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CBSE Questions for Class 12 Medical Physics Electric Charges And Fields Quiz 12 - MCQExams.com
CBSE
Class 12 Medical Physics
Electric Charges And Fields
Quiz 12
A charge $$\displaystyle \left ( 5\sqrt{2}+2\sqrt{5} \right )$$ coulomb is placed on the axis of an infinite disc at a distance a from the centre of disc. The flux of this charge on the part of the disc having inner and outer radius of a and 2a will be :
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$$\displaystyle \frac{3}{2\varepsilon _{0}}$$
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$$\displaystyle \frac{1}{2\varepsilon _{0}}$$
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$$\displaystyle \frac{2\left [ \sqrt{5}+\sqrt{2} \right ]}{\varepsilon _{0}}$$
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$$\displaystyle \frac{2\sqrt{5}+5\sqrt{2}}{2\varepsilon _{0}}$$
Explanation
As solid angle, $$w= 2\pi (1-cos\theta)$$ where $$\theta $$ is half planer angle.
From trignometry, $$tan \theta_1 = \dfrac{a}{a} = 1 \implies \theta_1 = 45^o$$
Also $$tan\theta_2 = \dfrac{2a}{a} = 2 \implies cos\theta_2 = \dfrac{1}{\sqrt{5}}$$
Now solid angle subtended by shaded region, $$w =w_2- w_1 $$
$$w= 2\pi (1-cos\theta_2) - 2\pi (1-cos\theta_1) $$
$$= 2\pi (1-\dfrac{1}{\sqrt{5}}) - 2\pi (1-\dfrac{1}{\sqrt{2}}) $$
$$w= 2\pi \bigg[\dfrac{1}{\sqrt{2}} - \dfrac{1}{\sqrt{5}}\bigg]$$
As flux through $$4\pi$$ steradian is equal to $$\dfrac{q}{\epsilon_o}$$
Thus flux through $$w$$ steradian, $$\phi = \dfrac{q}{\epsilon_o} \times \dfrac{w}{4\pi}$$
$$\phi = \dfrac{2\sqrt{5} + 5\sqrt{2}}{\epsilon_o} \times \dfrac{2\pi [\frac{1}{\sqrt{2}}-\frac{1}{\sqrt{5}}]}{4\pi}$$
$$\phi = \dfrac{\sqrt{10} +5 -2 - \sqrt{10}}{2\epsilon_o}$$
$$\implies \phi = \dfrac{3}{2\epsilon_o}$$
Three point charges +q, -2q and +q are placed at point (x =0,y=a, z=0),(x = 0, y=0, z=0) and (x=a, y =0, z =0) respectively. The magnitude and direction of the electric dipole moment vector of this charge assembly are :
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$$\displaystyle \sqrt { 2qa } $$ along + y direction
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$$\displaystyle \sqrt { 2qa } $$ along the line joining points
(x=0,y=0,z=0) and (x=a,y=a,z=0)
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$$\displaystyle qa$$ along the line joining points
(x=0,y=0,z=0) and (x=a,y=a,z=0)
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$$\displaystyle \sqrt { 2qa } $$ along + x direction
Explanation
The given charge assembly can be represented using the three co-ordinate axes
x, y and z as shown in figure.
The charge -2q is placed at the origin O. One + q charge is place at (a,0, 0) and the other +q charge is placed at(0, a, 0). Thus the system has two dipoles along x-axis and y-axis respectively.
As the electric dipole moment is directed from the negative to the positive charge hence the resultant dipole moment will be along OA
where co-ordinates of point A are (a, a, 0). The magnitude of each dipole moment,
, $$p = qa$$
So, the magnitude of resultant dipole moment is
$$P_R = \sqrt{p^2 + P^2} = \sqrt{(qa)^2+(qa)^2} = \sqrt{2}qa$$
A charged particle 'q' lies at 'P' and the line PC is perpendicular to the surface of ABC (part of disc). Find the flux passing through
the surface ABC.
