhere.

(i) Charges $$A$$ and $$C$$ are positive since lines of force emanate from them.

(ii) Charge $$C$$ has the largest magnitude since the maximum number of field lines is associated with it.

(iii) (i) near $$A$$.
Justification: There is no neutral point between a positive and a negative charge. A neutral point may exist between two like charges. From the figure, we see that a neutral point exists between charges $$A$$ and $$C$$. Also between two like charges the neutral point is closer to the charge with smaller magnitude. Thus, the electric field is zero near charge $$A$$.

Proton in a well. The above figure shows electric potential $$V$$ along an $$x$$ axis. The scale of the vertical axis is set by $$V_s = 10.0 \ V$$. A proton is to be released at $$x=3.5 \ cm$$ with initial kinetic energy $$4.00 \ eV$$.
(a) If it is initially moving in the negative direction of the axis, does it reach a turning point (if so, what is the $$x$$ coordinate of that point) or does it escape from the plotted region (if so, what is its speed at $$x=0$$)?
(b) If it is initially moving in the positive direction of the axis, does it reach a turning point (if so, what is the $$x$$ coordinate of that point) or does it escape from the plotted region (if so, what is its speed at $$x=6.0 \ cm$$)?
What are the
(c) magnitude $$F$$ and
(d)direction (positive or negative direction of the $$x$$ axis) of the electric force on the proton if the proton moves just to the left of $$x=3.0 \ cm$$?
What are
(e) $$F$$ and
(f) the direction if the proton moves just to the right of $$x=5.0 \ cm$$?

1773406_1774339_ans_a40ed2e259df403e936b531fdf967fc9.jpg

Electric Flux
A flat surface of area $$3.20 m^{2}$$ is rotated in a uniform electric field of magnitude $$E = 6.20 \times 10^{5} N/C$$. Deter-mine the electric flux through this area
(a) when the electric field is perpendicular to the surface and
(b) when the electric field is parallel to the surface.

For a uniform electric field passing through a plane surface,
$$\phi_{E}=\overset{\rightarrow}{E}\cdot \overset{\rightarrow}{A}=EA\cos \theta$$ where $$\theta$$ is the angle between the electric field and the normal to the surface
(a) The electric field is perpend lade to the surface so $$\theta =0^{o}$$
$$\phi_{E}=(6.20\times 10^{5}N/C)(3.20m^{2})\cos 0^{o}$$
$$\phi_{E}=1.98\times 10^{6}N\cdot m^{2}/C$$

(b) The electric field is parallel to the surface $$\theta =90^{o}$$, so $$\cos \theta=0$$ and the flux is zero

Electric Flux
Consider a closed triangular box resting within a horizontal electric field of magnitude $$E=7.80 \times 10^{4} N/C$$ as shown in Figure P24.Calculate the electric flux through
(a) the vertical rectangular surface,
(b) the slanted surface, and
(c) the entire surface of the box.

(a) For the vertical rectangular surface, the area (shown as $$A^{1}$$ is ANS Fig.P24.4)is
$$A^{1}=(10.0cm)(30.0cm)=300cm^{2}=0.030m^{2}$$
Since electric field is perpendicular to the surface and is directed inward, $$\theta=180^{o}$$ and
$$\phi_{E,A^{1}}=EA^{1}\cos \theta$$
$$\phi_{E,A^{1}}=(7.80\times 10^{4}N/C)(0.0300m^{2})\cos 180^{o}$$
$$\phi_{E,A^{1}}=-2.34kN\cdot m^{2}/C$$

(b) To find the area $$A$$ of the slanted surface, we note that the side for which dimensions are not given has length $$(10.0cm)=\omega \cos 60.0^{o}$$ so that
$$A=(30.0cm)(\omega)=(30.0cm)(\dfrac{10.0cm}{\cos 60.0^{o}})=600cm^{2}$$
$$=0.060m^{2}$$
The flux through this surface is then
$$\phi_{E,A}=EA\cos \theta=(7.80\times 10^{4})(A)\cos 60.0^{o}$$
$$=(7.80\times 10^{4}N/C)(0.060m^{2})\cos 60.0^{o}$$
$$=+2.34kN\cdot m^{2}/C$$

