$$4$$ charges are placed each at a distance '$$a$$' from origin. The dipole moment of configuration is:

$$2aq\left[ \hat { i } +\hat { j } \right] $$

If we use permittivity $$\varepsilon $$, resistance $$R$$, gravitational constant $$G$$ and voltage $$V$$ as fundamental physical quantities, then :

[force$$]={ \varepsilon }^{ 1 }{ R }^{ 0 }{ G }^{ 0 }{ V }^{ 2 }$$

[angular displacement]$$= { \varepsilon }^{ 0 }{ R }^{ 0 }{ G }^{ 0 }{ V }^{ 0 }$$

[velocity]$$={ \varepsilon }^{ -1 }{ R }^{ -1 }{ G }^{ 0 }{ V }^{ 0 }$$

[dipole moment]$$={ \varepsilon }^{ 1 }{ R }^{ 0 }{ G }^{ 0 }{ V }^{ 1 }$$

Two infinite sheets of uniform charge density $$+\sigma $$ and $$-\sigma $$ are parallel to each other as show in figure. Electric field at the

midpoint of the sheets is $${ \sigma }/{ { \varepsilon }_{ 0 } }$$ and is directed towards right.

points to the left or to the right of the sheets is zero.

midpoint between the sheets is zero.

midpoint of the sheets is $${ 2\sigma }/{ { \varepsilon }_{ 0 } }$$ and is directed towards right.

An electric dipole is placed in an electric field increasing at the rate of $$\lambda$$ per unit distance along the x-axis. The dipole may undergo :

rotational motion and translational motion along x-axis

rotational motion and translational motion along z-axis

only translational motion along x-axis

Two point charge $$+q$$ and $$-q$$ are held fixed at $$(-d, 0)$$ and $$(d, 0)$$ respectively of a $$x-y$$ coordinate system. Then

The dipole moment is $$2qd$$along the negative $$x-$$axis

The electric field $$E$$ at all points on the $$x-$$axis has the same direction

Work has to be done in bringing a test charge from $$\infty$$ to the origin

Electric field at all points on $$y-$$axis is along positive $$x-$$axis

Two thin infinite parallel sheets have uniform surface densities of charge $$+\sigma$$ and $$-\sigma$$. Electric field in the space between the two sheets is :

$$2\sigma /{ \varepsilon }_{ 0 }$$

What is the value of $$E$$ in the space outside the charged sheets?

$$E\ne 0$$, The value of $$E$$ is different at diffrerent distances from sheets.

$$2\sigma /{ \varepsilon }_{ 0 }$$

MCQ Questions for Class 12 Medical Physics Electric Charges And Fields Quiz 4

4

$4$ charges are placed each at a distance '

a

$a$ ' from origin. The dipole moment of configuration is:

0%
$2qa\hat { j }$
0%
$3qa\hat { j }$
0%
$2aq\left[ \hat { i } +\hat { j } \right]$
0%
$2aq (\hat i -\hat j)$

Explanation

Dipole moment for the given configuration is

\vec{p} = (- 2 q) (a \hat{i}) + (3 q) (a \hat{j}) + (- 2 q) (- a \hat{i}) + (q) (- a \hat{j}) = 2 q a \hat{j}

$\vec{p}=(-2q)(a\hat i)+(3q)(a\hat j)+(-2q)(-a\hat i)+(q)(-a\hat j)=2qa\hat j$
Ans:(A)

Three charges

q, q, - 2 q

$q, q ,-2q$ placed at the corners of equilateral triangle of side

l

$l$ .Find the net dipole moment.

0%
2q $l$
0%
$\sqrt{3}$ q $l$
0%
$\displaystyle \frac{\sqrt{3}ql}{2}$
0%
$\displaystyle \frac{ql}{2}$

Explanation

From figure, we split

- 2 q

$-2q$ charge into two

- q

$-q$ charges and calculate their dipole moment with

+ q

$+q$ charges and then do the vector sum of two dipole moments to get the resultant dipole moment.

p = q . l

$p=q.l$

Angle between two dipole moments is

60^{o}

$60^{o}$ .

p_{n e t} = \sqrt{p^{2} + p^{2} + 2. p . p \cos 60^{o}}

$p_{net}=\sqrt{p^2+p^2+2.p.p\cos60^{o}}$

⟹ p_{n e t} = \sqrt{3 p^{2}} = p \sqrt{3}

$\implies p_{net}=\sqrt{3p^2}=p\sqrt{3}$

⟹ p_{n e t} = \sqrt{3} q l

$\implies p_{net}=\sqrt{3}ql$

Answer-(B)

If we use permittivity

ε

$\varepsilon$ , resistance

R

$R$ , gravitational constant

G

$G$ and voltage

V

$V$ as fundamental physical quantities, then :

0%
[angular displacement] $= { \varepsilon }^{ 0 }{ R }^{ 0 }{ G }^{ 0 }{ V }^{ 0 }$
0%
[velocity] $={ \varepsilon }^{ -1 }{ R }^{ -1 }{ G }^{ 0 }{ V }^{ 0 }$
0%
[dipole moment] $={ \varepsilon }^{ 1 }{ R }^{ 0 }{ G }^{ 0 }{ V }^{ 1 }$
0%
[force $]={ \varepsilon }^{ 1 }{ R }^{ 0 }{ G }^{ 0 }{ V }^{ 2 }$

Explanation

ϵ^{- 1} = L^{3} M T^{- 2} Q^{- 2}, R = L^{2} M T^{- 1} Q^{- 2}, V = M L^{2} T^{- 2} Q^{- 1}, G = L^{3} M^{- 1} T^{- 2}

$\displaystyle \epsilon^{-1}=L^3MT^{-2}Q^{-2}, R=L^2MT^{-1}Q^{-2}, V=ML^2T^{-2}Q^{-1}, G=L^3M^{-1}T^{-2}$
As the angular displacement

(θ)

$(\theta)$ is dimensionless so
[displacement]

= ϵ^{0} R^{0} G^{0} V^{0}

$=\epsilon^0R^0G^0V^0$
Velocity,

[v] = L T^{- 1}

$[v]=LT^{-1}$
here

ϵ^{- 1} R^{- 1} G^{0} V^{0} = L^{3} M T^{- 2} Q^{- 2} L^{- 2} M^{- 1} T^{1} Q^{2} = L T^{- 1} = [v]

$\epsilon^{-1}R^{-1}G^0V^0=L^3MT^{-2}Q^{-2}L^{-2}M^{-1}T^{1}Q^{2}=LT^{-1}=[v]$
dipole moment,

[p] = Q L

$[p]=QL$
here

ϵ^{1} R^{0} G^{0} V^{1} = (L^{- 3} M^{- 1} T^{2} Q^{2}) (M L^{2} T^{- 2} Q^{- 1}) = Q L^{- 1} \neq [p]

$\epsilon^1R^0G^0V^1=(L^{-3}M^{-1}T^{2}Q^{2})(ML^2T^{-2}Q^{-1})=QL^{-1}\neq [p]$
force,