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$$\displaystyle \frac { q }{ 4{ \varepsilon }_{ 0 } } $$
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$$\displaystyle \frac { q }{ 16{ \varepsilon }_{ 0 } } $$
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$$\displaystyle \frac { q }{ 32{ \varepsilon }_{ 0 } } $$
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$$\displaystyle \frac { q }{ 48{ \varepsilon }_{ 0 } } $$
Explanation
$$E=\cfrac{q}{4\pi \epsilon_0 a^2}$$
Area of whole disk$$=\pi r^2=\pi(\sqrt{3}a)^2=3\pi a^2$$
Area of $$ABC=\pi r^2\left( \cfrac { 30 }{ 360 } \right) =\cfrac { 3\pi { a }^{ 2 } }{ 12 } $$
$$=\cfrac { 1 }{ 12 } $$ (area of whole disk)
The electric flux through the whole disk,
$${ \phi }_{ T }=EA=\cfrac {q} {4\pi \epsilon_0 a^2}\times 3\pi a^2=\cfrac {3q}{4\epsilon_0}$$
But electric flux, $$\phi=\cfrac {1}{12}\phi_T=\cfrac{1}{12}\times\cfrac{3q}{4\epsilon_0}$$
$$=\cfrac {q}{16\epsilon_0}$$
The spatial distribution of the electric field due to two charges (A, B) is shown in figure. Which one of the following statements is correct ?
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A is +ve and B is -ve; $$\displaystyle \left| A \right| >\left| B \right| $$
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A is -ve and B is +ve; $$\displaystyle \left| A \right| =\left| B \right| $$
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Both are +ve but $$\displaystyle A>B$$
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Both are -ve but $$\displaystyle A>B$$
Explanation
We know that the electric field lines are coming out for positive charge and coming in for negative charge. So A is +ve and B is -ve. As A has more field lines so $$|A|>|B|$$
An electrostatic field line is continuous curve. That is, a field line cannot have sudden breaks. Why not?
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Electric field vanishes only at the origin
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Electric field vanishes only at infinity
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Electric field never vanishes
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Electric field is imaginary at some points
Explanation
The direction of tangent at any point in a field shows the direction of field at that point.
At sudden breaks, we can draw infinite tangents and hence infinitely many directions of field at that point which is not possible since electric field lines show the direction of net field and hence have unique direction.
Hence, electric field line can not vanish suddenly and vanishes only at infinity.
Answer-(B)
There exists a non-uniform electric field along x-axis as shown in the figure below. 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 moves along positive x-axis and undergoes a clockwise rotation
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Dipole moves along negative x-axis and undergoes a clockwise rotation
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Dipole moves along positive x-axis and undergoes a anticlockwise rotation
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Dipole moves along negative x-axis and undergoes a anticlockwise rotation
Explanation
The dipole is placed in a non-uniform field, therefore a force as well as a couple acts on it. The force on the negative charge is more $$(F \propto E)$$
and is directed along negative x-axis. Thus the dipole moves along negative x-axis and rotates in an
anticlockwise direction.
The diagram shows the cross-section of a sheet that is infinitely long in the $$x$$ and $$z$$ directions. The sheet has a height $$2D$$ in the $$y$$-direction. The charge carried by the sheet is NOT uniform; the charge density varies according to the equation $$=By^2$$.
Which of the following expressions gives the electric field strength at point $$P$$, a point that lies outside the sheet and is located at $$y=d$$?
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$$\cfrac{BD^4}{4{\epsilon}_0}$$
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$$\cfrac{Bd^3}{3{\epsilon}_0}$$
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$$\cfrac{BD^3}{3{\epsilon}_0}$$
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$$\cfrac{Bd^4}{{\epsilon}_0}$$
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The electric field is undefined because the sheet carries an infinite amount of charge.
Coaxial cable (typically used for cable and satellite tv) has its signal run on a copper wire surrounded by an insulator which is surrounded by the ground wire, as opposed to the typical side by side configuration.
What is the most logical reason for this?
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The energy held in the electric field between the inside wire and the outside allow for a clearer signal.
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The outside wire prevents any magnetic field from the inside wire from leaking out of the wire.