(c) The bottom and the two triangular sides all lie parallel to $$\overset{\rightarrow}{E}$$ so $$\phi_{E}=0$$ for each of these. Thus,
$$\phi_{E,total}=-2.34kN\cdot m^{2}/C+2.34kN\cdot m^{2}/C+0+0+0=0$$

1807790_1860817_ans_b6ca1e04eea0425fbe52db53db4be69f.png

Electric Flux
A vertical electric field of magnitude $$2.00 \times 10^{4} N/C$$ exists above the Earths surface on a day when a thunderstorm is brewing. A car with a rectangular size of $$6.00 m$$ by $$3.00 m$$ is traveling along a dry gravel road-way sloping downward at $$10.0^{o}$$. Determine the electric flux through the bottom of the car.

The electric flux through the bottom of the car is given by
$$\phi_{E}=EA\cos \theta=(2.00\times 10^{4}N/C)(3.00m)(6.00m)\cos 10.0^{o}$$
$$= 355kN\cdot m^{2}/C$$

An electric field line emerges from a positive point charge $$+{q}_{1}$$ at an angle $$\alpha$$ to the straight line connecting it to a negative point charge $$-{q}_{2}$$ (figure).
At what angle $$\beta$$ will the field line enter the charge $$-{q}_{2}$$?

In the immediate proximity of each of point charges, the contribution from the other charge to the total field strength is negligibly small, and hence the electric field lines emerge from (enter) the charge in a spatially homogeneous bundle, their total number being proportional to the magnitude of the charge. Only fraction of these lines gest into a cone with an angle $$2\alpha$$ at the vertex near the charge $$+{q}_{1}$$. The ratio o the number of these lines to the total number of the lines emerging from the charge $$+{q}_{1}$$ is equal to the ratio of the areas of the corresponding spherical segments:
$$\cfrac { 2\pi RR\left( 1-\cos { \alpha } \right) }{ 4\pi { R }^{ 2 } } =\cfrac { 1 }{ 2 } \left( 1-\cos { \alpha } \right) $$
Since the electric field lines connect the two charges of equal magnitude, the number of lines merging from the charge $$+{q}_{1}$$ within the angle $$2\alpha$$ is equal to the number of lines entering the charge $$-{q}_{2}$$ at an angle $$2\beta$$. Consequently
$$\left| { q }_{ 1 } \right| \left( 1-\cos { \alpha } \right) =\left| { q }_{ 2 } \right| \left( 1-\cos { \beta } \right) $$
whence
$$\sin { \cfrac { \beta }{ 2 } } =\sin { \cfrac { \alpha }{ 2 } \sqrt { \cfrac { \left| { q }_{ 1 } \right| }{ \left| { q }_{ 2 } \right| } } } $$
If $$\sin { \cfrac { \alpha }{ 2 } \sqrt { \cfrac { \left| { q }_{ 1 } \right| }{ \left| { q }_{ 2 } \right| } } } > 1$$, an electric field line will not enter the charge $$-{q}_{2}$$

Application of Gausss Law to Various Charge Distributions
The charge per unit length on a long, straight filament is $$-90.0 \mu C/m$$. Find the electric field (a) $$10.0 cm$$,(b) $$20.0 cm$$, and (c) $$100 cm$$ from the filament, where distances are measured perpendicular to the length of the filament.

Note that the electric field in each case is directed radially inward, toward the filament. We use $$E=\dfrac{2k_{e}\lambda}{r}$$ and substitute numerical values.
(a) At $$r=10.0cm=0.100m$$
$$E=\dfrac{2k_{e}\lambda}{r}=\dfrac{2(8.99\times 10^{9}N\cdot m^{2}/C^{2})(90.0\times 10^{-6}C/m)}{0.100m}$$
$$=16.2MN/C$$
(b) At $$r=20.0cm=0.200m$$
$$E=\dfrac{2k_{e}\lambda}{r}=\dfrac{2(8.99\times 10^{9}N\cdot m^{2}/C^{2})(90.0\times 10^{-6}C/m)}{0.200m}$$
$$=8.09MN/C$$
(a) At $$r=100cm=1.00m$$
$$E=\dfrac{2k_{e}\lambda}{r}=\dfrac{2(8.99\times 10^{9}N\cdot m^{2}/C^{2})(90.0\times 10^{-6}C/m)}{1.00m}$$
$$=1.62MN/C$$

Objects	Electron exchange	Positive	Negative
Glass, Silk	From glass rod to silk	....a....	Silk
Ebonite, Wool	Wool to ebonite	...b...	...c...
Rubber rod, Wool	...d...	Wool	...e...