[f] = ϵ^{- 1} L^{- 2} Q^{2}

$[f]=\epsilon^{-1}L^{-2}Q^2$
here

ϵ^{1} R^{0} G^{0} V^{2} = (L^{- 3} M^{- 1} T^{2} Q^{2}) (M^{2} L^{4} T^{- 4} Q^{- 2}) = (L^{3} M T^{- 2} Q^{- 2}) L^{- 2} Q^{2} = ϵ^{- 1} L^{- 2} Q^{2} = [f]

$\epsilon^1R^0G^0V^2=(L^{-3}M^{-1}T^{2}Q^{2})(M^2L^4T^{-4}Q^{-2})=(L^3MT^{-2}Q^{-2})L^{-2}Q^2=\epsilon^{-1}L^{-2}Q^2=[f]$

Electric flux through a surface of area

100 m^{2}

$100 m^{ 2 }$ lying in the xy plane is (in Vm) if

\vec{E} = \hat{i} + \sqrt{2} \hat{j} + \sqrt{3} \hat{k}

$\overrightarrow { E } =\hat { i } +\sqrt { 2 } \hat { j } +\sqrt { 3 } \hat { k }$

0%
$100$
0%
$141.4$
0%
$173.2$
0%
$200$

Explanation

Electric flux,

ϕ = \int \vec{E} . \hat{n} d s

$\phi=\int\vec{E}.\hat{n}ds$

given,

\hat{n} = \hat{k}, d s = 100 m^{2} \vec{E} = \hat{i} + \sqrt{2} \hat{j} + \sqrt{3} \hat{k}

$\hat{n}=\hat{k},ds=100 m^2 \vec{E}=\hat { i } +\sqrt { 2 } \hat { j } +\sqrt { 3 } \hat { k }$

ϕ = (\hat{i} + \sqrt{2} \hat{j} + \sqrt{3} \hat{k}) . \hat{k} 100 = 100 \sqrt{3} = 173.2

$\phi=(\hat { i } +\sqrt { 2 } \hat { j } +\sqrt { 3 } \hat { k }).\hat{k}100=100\sqrt{3}=173.2$

Three points charges are placed at the corners of an equilateral triangle of side

L

$L$ shown in the figure.

0%
The potential at the centroid of the triangle is zero
0%
The electric field at the centroid of the triangle is zero
0%
The dipole moment of the system is $\sqrt { 2 } qL$
0%
The dipole moment of the system is $\sqrt { 3 } qL$

Explanation

For equilateral triangle AOB, $AO=OB=BA=a$ and $AC=OC=BC=r$

The potential at centroid C is $V=k\dfrac{2q}{AC}+k\dfrac{-q}{OC}+k\dfrac{-q}{BC}=k\dfrac{2q}{r}-k\dfrac{q}{r}-k\dfrac{q}{r}=0$

The net electric field at centroid C is $E=k\dfrac{2q}{r^2}-k\dfrac{q}{r^2}\cos30-k\dfrac{q}{r^2}\cos30=k(0.2)\dfrac{2q}{r^2} \neq 0$

As the total charge is zero so for dipole moment the choice of origin is independent. We assume O as origin.

dipole moment , $p=-q(0)-q(L\hat i)+2q(\dfrac{L}{2}\hat i+\dfrac{\sqrt{3}L}{2}\hat j)$

or $p=\sqrt{3}qL \hat j$

Two infinite sheets of uniform charge density

+ σ

$+\sigma$ and

- σ

$-\sigma$ are parallel to each other as show in figure. Electric field at the

0%
points to the left or to the right of the sheets is zero.
0%
midpoint between the sheets is zero.
0%
midpoint of the sheets is ${ \sigma }/{ { \varepsilon }_{ 0 } }$ and is directed towards right.
0%
midpoint of the sheets is ${ 2\sigma }/{ { \varepsilon }_{ 0 } }$ and is directed towards right.

Explanation

Electric field due to charge sheet for charge density $+\sigma$ is $E=\dfrac{+\sigma}{2\epsilon_0}$ and for $-\sigma$ is $E=\dfrac{-\sigma}{2\epsilon_0}$

outside the plates field is $E=\dfrac{+\sigma}{2\epsilon_0}+\dfrac{-\sigma}{2\epsilon_0}=0$

between the plates field is $E=\dfrac{\sigma}{2\epsilon_0}+\dfrac{\sigma}{2\epsilon_0}=\dfrac{\sigma}{\epsilon_0}$

Ans: (A),(C)

A point charge is brought in an electric field. The electric field at a nearly point :

0%
will increase if the charge is positive
0%
will decrease if the charge is negative
0%
may increase if the charge is positive
0%
may decrease if the charge is negative

Explanation

The electric field is proportional to the charge i.e

E = k q / r^{2}

$E=kq/r^2$ . So positive charge may increase the field and negative charge may decrease the field.
Ans: (C),(D)

The cube as shown in Fig. has sides of length

L = 10.0 c m

$L = 10.0\ cm$ . The electric field is uniform, has a magnitude

E = 4.00 \times 10^{3} N C^{- 1}

$E = 4.00\times 10^{3} NC^{-1}$ , and is parallel to the

x y - p l a n e

$xy - plane$ at an angle of

37^{\circ}

$37^{\circ}$ measured from the

+ x - a x i s

$+x-axis$ towards the

+ y - a x i s

$+y-axis$ .
Electric flux passing through surface

S_{6}

$S_{6}$ is

0%
$24 Nm^{2} C^{-1}$
0%
$-24 Nm^{2} C^{-1}$
0%
$32 Nm^{2} C^{-1}$
0%
$-32 Nm^{2} C^{-1}$

Explanation

Here, the electric field is given by

\vec{E} = E \cos 37 \hat{i} + E \sin 37 \hat{j}

$\vec{E}=E\cos37\hat i+E\sin37 \hat j$

Electric flux through back surface

S_{6}

$S_6$ is

ϕ = \vec{E} . \hat{n} S_{6}

$\phi=\vec E.\hat{n}S_6$

Here

\hat{n} = - \hat{i}

$\hat{n}=-\hat i$ , normal outward to the back surface

S_{1} = L^{2}

$S_1=L^2$

Thus,

ϕ = (E \cos 37 \hat{i} + E \sin 37 \hat{j}) . (- L^{2} \hat{i}) = - E L^{2} \cos 37 = - (4 \times 10^{3}) (0.1)^{2} \cos 37 = - 31.94 \sim - 32 N m^{2} C^{- 1}

$\phi=(E\cos37 \hat i+E\sin37\hat j).(-L^2\hat i)=-EL^2\cos37=-(4\times 10^3)(0.1)^2\cos37=-31.94 \sim -32 Nm^2C^{-1}$

An electric dipole is placed in an electric field generated by a point charge.

0%
The net electric force on the dipole must be zero.
0%
The net electric force on the dipole may be zero.
0%
The torque on the dipole due to the field must be zero.
0%
The torque on the dipole due to field may be zero.

Explanation

Hint: Find the net force on the dipole by using $F=qE$ to calculate the force on both charges of the dipole.
Explanation:
Step 1: Reasons for incorrect options:

For net electric force to be zero on the dipole, the electric field lines should be parallel to each other, and the dipole is placed in such a way that it is perpendicular to the electric field. But parallel field lines can be there as well for a point charge. So, options A and B are incorrect.
Now, torque on the dipole is given by the formula:

$\vec{ \tau} = \vec{P} \times \vec{E}$

If the dipole moment $P$ and electric field line $E$ is in the same direction, in that case, torque will be zero in another case, it will have some non-zero value. So, option C is incorrect.