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The outside wire prevents electric fields from interfering with the signal in the inside wire.
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The overall resistance of the wire is reduced using this configuration.
Explanation
Coaxial Cable is type of electric cable that has an inner conductor surrounded by a tubular insulating layer, surrounded by a tubular conducting shield. The energy held in the electric field between the inside wire and outside allow for a clearer signal.
$$4$$ charges are placed at follows:
$$-2q$$ at $$(a,0)$$
$$q$$ at $$(0,-a)$$
$$-2q$$ at $$(-a,0)$$
$$3q$$ at $$(0,a)$$
The dipole moment of the configuration is
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$$2qa\hat j$$
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$$3qa\hat j$$
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$$2qa[\hat i + \hat j]$$
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None of these
Explanation
Net dipole moment $$=\underset{i}{\Sigma} q_{i}r_{i}$$
$$=(a\times 3q)\widehat{j}+(-2qa)\widehat{i}+[-2(-a)\widehat{i}]+q(-a)\widehat{j}$$
$$\implies 2qa\widehat{j}$$
In a region, the intensity of an electric field is given by $$E = 2i + 3j + k$$ in $$NC^{-1}$$. The electric flux through a surface $$S = 10i \ m^{2}$$ in the region is:
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$$5\ Nm^{2} C^{-1}$$
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$$10\ Nm^{2} C^{-1}$$
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$$15\ Nm^{2} C^{-1}$$
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$$20\ Nm^{2} C^{-1}$$
Explanation
Given: $$E = (2i + 3j + k) NC^{-1}$$
and $$S = 10i \ m^{2}$$
We know that:
$$\phi = E\cdot S$$
$$\phi = (2i + 3j + k) \cdot (10i)$$
$$\phi = 20\ Nm^{2} C^{-1}$$
Consider three charged bodies A, B and C. If A and B repel each other and A attracts C, what is nature of the force between B and C?
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Attractive
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Repulsive
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Zero
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None of these
Explanation
Since A repel B, A and B have same nature of charge .
And as A attracts C, A and C have opposite nature of charge.
Hence, B and C also has opposite nature of charge.
And, hence nature of force between B and C will be attractive.
Answer-(A)
The figure shows electric field lines penetrating two surfaces. Then
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$$|E_a| > |E_a|$$
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$$|E_a| = |E_a|$$
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$$|E_a| < |E_a|$$
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None of these
Explanation
From figure we can see that number of electric field passing through per unit area is more for plate A than plate B.
Hence, magnitude of field at A is greater than B.
$$|E_A|>|E_B|$$
Mathematically, electric flux can $$\phi$$ be represented as:
$$\vec E = $$ electric field
$$\hat n=$$ surface normal vector
$$A=$$ surface area
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$$\vec \phi = E\hat nA $$
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$$\vec \phi = \vec E.\hat nA $$
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$$\phi =\vec E.\hat nA $$
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$$\vec \phi = E nA $$
Explanation
$$\phi=\vec{E}.\vec{A}$$
Direction of area vector is taken along the normal to the surface.
Hence, $$\phi=\vec{E}.\hat{n}A$$ where $$\hat{n}$$ is direction normal to surface.
Answer-(C)
Calculate the flux through the base of the cone of radius $$r$$.
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$$2\pi r^2 E$$
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$$4\pi r^2 E$$
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$$0.5\pi r^2 E$$
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$$\pi r^2 E$$
Explanation
We know that electric flux is given bt
$$\phi=\oint\vec{E}.\vec{dS}$$
where $$\vec{dS}$$ is area vector directed along normal to area element.
For base of cone we can see that field is perpendicular to base surface and hence area vector is along field.
Hence,$$\phi=\oint EdS=E\oint dS$$
$$\implies \phi=E(\pi r^2)$$
$$\implies \phi=\pi r^2E$$
Answer-(D)
Electric field lines
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can cross each other.
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can only bisect one another.
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can only intersect at right angles.
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should never cross one another.
Explanation
If two electric field lines crosses each other , than at the point of intersection there will be two tangents showing two different directions of net electric field which is not possible for a point in space.