Electric Charges And Fields - Class 12 Medical Physics - Extra Questions

Find an expression for the oscillation frequency of an electric dipole of dipole moment $$\vec{p}$$ and rotational inertia $$I$$ for small amplitudes of oscillation about its equilibrium position in a unit form electric field of magnitude $$E$$.

A point charge Q is placed at point O as shown in the figure. The potential difference VA-VB is positive. Is the charge Q negative or positive

The figure shows the electric field lines around three point charges $$A$$, $$B$$ and $$C$$. Which charges are positive?

During dry weather, clothes made of synthetic fiber often stick to the skin. Which type of force is responsible for this phenomenon?

Why does a plastic comb rubbed with dry hair attract bits of paper?

What are the two kinds of charges?

Name the physical quantity of which the units are:Coulomb

Describe an experiment to demonstrate that there are two $$ r $$ kinds of charges.

How will you show that like charges repel and unlike charges attract each other?

What is the condition of an ideal electric dipole?

On what factors does the constant $$k =\dfrac{1}{4\pi \epsilon_0}$$ depends?

State the Coulombs law for the force acting between two definite point charges.

Can electric potential be zero at a point in a vacuum, the electric field at that point is not zero. Give example.

To determine the electric field due to a point charge, the test charge should be small. Explain why?

An electric dipole is placed in a uniform electric field. Show that it will not accelerate.

What is the electric field lines? Write any two of its properties.

In given fig., an equilateral triangle ABC of side a has charges $$+2q, - q$$ and $$-q$$ at the vertices A, B, and C respectively. Calculate the dipole moment of this system.

Explain why two field lines never cross each other at any point?

The electric field components due to a charge inside the cube of side 0.1 m are shown in figure.where, $$E_x = \alpha$$, where $$\alpha$$ = 500 N/C-m, $$E_y$$ = 0, $$E_z = 0$$.Calculate(a) the flux through the cube and (b) the charge inside the cube.

A hollow cylinder has a charge q coulomb within it. If $$\phi $$ is the electric flux in units of volt-meter associated with the curved surface B, the flux linked with the plane surface A in units of volt-meter will be?

The figure shows the electric field lines around a conductor.(a) At which of the four points is the field strongest?(b) At which point is there a negative charge?

Given a uniform electric field $$\overrightarrow E = 2 \times 10^3 \widehat i$$N/C. Find the flux of this field through a square of side 20 cm, whose plane is parallel to the y-z plane. What would be the flux through the same square, if the plane makes an angle of $$30^o$$ with the x-axis?

State Guass law in electrostatics. Derive an expression of electric field due to an infinitely long straight uniformly charged wire. Draw necessary diagram.

What is an electric dipole? Define electric dipole moment?

Obtain the expression for the intensity of electric field near a uniformly charged straight wire of infinite length with the help of Gauss theorem.

Write Gauss's theorem of electrostatics. Obtain an expression for the intensity of electric field near a uniformly charged straight wire of infinite length with its help.

A charge $$Q$$ is uniformly distributed over a rod of length $$l$$. Consider a hypothetical cube of edge $$l$$ with the centre of the cube at one end of the rod. Find the minimum possible flux of the electric field through the entire surface of the cube.

Figure shows tracks of three charged particles in a uniform electrostatic field. Given the signs of the three charges. Which particle has the highest charge to mass ratio.

State and explain coulomb's inverse square law

Valence band

An infinite line charge produces an electric field of $$9.0\times 10^4$$N $$C^{-1}$$ at a distance of $$2.0$$cm. What is the linear charge density?

A small test charge is released at rest at a point in an electrostatic field configuration. Will it travel along the line of force?

If an electric field is given by $$10\hat i + 3\hat j + 4\hat k,$$ calculate the electric flux through a surface of area $$10$$ units lying in $$yz$$ plane

Find the dimensions of permittivity of vacuum(\in_0).

In electric dipole what is the locus of Zero potential?