Answer:

Hence, option D is the correct answer.

The surface of a polar liquid such as water, can be viewed as a series of dipole strung together in the stable arrangement in which dipole moment vectors are parallel to the surface and all point in the same direction. Now suppose that something presses downwards (inwards) on the surface, distorting the dipoles as shown in figure.

Then the two slanted dipoles will exert a net force in which direction ?

0%
downward
0%
upward
0%
horizontal
0%
none of these

A charged particle moves with a speed v in a circular path of radius r around a long uniformly charged conductor then :

0%
$v \propto r$
0%
$\displaystyle v \propto \dfrac{1}{r}$
0%
$\displaystyle v \propto \dfrac{1}{\sqrt{2}}r$
0%
$v$ is independent of r

Explanation

Consider a Gaussian cynlinder of radius $r$ and length $l$ . Using Gauss's law the electric field at distance $r$ is $\displaystyle E.(2\pi rl) =\dfrac{\lambda l}{\epsilon_0} \Rightarrow E=\dfrac{\lambda}{2\pi\epsilon_0 r}$

Here the electrostatic force is equal to centrifugal force. i.e,

$\displaystyle qE=\dfrac{mv^2}{r}$

$\displaystyle q\dfrac{\lambda}{2\pi \epsilon_0 r}=\dfrac{mv^2}{r}$

or $\displaystyle v^2=\dfrac{q\lambda}{2\pi\epsilon_0 m}$

so $v$ is independent of $r$ .

Which of the following is not correct?

0%
Dipole moment of bad conductors of electricity is zero but when placed in an electric field becomes dipole.
0%
Electric dipole moment is a scalar quantity.
0%
Electric line of force always emerge out perpendicular to charged surface.
0%
Bringing a proton from one point to another on a equipotential surface, no energy is lost.

The electric field at a point

5 c m

$5 cm$ from a long line charge of density

2.5 \times 10^{- 6} c m^{- 1}

$2.5 \times 10^{-6} cm^{-1}$ is :

0%
$9 \times 10^3 NC^{-1}$
0%
$9 \times 10^4 NC^{-1}$
0%
$9 \times 10^5 NC^{-1}$
0%
$9 \times 10^6 NC^{-1}$

Explanation

The electric field at a distance r from the long wire is

$\displaystyle E=\dfrac{1}{4\pi \epsilon_0} \dfrac{2\lambda}{r}=9\times 10^9\times \dfrac{2\times 2.5\times 10^{-6}}{5\times 10^{-2}}=9\times 10^5 NC^{-1}$

An electric dipole is placed in a uniform electric field. The net electric force on the dipole

0%
is always zero
0%
depends on the orientation of the dipole
0%
depends on the dipole moment
0%
is always finite but not zero

Explanation

Hint: Find the net force on the dipole by using $F=qE$ to calculate the force on both charges of the dipole.
Explanation:
Step 1: Concept used:
Dipole consists of 2 equal and opposite charges $+q \ and -q$ .

Consider a dipole, as shown below in the figure.

In uniform electric field E, if we place any dipole at that time,

Force on $+q$ is $q E$ and force on charge $-q$ is $-q E$ .

So, the total force will be sum of these two forces, and it will result in zero.

Answer:

Hence, option A is the correct answer.

An electric dipole is placed in an electric field increasing at the rate of

λ

$\lambda$ per unit distance along the x-axis. The dipole may undergo :

0%
rotational motion only
0%
rotational motion and translational motion along z-axis
0%
rotational motion and translational motion along x-axis
0%
only translational motion along x-axis

The field at a distance r from a long string of charge per unit length

λ

$\lambda$ is :

0%
$\displaystyle k \dfrac{\lambda}{r^2}$
0%
$\displaystyle k \dfrac{\lambda}{r}$
0%
$\displaystyle k \dfrac{\lambda}{2r}$
0%
$\displaystyle k \dfrac{2 \lambda}{r}$

Explanation

Consider a surface of a cylinder of radius $r$ and length $l$ as Gaussian surface. Apply Gauss's law, $\displaystyle \vec{E}.\vec{ds}=\dfrac{Q_{en}}{\epsilon_0}$ where $Q_{en}$ is the charge enclosed by the Gaussian surface. here $Q_{en}=\lambda l$

thus, $\displaystyle E.(2\pi rl)=\dfrac{\lambda l}{\epsilon_0}$

$\displaystyle \therefore E=\dfrac{\lambda }{2\pi \epsilon_0 r}=k\dfrac{2\lambda }{r}$ where $\displaystyle k=\dfrac{1 }{4\pi \epsilon_0 r}$

A charged particle having a charge

- 2 \times 10^{- 6} C

$-2\times 10^{-6} C$ is placed close to the non-conducting plate having a surface charge density

4 \times 10^{- 6} C m^{- 2}

$4\times 10^{-6} Cm^{-2}$ . The force of attraction between the particle and the plate is nearly :

0%
$0.9 N$
0%
$0.71 N$
0%
$0.62 N$
0%
$0.45 N$

Explanation

The electric field due to the plate is $E=\dfrac{\sigma}{2\epsilon_0}$

Here the force between the charged particle of charge q and the plate is

$F=qE=\dfrac{q\sigma}{2\epsilon_0}=\dfrac{2\times 10^{-6}\times 4\times 10^{-6}}{2\times 8.854\times 10^{-12}}=0.45 N$

The total flux passing through the cube is:

0%
$(B+C+D)L^2$
0%
$2(B+C+D)L^2$
0%
$6(B+C+D)L^2$
0%
zero

Explanation

{\hat{n}}_{S_{1}} = + \hat{j} \Rightarrow Φ_{1} = \vec{E} \cdot A {\hat{n}}_{S_{4}} = + C L^{2}

$\hat{n}_{S_1} = +\hat{j}\Rightarrow \Phi _1=\vec{E}\cdot A\hat{n}_{S_4}=+CL^2$

{\hat{n}}_{S_{2}} = - \hat{j} \Rightarrow Φ_{2} = \vec{E} \cdot A {\hat{n}}_{S_{5}} = - C L^{2}

$\hat{n}_{S_2} = -\hat{j}\Rightarrow \Phi _2=\vec{E}\cdot A\hat{n}_{S_5}=-CL^2$

{\hat{n}}_{S_{3}} = + \hat{k} \Rightarrow Φ_{3} = \vec{E} \cdot A {\hat{n}}_{S_{6}} = - D L^{2}

$\hat{n}_{S_3} = +\hat{k}\Rightarrow \Phi _3=\vec{E}\cdot A\hat{n}_{S_6}=-DL^2$

{\hat{n}}_{S_{4}} = - \hat{k} \Rightarrow Φ_{4} = \vec{E} \cdot A {\hat{n}}_{S_{4}} = + D L^{2}

$\hat{n}_{S_4} = -\hat{k}\Rightarrow \Phi _4=\vec{E}\cdot A\hat{n}_{S_4}=+DL^2$

{\hat{n}}_{S_{5}} = + \hat{i} \Rightarrow Φ_{5} = \vec{E} \cdot A {\hat{n}}_{S_{5}} = - B L^{2}

$\hat{n}_{S_5} = +\hat{i}\Rightarrow \Phi _5=\vec{E}\cdot A\hat{n}_{S_5}=-BL^2$

{\hat{n}}_{S_{6}} = - \hat{i} \Rightarrow Φ_{6} = \vec{E} \cdot A {\hat{n}}_{S_{6}} = + B L^{2}

$\hat{n}_{S_6} = -\hat{i}\Rightarrow \Phi _6=\vec{E}\cdot A\hat{n}_{S_6}=+BL^2$
Total flux is

\sum_{i = 1}^{6} Φ_{1} = 0

$\sum_{i=1}^{6} \Phi _1=0$

Find the electric flux crossing the wire frame ABCD of length l, width b, and center at a distance OP=d from an infinite line of charge with linear charge density

λ

$\lambda$ . Consider that the plane of the frame is perpendicular to the line OP Fig.