Hence, electric field lines should not cross each other.
Answer-(D)
The density of electric field lines at a specific location in space reveals information
about
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the strength of the field at that location.
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the strength of the charge at that location.
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the strength of the potential at that location.
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All of the above
Explanation
More is the density of electric field lines at a specific location in space, it means more stronger is the field there.
The density of electric field lines shows the strength of electric field .
Answer-(A)
From the schematic electric field lines, we can infer that
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the left charge is higher in magnitude than the right one
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the left charge is lower in magnitude than the right one
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the left charge is equal in magnitude to the right one
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None of these
Explanation
Since, number of field lines denstiy and hence electric field is lower near left positive charge. Hence, magnitude of charge is lower on left charge than right one.
Answer-(B)
From the schematic electric field lines, we can infer that the
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positive charge is higher in magnitude than the negative one
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positive charge is lower in magnitude than the negative one
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positive charge is equal in magnitude to the negative one
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None of these
Explanation
The electric field line density at positive charge is higher than at negative charge.
Hence, positive charge is of higher magnitude than the negative charge.
Answer-(A)
In the given distribution of electric field lines, it can be ascertained that
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$$E_x < E_y$$
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$$E_x = E_y$$
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$$E_x > E_y$$
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$$E_x . E_y = 1$$
Explanation
Since field lines are closer at X than at Y, hence field line density is more at point-X than at point-Y.
Hence $$E_x>E_y$$
Answer-(C)
Consider an area element $$dS$$ at a distance $$r$$ from a point P. Let $$\hat r$$ be the unit vector along the outward normal to $$dS$$.
If $$\alpha$$ is the angle between $$\hat r$$ and $$dS$$, the element of the solid angle subtended by the area element at P is defined as
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$$d\Omega = \dfrac{dS\ \sin \alpha}{r^2}$$
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$$d\Omega = \dfrac{dS\ \tan \alpha}{r^2}$$
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$$d\Omega = \dfrac{dS\ \cos \alpha}{r^2}$$
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$$d\Omega = \dfrac{dS\ \alpha}{r^2}$$
Explanation
Solid angle is defined as the component of area normal to radius per unit $$r^2$$.
$$d\Omega=\dfrac{dS\cos\alpha}{r^2}$$
Answer-(C)
The SI unit of solid angle is
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radians
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steradians
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degrees
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All of the above
Explanation
The S.I unit of solid angle is steradian, or square radian (symbol: $$sr$$).
It is used in three-dimensional geometry, and is analogous to the radian, which qualifies planar angles.
Two charges $$q_{1}$$ and $$q_{2}$$ are kept on the x-axis and the variation of electric field strength at different points on the x-axis is described in the adjacent figure graphically. Choose correct statement nature are magnitude of $$q_{1}$$ and $$q_{2}$$.
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$$q_{1} + ve, q_{2} + ve, |q_{1}| > |q_{2}|$$
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$$q_{1} + ve, q_{2} - ve, |q_{1}| > |q_{2}|$$
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$$q_{1} + ve, q_{2} - ve, |q_{1}| < |q_{2}|$$
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$$q_{1} - ve, q_{2} + ve, |q_{1}| < |q_{2}|$$
Find the charge on an iron particle of mass $$2.24 mg$$, if $$0.02$$ % of the electrons are removed from it.
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$$-0.01996 C$$
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$$0.01996 C$$
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$$0.00077 C$$
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$$2.0 C$$
Two charges of $$-2Q$$ and $$Q$$ are located at points $$(a,0)$$ and $$(4a,0)$$ respectively. What is the electric flux through a sphere of radius $$3a$$ centred at the origin?
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$$\dfrac{Q}{ \epsilon_0}$$
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$$\dfrac{-Q}{ \epsilon_0}$$
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$$\dfrac{-2Q}{ \epsilon_0}$$
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$$\dfrac{3Q}{ \epsilon_0}$$
Explanation
We know that flux$$=\phi=\dfrac{q}{\epsilon_o}$$ where $$q$$ is charge inside the closed surface area and flux due to charge outside is zero.