What is the direction of electric field $$E$$ at point $$P$$ in the following diagram

Why do electric lines of force never intersect each other?

What is electric dipole moment state its SI unit and dimension.

A person standing on an insulating stool touches a charge insult conductor . will the conductor get complete discharged ?

State Coulomb's law of force in electrostatics.

Depict the orientation of the dipole moment. Is it a scalar or a vector quantity? What are its $$SI$$ unit?

Find the magnitude of the electric field at a point $$4 cm$$ away from a line charge of density $$2 \times 10^{-6} Cm^{-1}$$.

Calculate the number of coulombs of positive charge in $$250\ cm^{3}$$ of (neutral) water. (Hint: A hydrogen atom contains one pro eight protons.)

Calculate the electric dipole moment of an electron and a proton $$4.30\ nm$$ apart.

If a cat repeatedly rubs against your cotton slacks on a dry day, the charge transfer between the cat hair and the cotton can leave you with an excess charge of $$-2.00 \mu C$$.Just before the spark appears, do you induce positive or negative charge in the faucet?

An electric dipole consisting of charges of magnitude $$1.50\ nC$$ separated by $$6.20\ \mu m$$ is in an electric field of strength $$1100\ N/C$$. What arethe magnitude of the electric dipole moment

Calculate the solid angle subtends by the periphery of an area $$1 cm^2$$ at a point situated symmetrically at a distance of $$5 cm$$ from the area.

Answer the following questions.(i) Write $$SI$$ unit of electric field intensity.(ii) Write $$SI$$ unit of electric dipole moment.

A hollow half cylinder surface of radius R and length l is placed in a uniform electric field $$\vec{E}$$. Electric field is acting perpendicular on the ABCD. If the flux through the curved surface of the hollow cylindrical surface is $$E\times X Rl$$. Find X?

$$Match \: the \: statements \: in \: Column \: A, \: with \: those \: in \: Column \: B.$$

If the electric field to the left of two sheets is $$K\sigma/\varepsilon_0$$. Find K?

Figure shows some of the electric field lines due to three point charges $$q_1, q_2$$ and $$q_3$$ of equal magnitude. What are the signs of each of the three charges?

State and explain Coulomb's law in electrostatics.

Write any three properties of electric lines of force.

If $$\vec { E } =3\hat { i } +4\hat { j } -5\hat { k } $$, calculate the electric flux through a surface of area 50 units in z-x plane

A metallic sphere is placed in a uniform electric field as shown in the figure. Which path is followed by electric field line and why?

A,B and C are three charged bodies. If A and B repel each other and A attracts C, what is the nature of the force between B and C?

An electric field $$\vec E - \vec iAx$$ exists in the space, where $$A = V/{m^2}$$. take the potential at $$(10m, 20m)$$ to be zero. Find the potential at the origin.

An electric dipole is formed by charge $$\pm\ 4\mu C$$, separated by $$5\ mm$$. Calculate dipole moment and its direction.

Two point dipoles of dipole moment $$ \overline { { P }_{ 1 } } and\overline { { P }_{ 2 } } $$ are at a distance x from each other and $$ \overline { { P }_{ 1 } } ||\overline { { P }_{ 2 } } $$The force between the dipoles is

A system has two charges $$qA = 2.5\times 10^{-7}C$$ and $$qB = -2.5\times 10^{-7}C$$ located at point $$A(0, 0, -15) cm$$ and $$B(0, 0, 15) cm$$ respectively. What are the total charge and electric dipole moment vector of the system?

Complete the table:ObjectsElectron exchangePositiveNegativeGlass, SilkFrom glass rod to silk....a....SilkEbonite, WoolWool to ebonite...b......c...Rubber rod, Wool...d...Wool...e...

From which one to which does electronic transfer occur when the pair of substances given below are rubbed against each other?Glass rod Silk cloth.

Why must hospital personnel wear special conducting shoes while working around oxygen in an operating room? What might happen if the personnel wore shoes with rubber soles?

From which one to which does electronic transfer occur when the pair of substances given below are rubbed against each other?Ebonite Wool.

Conceptual question:On the basis of the repulsive nature of the force between like charges and the freedom of motion of charge within a conductor, explain why excess charge on an isolated conductor must reside on its surface.