0%
$\displaystyle\frac{2N}{\pi \varepsilon_0}tan^{-1}\left ( \frac{b}{2d} \right )$
0%
$\displaystyle\frac{N}{\pi \varepsilon_0}tan^{-1}\left ( \frac{b}{2d} \right )$
0%
$\displaystyle\frac{3N}{\pi \varepsilon_0}tan^{-1}\left ( \frac{7b}{2d} \right )$
0%
$\displaystyle\frac{N}{2\pi \varepsilon_0}tan^{-1}\left ( \frac{b}{2d} \right )$

Explanation

The flux passing through the strip of area dA is

d ϕ = E d A c o s θ

$d \phi= EdA cos \theta$

= \frac{λ}{2 π ε_{0} \sqrt{d^{2} + x^{2}}} (l d x) \frac{d}{\sqrt{d^{2} + x^{2}}}

$=\dfrac{\lambda}{2 \pi \varepsilon_0\sqrt{d^2+x^2}}(l dx) \dfrac{d}{\sqrt{d^2+x^2}}$

d ϕ = \frac{λ d l}{2 π ε_{0}} \frac{d}{(\sqrt{d^{2} + x^{2}})}

$d\phi=\dfrac{\lambda dl}{2 \pi \varepsilon_0} \dfrac{d}{(\sqrt{d^2+x^2})}$

∴ ϕ = \int d ϕ = \frac{λ d l}{2 π ε_{0}} \int_{- b / 2}^{b / 2} \frac{d}{d^{2} + x^{2}}

$\therefore \phi =\int d \phi=\dfrac{\lambda dl}{2 \pi \varepsilon_0} \int_{-b/2}^{b/2}\dfrac{d}{d^2+x^2}$

= \frac{λ d l}{2 π ε_{0}} \frac{1}{d} {[t a n^{- 1} \frac{x}{d}]}_{- b / 2}^{b / 2}

$=\dfrac{\lambda dl}{2 \pi \varepsilon_0} \dfrac{1}{d}\left [ tan^{-1} \dfrac{x}{d} \right ]^{b/2}_{-b/2}$

= \frac{N}{2 π ε_{0}} \times 2 t a n^{- 1} (\frac{b}{2 d})

$=\dfrac{N}{2\pi \varepsilon_0}\times 2tan^{-1}\left ( \dfrac{b}{2d} \right )$

ϕ = \frac{N}{2 π ε_{0}} t a n^{- 1} (\frac{b}{2 d})

$\phi=\dfrac{N}{2\pi \varepsilon_0} tan^{-1}\left ( \dfrac{b}{2d} \right )$

In a certain region of space, there exists a uniform electric field of

2 \times 10^{3} \hat{k} V m^{- 1}

$2\times 10^{3} \hat {k}\ Vm^{-1}$ . A rectangular coil of dimension

10 c m \times 20 c m

$10\ cm\times 20\ cm$ is placed in

x - y

$x - y$ plane. The electric flux through the coil is

0%
Zero
0%
$30\ Vm$
0%
$40\ Vm$
0%
$50\ Vm$

Explanation

\vec{E} = 2000 \vec{k}, \vec{A} = 10 \times 20 \times 10^{- 4} \vec{k}

$\vec{E}=2000\vec{k},\vec{A}= 10 \times 20 \times 10^{-4} \vec{k}$

ϕ = \vec{E} \cdot \vec{A} = 2000 \times 10 \times 20 \times 10^{- 4} V m = 40 V m

$\phi= \vec{E}\cdot \vec{A}=2000\times 10 \times 20 \times 10^{-4}Vm=40Vm$

The sign of each charge

Q_{1}

$Q_1$ and

Q_{2}

$Q_2$ is:

0%
+, -
0%
-, +
0%
+, +
0%
-, -

Explanation

Electric field at (2) tends to

- \infty

$- \infty$ , hence the charge at (2) should be negative. There is a neutral point to the right of charges. This is possible only when the charge at (1) should be positive. Hence,

Q_{1}

$Q_1$ is positive and

Q_{2}

$Q_2$ is negative. At neutral point,

{\vec{E}}_{1} = {\vec{E}}_{2}

$\vec E_1 = \vec E_2$

\frac{1}{4 π ε_{0}} \frac{Q}{(a + l)^{2}} = \frac{1}{4 π ε_{0}} \frac{Q_{2}}{a^{2}} o r \frac{Q_{1}}{Q_{2}} = {(\frac{a + l}{a})}^{2}

$\displaystyle \frac{1}{4 \pi \varepsilon_0} \frac{Q}{(a + l)^2} = \frac{1}{4 \pi \varepsilon_0} \frac{Q_2}{a^2} or \frac{Q_1}{Q_2} = \left ( \frac{a + l}{a} \right )^2$
Electric field at any position on the right of charges is

E = E_{1} - E_{2} = \frac{1}{4 π ε_{0}} \frac{Q}{(l + x)^{2}} - \frac{1}{4 π ε_{0}} \frac{Q_{2}}{x^{2}}

$\displaystyle E = E_1 - E_2 = \frac{1}{4 \pi \varepsilon_0} \frac{Q}{(l + x)^2} - \frac{1}{4 \pi \varepsilon_0} \frac{Q_2}{x^2}$
For the maximum value of E,

\frac{d E}{d x} = 0 o r \frac{d}{d x} [\frac{1}{4 π ε_{0}} \frac{Q_{1}}{(l + x)^{2}} - \frac{1}{4 π ε_{0}} \frac{Q_{2}}{x^{2}}] = 0

$\displaystyle \frac{dE}{dx} = 0 or \frac{d}{dx} \left [ \frac{1}{4 \pi \varepsilon_0} \frac{Q_1}{(l + x)^2} - \frac{1}{4 \pi \varepsilon_0} \frac{Q_2}{x^2} \right ] = 0$
or

Q_{1} \frac{(- 2)}{(1 + x)^{3}} - Q_{2} \frac{(- 2)}{x^{3}} = 0

$\displaystyle Q_1 \frac{(-2)}{(1 + x)^3} - Q_2 \frac{(-2)}{x^3} = 0$
or

\frac{Q_{1}}{(l + x)^{3}} = \frac{Q_{2}}{x^{3}} o r {(\frac{l + x}{x})}^{3} = \frac{Q_{1}}{Q_{2}}