Here, charge inside$$=q=-2Q$$
Hence, flux$$=\phi=\dfrac{-2Q}{\epsilon_o}$$
Answer-(C)
Find the charge on the sphere
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$$1.45\mu C$$
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$$1.45mC$$
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$$1.45C$$
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$$1.45kC$$
Explanation
Charge,$$q$$=surface charge density $$(\sigma) \times $$ area $$(A)$$
$$\implies q=\sigma\times 4\pi r^2$$
$$\implies q=\sigma\times 4\pi \left(\dfrac{d}{2}\right)^2$$ where $$d=diameter$$
$$\implies q=80\times 10^{-6}\times 4\pi\times \left(\dfrac{2.4}{2}\right)^2$$
$$\implies q=1.45\times 10^{-3}C=1.45mC$$
Answer-(B)
An electric charge $$q$$is placed at the centre of a cube of side $$a$$ The electric flux through one of its faces is
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$$\dfrac{q}{6\epsilon_0}$$
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$$\dfrac{q}{6\epsilon_0a^2}$$
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$$\dfrac{q}{6\pi \epsilon_0a^2}$$
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$$\dfrac{q}{\epsilon_0}$$
Explanation
Flux through the cube $$=\phi=\dfrac{q}{\epsilon_o}$$
Since, charge is placed symmetrically and is equivalent to all 6 faces , hence flux through all 6 faces will be equal and the sum of flux through all six face equals the total flux.
Hence, flux through each face $$=\dfrac{q}{6\epsilon_o}$$
Answer-(A)
A charge $$Q$$ is located at the centre of a sphere of radius $$R$$. Calculate the flux going out through the surface
of the sphere.
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$$\dfrac{Q}{4\pi \epsilon_0 R^2}$$
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$$\dfrac{Q}{4\pi \epsilon_0 R}$$
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$$\dfrac{Q}{4\pi R^2}$$
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$$\dfrac{Q}{\epsilon_0}$$
Explanation
Flux going out of a closed surface is given by charge inside closed surface divided by permittivity of free space.
$$\phi=\dfrac{Q}{\epsilon_o}$$
Answer-(D)
A rod of length l having charge q uniformly distributed moves towards right with constant speed v. At $$t=0$$, it enters in an imaginary cube of edge $$I/2$$, sketch variation of electric flux passing through the cube with respect to time.
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Explanation
$$t=0$$ to $$t=\displaystyle\frac{I}{2v}$$
At any time t, length of rod inside cube
$$=vt$$
Charge inside cube $$=\displaystyle\frac{q}{I}vt$$
$$\therefore \phi =\displaystyle\frac{q_{In}}{\varepsilon_0}=\frac{qv}{\varepsilon_0I}t, \phi \propto t$$
Graph will be straight line of slope $$qv/\varepsilon_0I$$
$$t=0, \phi =0$$
$$t=\displaystyle\frac{I}{2v}, \phi =\frac{q}{2\varepsilon_0}$$
$$t=I/2v$$ to $$t=I/v$$, half of rod inside cube, $$\phi =\displaystyle\frac{q}{2\varepsilon_0}$$
$$t=\displaystyle\frac{I}{v}$$ to $$t=\displaystyle\frac{3I}{2v}$$
Charge inside the cube,
$$q_{In}=q-\frac{q}{I}\left(vt -\frac{I}{2}\right)$$
$$=\displaystyle\frac{3q}{2}-\frac{qv}{I}$$t
$$\phi =\frac{\displaystyle \left(\frac{3q}{2}-\frac{qv}{I}t\right)}{\varepsilon_0}$$
$$=\displaystyle\frac{3q}{2\varepsilon_0}-\frac{qv}{\varepsilon_0I}t$$
The graph will be straight line of negative slope and positive intercept.
$$t=\displaystyle\frac{I}{v}$$, $$\phi=\displaystyle\frac{q}{2\varepsilon_0}$$
$$t=\displaystyle\frac{3I}{2v}$$, $$\phi =0$$.