Answer to conceptual questions.A student who grew up in a tropical country and is studying in the United States may have no experience with static electricity sparks and shocks until his or her first American winter. Explain.

Conceptual Questions :A person is placed in a large, hollow, metallic sphere that is insulated from ground. If a large charge is placed on the sphere, will the person be harmed upon touching the inside of the sphere?

You might have seen iron chains, suspended from the body of lorries and trucks touching the earth. What is this for?

Application of Gausss Law to Various Charge DistributionsA large, flat, horizontal sheet of charge has a charge per unit area of $$9.00 \mu C/m^{2}$$. Find the electric field just above the middle of the sheet.

Find the initial acceleration of the rod

Define relative premittivity. Write its unit.

What are electric lines of force? State the properties of lines of force.

Do we have to include the signs (positive or negative)of the charges while solving problems?

No two electric lines of intersect each other. Why ?

Why do the electrostatic filed lines not from clossed loop ?

An electrostatic field line is a continuous curve. That is, a field line cannot have sudden breaks. Why not ?

Define one coulomb charge.

Define Permitivity of vacuum $$ (E_0) $$

Name the physical quantity of which the units are:
Coulomb

The electric field components due to a charge inside the cube of side 0.1 m are shown in figure.
where, $$E_x = \alpha$$, where $$\alpha$$ = 500 N/C-m, $$E_y$$ = 0, $$E_z = 0$$.
Calculate
(a) the flux through the cube and
(b) the charge inside the cube.

The figure shows the electric field lines around a conductor.
(a) At which of the four points is the field strongest?
(b) At which point is there a negative charge?

If a cat repeatedly rubs against your cotton slacks on a dry day, the charge transfer between the cat hair and the cotton can leave you with an excess charge of $$-2.00 \mu C$$.
Just before the spark appears, do you induce positive or negative charge in the faucet?

An electric dipole consisting of charges of magnitude $$1.50\ nC$$ separated by $$6.20\ \mu m$$ is in an electric field of strength $$1100\ N/C$$. What are
the magnitude of the electric dipole moment

Answer the following questions.
(i) Write $$SI$$ unit of electric field intensity.
(ii) Write $$SI$$ unit of electric dipole moment.

Two point dipoles of dipole moment $$ \overline { { P }_{ 1 } } and\overline { { P }_{ 2 } } $$ are at a distance x from each other and $$ \overline { { P }_{ 1 } } ||\overline { { P }_{ 2 } } $$
The force between the dipoles is

Complete the table:
Objects
Electron exchange
Positive
Negative
Glass, Silk
From glass rod to silk
....a....
Silk
Ebonite, Wool
Wool to ebonite
...b...
...c...
Rubber rod, Wool
...d...
Wool
...e...

From which one to which does electronic transfer occur when the pair of substances given below are rubbed against each other?
Glass rod Silk cloth.

From which one to which does electronic transfer occur when the pair of substances given below are rubbed against each other?
Ebonite Wool.

Conceptual question:
On the basis of the repulsive nature of the force between like charges and the freedom of motion of charge within a conductor, explain why excess charge on an isolated conductor must reside on its surface.

Answer to conceptual questions.
A student who grew up in a tropical country and is studying in the United States may have no experience with static electricity sparks and shocks until his or her first American winter. Explain.

Conceptual Questions :
A person is placed in a large, hollow, metallic sphere that is insulated from ground. If a large charge is placed on the sphere, will the person be harmed upon touching the inside of the sphere?

Application of Gausss Law to Various Charge Distributions
A large, flat, horizontal sheet of charge has a charge per unit area of $$9.00 \mu C/m^{2}$$. Find the electric field just above the middle of the sheet.

Tick the correct one:
Object with same charge
(attract / repel)

Tick the correct one:
Bodies with different charges
(attract / repel)

Tick the correct one:
The charged body ............ the neutral body.
(attract / repel)

Answer the following questions.
(i) An electrostatic field line is a continuous curve. That is, a field line cannot have sudden breaks. Why is it so?
(ii) Explain why two field lines never cross each other at any point.

Find out the magnitude of electric field intensity and electric potential due to a dipole of dipole moment $$\bar P = \hat i + \sqrt 3 \hat j$$ kept at origin at following points.
(i) (2, 0, 0)
(ii) (-1, $$\sqrt 3 $$, 0)