$\displaystyle \frac{Q_1}{(l + x)^3} = \frac{Q_2}{x^3} or \left ( \frac{l + x}{x}\right )^3 = \frac{Q_1}{Q_2}$
or

\frac{l + x}{x} = {(\frac{Q_{1}}{Q_{2}})}^{1 / 3} o r \frac{l}{x} = {(\frac{Q_{1}}{Q_{2}})}^{1 / 3} - 1

$\displaystyle \frac{l + x}{x} = \left ( \frac{Q_1}{Q_2} \right )^{1/3} or \frac{l}{x} = \left ( \frac{Q_1}{Q_2} \right )^{1/3} - 1$
or

x = \frac{l}{{(\frac{Q_{1}}{Q_{2}})}^{1 / 3} - 1}

$\displaystyle x = \frac{l}{\left ( \frac{Q_1}{Q_2} \right )^{1/3} - 1}$

In a region of space, the electric field is given by

\vec{E} = 8 \hat{i} + 4 \hat{j} + 3 \hat{k}

$\vec{E}=8\hat{i}+4\hat{j}+3\hat{k}$ . The electric flux through a surface of area

100

$100$ units in the xy plane is :

0%
$800$ units
0%
$300$ units
0%
$400$ units
0%
$1500$ units

Explanation

\vec{E} = 8 \hat{i} + 4 \hat{j} + 3 \hat{k}, \vec{A} = 100 \hat{k}

$\vec{E}=8\hat{i}+4\hat{j}+3\hat{k},\vec{A}=100\hat{k}$
S0

ϕ = \vec{E} \cdot \vec{A} = 300 u n i t s

$\phi=\vec{E}\cdot \vec{A}=300 units$

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

0%
$\displaystyle\frac{q}{\varepsilon_0}\left [ 5-\frac{1}{\sqrt{1+(R/a)^2}} \right ]$
0%
$\displaystyle\frac{q}{\varepsilon_0}\left [ 3-\frac{1}{\sqrt{1+(2R/a)^2}} \right ]$
0%
$\displaystyle\frac{8q}{\varepsilon_0}\left [ 1-\frac{3}{\sqrt{1+(R/a)^2}} \right ]$
0%
$\displaystyle\frac{q}{\varepsilon_0}\left [ 1-\frac{1}{\sqrt{1+(R/a)^2}} \right ]$

Explanation

d ϕ = 2 E c o s θ 2 π y d y

$d \phi=2 E cos \theta 2 \pi ydy$

2 \frac{1}{4 π ε_{0}} \frac{q}{(a^{2} + y^{2})} \frac{q 2 π y d y}{(a^{2} + y^{2})^{1 / 2}}

$\displaystyle2\frac{1}{4 \pi \varepsilon_0} \frac{q}{(a^2+y^2)} \frac{q 2 \pi ydy}{(a^2+y^2)^{1/2}}$

ϕ = \int d ϕ = \frac{q a}{ε_{0}} \int_{0}^{R} \frac{y d y}{(a^{2} + y^{2})^{3 / 2}}

$\displaystyle\phi=\int d \phi=\frac{qa}{\varepsilon_0}\int_{0}^{R} \frac{ydy}{(a^2+y^2)^{3/2}}$

\frac{q}{ε_{0}} [1 - \frac{1}{\sqrt{1 + (R / a)^{2}}}]

$\displaystyle\frac{q}{\varepsilon_0}\left [ 1-\frac{1}{\sqrt{1+(R/a)^2}} \right ]$

A cylinder of length

L

$L$ and radius

b

$b$ has its axis coincident with the x-axis. The electric field in this region is

\vec{E} = 200 \hat{i}

$\vec {E} = 200\hat {i}$ . Find the flux through the left end of cylinder.

0%
$0$
0%
$200\pi b^{2}$
0%
$100\pi b^{2}$
0%
$-200\pi b^{2}$

Explanation

\vec{E} = 200 \hat{i}

$\vec E=200\hat{i}$
Area vector for the left end is

\vec{A} = - A \hat{i}

$\vec A=-A\hat{i}$

ϕ = \vec{E} . \vec{A} = - E A = - E π b^{2} = - 200 π b^{2}

$\phi=\vec E.\vec A =-EA=-E \pi b^2 =- 200 \pi b^2$

Two lithium nuclei in a lithium vapour at room temperature do not combine to form a carbon nucleus- because:

0%
a lithium nucleus is more tightly bound than a carbon nucleus
0%
carbon nucleus is unstable particle
0%
it is not energetically favourable
0%
coulomb repulsion does not allow the nuclei to come very close

The electric field in a region is given by

\vec{E} = \frac{3}{5} E_{0} \hat{j}

$\overrightarrow{E} = \dfrac{3}{5} E_0 \hat{j}$ with

E_{0} = 2 \times 10^{3} N C^{- 1}

$E_0=2 \times 10^3 NC^{-1}$ Find the flux of this field through a rectangular surface of area

0.2 m^{2}

$0.2 m^2$ parallel to the

Y - Z

$Y-Z$ plane.

0%
$320 Nm^2 C^{-1}$
0%
$240 Nm^2 C^{-1}$
0%
$400 Nm^2 C^{-1}$
0%
none of these

Explanation

The electric flux ,

ϕ = \vec{E} . \hat{n} d S

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

Since the surface is parallel to

Y - Z

$Y-Z$ plane so the normal unit vector will be along x-axis i.e

\hat{n} = \hat{i}

$\hat n=\hat i$

Thus,

ϕ = (3 / 5) E_{0} \hat{j} . \hat{i} (0.2) = 0

$\phi=(3/5)E_0 \hat j. \hat i (0.2)=0$ (as

\hat{j} . \hat{i} = 0

$\hat j.\hat i=0$ )

Thus, option D will correct.

A cube of side

10

$10$ cm enclose a charge of

0.1 μ C

$0.1 \mu C$ at its center. Calculate the number of lines of force through each face of the cube.

0%
$1.113\times 10^{11}$
0%
$1.13\times 10^{4}$
0%
$1.13\times 10^{9}$
0%
$1883$

Explanation

Using Gauss's law, the electric flux through the cube is $\displaystyle \phi=\dfrac{Q_{en}}{\epsilon_0}$ where $Q_{en}$ is the charge enclosed by cube.

As cube has six faces so the flux through each face is $\displaystyle \phi_e=\dfrac{\phi}{6}=\dfrac{Q_{en}}{6\epsilon_0}$ .

The flux is also the number of lines of force passing through the surface of cube. So

the number of lines of force through each face of the cube is $\displaystyle n=\phi_e=\dfrac{Q_{en}}{6\epsilon_0}=\dfrac{0.1\times 10^{-6}}{6\times 8.85 \times 10^{-12}} \sim 1883$

The flux passing through the surface

S_{5}

$S_5$ will be

0%
$-0.135 Nm^2C^{-1}$
0%
$-0.054 Nm^2C^{-1}$
0%
$0.081 Nm^2C^{-1}$
0%
$0.054 Nm^2C^{-1}$

Explanation

At surface

S_{5}

$S_5$ , component of fields perpendicular to the surface=

- 1.5 \hat{i}

$-1.5 \hat{i}$

Flux

= A r e a S_{5} \times 1.5 = {0.3}^{2} \times - 1.5 = - 1.35 N m^{2} C^{- 1}

$= Area\ S_5 \times 1.5 = 0.3^2 \times -1.5 = -1.35Nm^2C^{-1}$

Two point charge

+ q

$+q$ and

- q

$-q$ are held fixed at

(- d, 0)

$(-d, 0)$ and

(d, 0)

$(d, 0)$ respectively of a

x - y

$x-y$ coordinate system. Then

0%
The electric field $E$ at all points on the $x-$ axis has the same direction
0%
Work has to be done in bringing a test charge from $\infty$ to the origin
0%
Electric field at all points on $y-$ axis is along positive $x-$ axis
0%
The dipole moment is $2qd$ along the negative $x-$ axis

Explanation

To the left of

+ q,

$+q,$ electric field due to

+ q

$+q$ will dominate and

\vec{E}

$\vec { E }$ will be in -ve

x

$x$ direction.