The total solid angle subtended by the sphere is
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$$0.25\pi$$
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$$0.5\pi$$
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$$4\pi$$
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$$2\pi$$
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$$1\pi$$
Explanation
Solid angle is given by area subtended divided by $$r^2$$.
$$\Omega=\dfrac{\Delta S}{r^2}$$
Now, area subtended by a sphere is the whole surface area of sphere=$$4\pi r^2$$
$$\implies \Omega=\dfrac{4\pi r^2}{r^2}$$
$$\implies \Omega=4\pi$$
Answer-(E)
Which of the following curves shown cannot possibly represent electrostatic field lines?
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Explanation
Electrostatic field lines start or end only at $${90}^{o}$$ to the surface of the conductor. So, curve (b) cannot represent electrostatic field lines are closer.
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 $${ 10 }^{ 5 }N\quad { C }^{ -1 }{ m }^{ -1 }$$. The force experienced by the system having a total dipole moment equal to $${ 10 }^{ -7 }C$$ $$m$$ in the negative z-direction is
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$$-{ 10 }^{ -2 }N$$
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$${ 10 }^{ -2 }N$$
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$${ 10 }^{ -4 }N$$
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$$-{ 10 }^{ -4 }N$$
Explanation
Consider an electric dipole dipole with $$-q$$ charge at $$A$$ and $$+q$$ charge at $$B$$, placed along z-axis, that its dipole moment is in negative $$z$$ direction
i.e, $${ p }_{ z }=-{ 10 }^{ -7 }Cm$$
The electric field is along positive direction of z-axis, such that
$$\cfrac { dE }{ dz } ={ 10 }^{ 5 }N{ C }^{ -1 }{ m }^{ -1 }\quad $$
From $$F=qdE=\left( q\times dz \right) \times \cfrac { dE }{ dz } =p\cfrac { dE }{ dz } $$
$$\therefore F=-{ 10 }^{ -7 }\times { 10 }^{ 5 }=-{ 10 }^{ -2 }N\quad $$
An electric dipole consists of charges $$\pm 2.0\times { 10 }^{ -8 }C$$ separated by a distance of $$2.0\times { 10 }^{ -3 }m\quad $$.It is placed near a long line charge of linear charge density $$4.0\times { 10 }^{ -4 }C\quad { m }^{ -1 }$$ as shown in the figure, such that the negative charge is at a distance of $$2.0cm$$ from the line charge. The force acting on the dipole will be
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$$7.2N$$ towards the line charge
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$$6.6N$$ away from the line charge
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$$0.6N$$ away from the line charge
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$$0.6N$$ towards the line charge
Explanation
The electric field at a distance $$r$$ from the line charge of linear density $$\lambda$$ is given by
$$E=\cfrac { \lambda }{ 2\pi { \varepsilon }_{ 0 }r } $$
Hence, the field at the negative charge
$${ E }_{ 1 }=\cfrac { \left( 4.0\times { 10 }^{ -4 } \right) \left( 2\times 9\times { 10 }^{ 9 } \right) }{ 0.02 } =3.6\times { 10 }^{ 8 }N{ C }^{ -1 }\quad $$
The force on the negative charge
$${ F }_{ 1 }=\left( 3.6\times { 10 }^{ 8 } \right) \left( 2.0\times { 10 }^{ -8 } \right) =7.2N$$ towards the line charge
Similary, the field at the positive charge
i.e., at $$r=0.022m$$ is $$\quad { E }_{ 2 }=3.3\times { 10 }^{ 8 }N{ C }^{ -1 }$$
The force on the positive charge,
$${ F }_{ 2 }=\left( 3.3\times { 10 }^{ 8 } \right) \left( 2.0\times { 10 }^{ -8 } \right) =6.6N\quad $$ away from the line charge
hence, the net force on the dipole $$=7.2N-6.6N=0.6N$$ towards the line charge.
A molecule of a substance has a permanent electric dipole moment of magnitude $${10}^{-30}cm$$. A mole of this substance is polarised by applying a strong electrostatic field of magnitude $${ 10 }^{ 7 }V{ m }^{ -1 }$$. The direction of field is changed by an angle $${60}^{o}$$. The heat released by the substance in aligning its dipole along the new direction of the field is
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$$-6J$$
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$$-3J$$
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$$3J$$
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$$6J$$
Explanation
Dipole moment of each molecules $$={ 10 }^{ -30 }cm$$.