To the right of

- q,

$-q,$ electric field due to

- q

$-q$ will dominate and

\vec{E}

$\vec { E }$ will be in -ve

x

$x$ direction.

And in between

+ q

$+q$ and

- q,

$-q,$ when

x \in (- d, 0),

$x \in \left(-d,0\right),$

\vec{E}

$\vec { E }$ will in +ve x-direction and when

x \in (0, d)

$x\in \left(0,d\right)$

\vec{E}

$\vec { E }$ will be in +ve x direction. So, (A) is incorrect.

\Rightarrow

$\Rightarrow$ Potential at origin

= \frac{k (+ 9)}{2} + \frac{k (- 9)}{2} = 0

$=\dfrac{k\left(+9\right)}{2}+\dfrac{k\left(-9\right)}{2}=0$

So, no work will be done in bringing a test change from infinity to 0.

As we can see that, on y-axis vertical component of

\vec{E}

$\vec { E }$ due to

+ q

$+q$ and

- q

$-q$ will be cancelled and net

\vec{E}

$\vec { E }$ will be in +ve x direction. Option (C) is correct.

\Rightarrow

$\Rightarrow$ Dipole moment

=

$=$ ( change ) ( distance between

+ q

$+q$ and

- q

$-q$ )

= (q) (2 d)

$=\left(q\right)\left(2d\right)$

= 2 q d

$=2qd$ and

Direction will be from

- q

$-q$ to

+ q .

$+q.$

Hence, both the options C and D are correct.

The cube as shown in Fig. has sides of length

L = 10.0 c m

$L = 10.0\ cm$ . The electric field is uniform, has a magnitude

E = 4.00 \times 10^{3} N C^{- 1}

$E = 4.00\times 10^{3} NC^{-1}$ , and is parallel to the

x y - p l a n e

$xy - plane$ at an angle of

37^{\circ}

$37^{\circ}$ measured from the

+ x - a x i s

$+x-axis$ towards the

+ y - a x i s

$+y-axis$ .
The total net electric flux through all faces of the cube is

0%
$8 Nm^{2} C^{-1}$
0%
$-8 Nm^{2} C^{-1}$
0%
$24 Nm^{2} C^{-1}$
0%
Zero

Explanation

Here, the electric field is given by

\vec{E} = E \cos 37 \hat{i} + E \sin 37 \hat{j}

$\vec{E}=E\cos37\hat i+E\sin37 \hat j$

Electric flux through any surface

S

$S$ is

ϕ = \vec{E} . \hat{n} S

$\phi=\vec E.\hat{n}S$

Where

\hat{n}

$\hat{n}$ is the normal outward to the surface

S

$S$

For surface

S_{1}

$S_1$ :

\hat{n} = - \hat{j}, S_{1} = L^{2} = (0.1)^{2} m^{2}

$\hat n=-\hat j, S_1=L^2=(0.1)^2 m^2$

Thus,

ϕ_{1} = \vec{E} . \hat{n} S_{1} = (E \cos 37 \hat{i} + E \sin 37 \hat{j}) . (- L^{2} \hat{j}) = - E L^{2} \sin 37 = - (4 \times 10^{3}) (0.1)^{2} \sin 37 = - 24 N m^{2} C^{- 1}

$\phi_1=\vec{E}.\hat n S_1=(E\cos37 \hat i+E\sin37\hat j).(-L^2\hat j)=-EL^2\sin37=-(4\times 10^3)(0.1)^2\sin37=-24 Nm^2C^{-1}$

For surface

S_{2}

$S_2$ :

\hat{n} = \hat{k}, S_{2} = L^{2} = (0.1)^{2} m^{2}

$\hat n=\hat k, S_2=L^2=(0.1)^2 m^2$

Thus,

ϕ_{2} = \vec{E} . \hat{n} S_{2} = (E \cos 37 \hat{i} + E \sin 37 \hat{j}) . (L^{2} \hat{k}) = 0

$\phi_2=\vec{E}.\hat n S_2=(E\cos37 \hat i+E\sin37\hat j).(L^2\hat k)=0$

For surface

S_{3}

$S_3$ :

\hat{n} = \hat{j}, S_{3} = L^{2} = (0.1)^{2} m^{2}

$\hat n=\hat j, S_3=L^2=(0.1)^2 m^2$

Thus,

ϕ_{3} = \vec{E} . \hat{n} S_{3} = (E \cos 37 \hat{i} + E \sin 37 \hat{j}) . (L^{2} \hat{j}) = E L^{2} \sin 37 = (4 \times 10^{3}) (0.1)^{2} \sin 37 = 24 N m^{2} C^{- 1}

$\phi_3=\vec{E}.\hat n S_3=(E\cos37 \hat i+E\sin37\hat j).(L^2\hat j)=EL^2\sin37=(4\times 10^3)(0.1)^2\sin37=24 Nm^2C^{-1}$

For surface

S_{4}

$S_4$ :

\hat{n} = - \hat{k}, S_{4} = L^{2} = (0.1)^{2} m^{2}

$\hat n=-\hat k, S_4=L^2=(0.1)^2 m^2$

Thus,

ϕ_{4} = \vec{E} . \hat{n} S_{4} = (E \cos 37 \hat{i} + E \sin 37 \hat{j}) . (- L^{2} \hat{k}) = 0

$\phi_4=\vec{E}.\hat n S_4=(E\cos37 \hat i+E\sin37\hat j).(-L^2\hat k)=0$

For surface

S_{5}

$S_5$ :

\hat{n} = \hat{i}, S_{5} = L^{2} = (0.1)^{2} m^{2}

$\hat n=\hat i, S_5=L^2=(0.1)^2 m^2$

Thus,

ϕ_{5} = \vec{E} . \hat{n} S_{5} = (E \cos 37 \hat{i} + E \sin 37 \hat{j}) . (L^{2} \hat{i}) = E L^{2} \cos 37 = (4 \times 10^{3}) (0.1)^{2} \cos 37 = 32 N m^{2} C^{- 1}

$\phi_5=\vec{E}.\hat n S_5=(E\cos37 \hat i+E\sin37\hat j).(L^2\hat i)=EL^2\cos37=(4\times 10^3)(0.1)^2\cos37=32 Nm^2C^{-1}$

For surface

S_{6}

$S_6$ :

\hat{n} = - \hat{i}, S_{6} = L^{2} = (0.1)^{2} m^{2}

$\hat n=-\hat i, S_6=L^2=(0.1)^2 m^2$

Thus,

ϕ_{6} = \vec{E} . \hat{n} S_{6} = (E \cos 37 \hat{i} + E \sin 37 \hat{j}) . (- L^{2} \hat{i}) = - E L^{2} \cos 37 = - (4 \times 10^{3}) (0.1)^{2} \cos 37 = - 32 N m^{2} C^{- 1}

$\phi_6=\vec{E}.\hat n S_6=(E\cos37 \hat i+E\sin37\hat j).(-L^2\hat i)=-EL^2\cos37=-(4\times 10^3)(0.1)^2\cos37=-32 Nm^2C^{-1}$

Thus, net flux

= ϕ_{1} + ϕ_{2} + ϕ_{3} + ϕ_{4} + ϕ_{5} + ϕ_{6} = - 24 + 0 + 24 + 0 + 32 - 32 = 0

$=\phi_1+\phi_2+\phi_3+\phi_4+\phi_5+\phi_6=-24+0+24+0+32-32=0$

The total flux passing through the cube is

0%
$0.081 Nm^2C^{-1}$
0%
$-0.135 Nm^2C^{-1}$
0%
$-0.054 Nm^2C^{-1}$
0%
zero

Explanation

\vec{E} = (- 5.00 N C^{- 1}) x \hat{i} + (3.00 N C^{- 1}) z \hat{k}

$\overrightarrow{E}=(-5.00NC^{-1})x\hat{i}+(3.00NC^{-1})z\hat{k}$
Area of any side,

A = 0.3 \times 0.3 = 0.09 m^{2}

$A=0.3 \times 0.3=0.09m^2$
Here for

S_{1}

$S_1$ ,

S_{3}

$S_3$ ,

S_{5}

$S_5$ and

S_{6}

$S_6$

ϕ = \vec{E} . \vec{A} = 0

$\phi=\overrightarrow{E}.\overrightarrow{A}=0$
So, only contribution will come from

S_{2}

$S_2$ and

S_{4}

$S_4$

ϕ_{t o t a l} = 3.00 \times 0.09 \times 0.3 + (- 5.00) \times 0.09 \times 0.3

$\phi_{total}=3.00 \times 0.09\times 0.3+(-5.00) \times 0.09 \times 0.3$

= 0.081 - 0.135

$=0.081-0.135$

= - 0.054 N m^{2} C^{- 1}

$=-0.054 Nm^2C^{-1}$

The surface that have zero flux are

0%
$S_2, S_4,$ and $S_5$
0%
$S_1, S_3, S_4$ and $S_6$
0%
$S_1, S_2,$ and $S_3$
0%
$S_2, S_3,$ and $S_4$

Explanation

\vec{E}

$\vec{E}$ lies along the x-z plane. therefore surface

S_{1}

$S_1$ and

S_{3}

$S_3$ have zero flux as both these surface lie on the x-z plane and hence no field lines component cross them perpendicularly.
At surface

S_{6}

$S_6$ the magnitude of field

\vec{E}

$\vec{E}$ is zero, hence flux is zero.
At surface

S_{4}

$S_4$ the z component of field is zero. Therefore flux is zero

A cube of side

a

$a$ is placed such that the nearest face which is parallel to the

y - z

$y - z$ plane is at distance

^{'} a^{'}

$'a'$ from the origin. The electric field components are

E_{x} = α x^{1 / 2}, E_{y} = E_{z} = 0

$E_{x} = \alpha x^{1/2}, E_{y} = E_{z} = 0$ .
The flux

ϕ_{E}

$\phi_{E}$ through the cube is

0%
$\sqrt {2} \alpha a^{5/2}$
0%
$-\alpha a^{5/2}$
0%
$(\sqrt {2} - 1) \alpha a^{5/2}$
0%
Zero

Explanation

Given that nearest distance of cube from origin is a,then
Flux from the nearest surface

ϕ_{1} = - E a^{2} = - α^{1 / 2} a^{2} = - α a^{5 / 2}

$\phi _1=-E a^2=-\alpha ^{1/2} a^2=-\alpha a^{5/2}$
Flux from surface x=2a

ϕ_{2} = E a^{2} = - α (2 a)^{1 / 2} a^{2} = \sqrt{2} α a^{5 / 2}

$\phi _2=E a^2=-\alpha(2a) ^{1/2} a^2=\sqrt 2 \alpha a^{5/2}$
In y and z direction, there is no electric field present. So flux will be zero in that direction

ϕ_{n e t 2} = (\sqrt{2} - 1) α a^{5 / 2}

$\phi _{net2}=(\sqrt 2-1)\alpha a^{5/2}$

Coulomb's Law is true for

0%
atomic distances ( $={10}^{-11}m$ )
0%
nuclear distances ( $={10}^{-15}m$ )
0%
charged as well as uncharged particles
0%
all the distances

Explanation

Coulomb's law is true for all distances whether it is small and large. Hence it is called a long range force.

Coulomb's force

F = \frac{q_{1} q_{2}}{4 π ϵ_{o} r^{2}}

$F = \dfrac{q_1 q_2}{4\pi \epsilon_o r^2}$

⟹

$\implies$

F \propto \frac{1}{r^{2}}

$F \propto \dfrac{1}{r^2}$

A charged body is held near the disc of positively charged GLE. It is found that leaves of GLE diverge more. The nature of charge on the charged body is:

0%
positive
0%
negative
0%
can be any of two
0%
none of these

When a negatively charged body is placed in contact with the negatively charged disc of G.L.E. its leaves ................ .

0%
excess
0%
diverge
0%
induction
0%
rubbing

An electric dipole will experience a net force when it is placed in

0%
a uniform electric field
0%
a non-uniform electric field
0%
both a and b
0%
None of these

If a body is positively charged, then it has

0%
excess of electrons
0%
excess of protons
0%
deficiency of electrons
0%
deficiency of neutrons

Electric lines of force about a negative point charge are :

0%
circular anticlockwise
0%
circular clockwise
0%
radial and inwards
0%
radial and outwards

Explanation

Solution : Electric field lines for a negative charge is given by

$\textbf{Solution : Electric field lines for a negative charge is given by }$

When a negatively charged ebonite rod is brought near a negatively charged and freely suspended ball, then the ball :

0%
gets repelled
0%
attracted
0%
is not affected
0%
none of these

The

E - r

$E-r$ curve for an infinite linear charge distribution will be :

Explanation

The field due to infinite linear charge distribution

E = \frac{1}{4 π ε_{0}} \int \frac{d q}{r} \Rightarrow E \propto \frac{1}{r}

$E=\cfrac { 1 }{ 4\pi { \varepsilon }_{ 0 } } \int { \cfrac { dq }{ r } } \Rightarrow \quad E\propto \cfrac { 1 }{ r }$
So, hyperbola

Two thin infinite parallel sheets have uniform surface densities of charge

+ σ

$+\sigma$ and

- σ

$-\sigma$ . Electric field in the space between the two sheets is :

0%
$\sigma /{ \varepsilon }_{ 0 }$
0%
$\sigma /{ 2\varepsilon }_{ 0 }$
0%
$2\sigma /{ \varepsilon }_{ 0 }$
0%
zero

Explanation

E = E_{1} - E_{2} = \frac{σ}{2 ε_{0}} - (\frac{- σ}{2 ε_{0}}) = \frac{σ}{2 ε_{0}} = σ / ε_{0}

$E={ E }_{ 1 }-{ E }_{ 2 }=\cfrac { \sigma }{ 2{ \varepsilon }_{ 0 } } -\left( \cfrac { -\sigma }{ 2{ \varepsilon }_{ 0 } } \right) =\cfrac { \sigma }{ 2{ \varepsilon }_{ 0 } } =\sigma /{ \varepsilon }_{ 0 }$
The field intensity in between sheets having equal and opposite uniform surface densities of charge becomes constant, i.e, an uniform electric field is produced and it is independent of the distance between the sheets.

What is the value of

E

$E$ in the space outside the charged sheets?

0%
$\sigma /{ \varepsilon }_{ 0 }$
0%
$\sigma /{2 \varepsilon }_{ 0 }$
0%
$E\ne 0$ , The value of $E$ is different at diffrerent distances from sheets.
0%
$2\sigma /{ \varepsilon }_{ 0 }$

Explanation

Inside a charged conducting surface

E = 0

$E=0$ , but on or outside the surface

E \neq 0

$E\ne 0$
Therefore, Electric intensity is discontinuous across a charged conducting surface

A cylinder of radius

R

$R$ and length

l

$l$ is placed in a uniform electric field

E

$E$ parallel to the axis of the cylinder. The total flux over the curved surface of the cylinder is :

0%
Zero
0%
$\pi R^{2}E$
0%
$2\pi R^{2}E$
0%
$E/\pi R^{2}$

Explanation

For the curved surface,

At any point on the curved surface,

E

$E$ and area vector are perpendicular to each other.

θ = 90^{o}

$\quad \theta ={ 90 }^{ o }$

∴ ϕ = E d s \cos 90^{o} = 0

$\therefore \phi =E\quad ds\cos { { 90 }^{ o } } =0$

A soap bubble is given negative charge, its radius will :

0%
increase
0%
decrease
0%
remain unchanged
0%
fluctuate

Explanation

Hint: Like charges repel each other and unlike charges attract each other.

Part 1 : Figure for given situation

Consider that the soap is in negative charge,

$\textbf{Part2: Explanation}$

$\bullet$ The charge will be distributed uniformly over the surface of the bubble by symmetry.

$\bullet$ As we know, they want to step much farther apart, as charges repel, the only path outward, taking with them the soap surface.

$\bullet$ At the same point, the rise in the restored force of the soap film (surface tension) would be equal and contrary to the electrostatic force, which will result in a new (larger) radius of equilibrium.

$\bullet$ Because of the ionic similarities, this can happen to both positive and negatively charged bubbles.Therefore, when a negative charge is given to the soap bubble, then its radius will increase.

Hence option $A$ correct.

If the flux of the electric field through a closed surface is zero, then (more than option may be correct)

0%
the electric field must be zero everywhere on the surface
0%
the electric field may be zero everywhere on the surface
0%
the charge inside the surface is zero.
0%
the charge in the vicinity of the surface must be zero

Explanation

If the flux of the electric field through a closed surface is zero, according to Gauss's law the inside the surface must be zero.

(ϕ = \frac{Q_{e n}}{ϵ_{0}})

$(\phi=\dfrac{Q_{en}}{\epsilon_0})$
Thus the flux through a surface depends only on the charge which is inside the surface. So the charge in the vicinity of the surface may not zero.
The electric field is the resultant field effect due to all charge. So It must not be zero but sometimes it may zero on the surafce depend on the charge distribution.

An infinite parallel plane sheet of a metal is charged to charge density

σ

$\sigma$ coulomb per square metre in a medium of dielectric constant

K

$K$ . Intensity of electric field near the metallic surface will be :

0%
$E=\cfrac { \sigma }{ { \varepsilon }_{ 0 }K }$
0%
$E=\cfrac { \sigma }{ {2 \varepsilon }_{ 0 } }$
0%
$E=\cfrac { \sigma }{ {2 \varepsilon }_{ 0 } K}$
0%
$E=\cfrac { K\sigma }{ {2 \varepsilon }_{ 0 } }$

Explanation

Electric field intensity due to thin infinite plane sheet of charge,

E = \frac{σ}{2 ε} = \frac{σ}{2 ε_{0} K}

$\quad E=\cfrac { \sigma }{ 2\varepsilon } =\cfrac { \sigma }{ 2{ \varepsilon }_{ 0 }K }$

Where dielectric constant,

K = \frac{ε}{ε_{0}}

$\quad K=\cfrac { \varepsilon }{ { \varepsilon }_{ 0 } }$

Find the dipole moment of a system where the potential

2.0 \times 10^{- 5} V

$2.0\times 10^{-5}V$ at a point P,

0.1 m

$0.1 m$ from the dipole is

3.0 \times 10^{4}

$3.0\times 10^4$ . (Use

θ = 30^{o})

$\theta=30^o)$ .

0%
$2.57\times 10^{-17}Cm$
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$1.285\times 10^{-15}Cm$
0%
$1.285\times 10^{-17}Cm$
0%
$2.57\times 10^{-15}Cm$

Explanation

The potential at a distance r due to dipole is $\displaystyle V=\dfrac{1}{4\pi\epsilon_0} \dfrac{\vec{p}.\hat{r}}{r^2}$ where $\vec{p}=$ dipole moment.

or $\displaystyle V=\dfrac{1}{4\pi\epsilon_0} \dfrac{p\cos\theta}{r^2}$

or $\displaystyle p=\dfrac{4\pi\epsilon_0 Vr^2}{\cos\theta}=\dfrac{4\pi\times 8.85\times 10^{-12}\times 2\times 10^{-5}\times (0.1)^2}{\cos30}=2.57 \times 10^{-17} Cm$

There exists a non-uniform electric field along x-axis as shown in the figure.The field increases at a uniform rate along

+

$+$ ve x-axis. a dipole is placed inside the field as shown. Which one of the following is correct for the dipole?

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Dipole will move along positive x-axis and will undergo a clockwise rotation
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Dipole will move along negative x-axis and will undergo a clockwise rotation
0%
Dipole will move along positive x-axis and will undergo an anticlockwise rotation
0%
Dipole will move along negative x-axis and will undergo an anticlockwise rotation

Explanation

the dipole is lapsed in a non-uniform electric field, therefor a force are well as couple acts on it. the force on the negative charge is more

(F ⋉ E)

$(F\ltimes E)$ , and is directed along

- v e

$-ve$ x-axis. does, the dipole moves along -ve x-axis and rotets in an\quad anti clock wise direction.

Identify the wrong statement in the following. Coulomb's law correctly describes the electric force that

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binds the electrons of an atom to its nucleus
0%
binds the protons and neutrons in the nucleus of an atom
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binds atoms together to form molecules
0%
binds atoms and molecules together to form solids

CBSE Questions for Class 12 Medical Physics Electric Charges And Fields Quiz 4 - MCQExams.com

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Practice Class 12 Medical Physics Quiz Questions and Answers

CBSE Questions for Class 12 Medical Physics Electric Charges And Fields Quiz 4 - MCQExams.com

0:0:4

Practice Class 12 Medical Physics Quiz Questions and Answers

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