As $$1$$ mole of the substance contains $$6\times { 10 }^{ 23 }\quad $$ molecules
Total dipole moment of all the molecules
$$p=6\times { 10 }^{ 23 }\times { 10 }^{ -30 }Cm=6\times { 10 }^{ -7 }Cm$$
Initial potential energy $${ U }_{ i }=-pE\cos { \theta } =-6\times { 10 }^{ -7 }\times { 10 }^{ 7 }\cos { { 0 }^{ o } } =-6J$$
Final potential energy $${ U }_{ f }=-6\times { 10 }^{ -6 }\times { 10 }^{ 6 }\times \cos { { 60 }^{ o } } =-3J$$
Work done $$=-$$ (Change in potential energy)
$$=-\left[ (-3)-(-6) \right] =-3J\quad $$
A field of $$100V{ m }^{ -1 }$$ is directed at $${30}^{o}$$ to positive x-axis. Find $$\left( { V }_{ A }-{ V }_{ B } \right) $$ if $$OA=2m$$ and $$OB=4m$$.
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$$100\left( \sqrt { 3 } -2 \right) V$$
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$$-100\left( 2+\sqrt { 3 } \right) V\quad $$
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$$100\left( 2-\sqrt { 3 } \right) V$$
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$$200\left( 2+\sqrt { 3 } \right) V$$
Explanation
A field of $$=100v/m$$ directed $$={ 30 }^{ 0 }$$
$${ V }_{ A }-{ V }_{ B }=?$$
$$OA=2m$$
$$OB=4m$$
Exaplanation
$$\overline { E } =\left( 100\cos{ 30 }^{ 0 }\hat { i } +100\sin{ 30 }^{ 0 }\hat { j } \right) $$
$$\overline { F } =50\sqrt { 3 } \hat { i } +50\hat { j } $$
$$A=\left( 2m,0 \right) $$
$$B=\left( 0,4m \right) $$
$${ V }_{ A }-{ V }_{ B }=-\int _{ B }^{ A }{ \overline { E } .d\overline { r } } $$
$$=\int _{ \left( -2,0 \right) }^{ \left( 0,4 \right) }{ 50\sqrt { 3 } dx } +50dy$$
$$={ \left[ 50\sqrt { 3 } x+50y \right] }_{ \left( 2,0 \right) }^{ \left( 0,4 \right) }$$
$$=-100\left( 2-\sqrt { 3 } \right) $$
The electric field at a point $$2$$cm from an infinite line charge of linear charge density $$10^{-7}$$ $$Cm^{-1}$$ is?
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$$4.5\times 10^4NC^{-1}$$
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$$9\times 10^4NC^{-1}$$
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$$9\times 10^2NC^{-1}$$
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$$18\times 10^4NC^{-1}$$
Explanation
$$\textbf{Step 1: Electric field due to infinitely long charged wire}$$
Linear charge density of wire, $$\lambda= 10^{-7}Cm^{-1}$$
Electric field at a distance $$r$$ from infinitely long charged wire is given by,
$$E = \dfrac{\lambda }{2\pi\epsilon_o r}$$ $$......(1)$$
$$\textbf{Step 2: Substituting the values in equation (1)}$$
$$ \ E =\dfrac{10^{-7}}{2\pi \times 8.85\times 10^{-12}\times 0.02} =9\times 10^4 \ NC^{-1} $$
Hence, Option (B) is correct.
Two non-conducting plates $$A$$ and $$B$$ of radii $$2R$$ and $$4R$$ respectively are kept at distances $$x$$ and $$2x$$ from the point charge $$q$$. A surface cutout of a non conducting shell $$C$$ is kept such that its centre coincides with the point charge. Each plate and the spherical surface carries a surface charge density $$\sigma$$. If $$\phi_{1}$$ is flux through surface of $$(B)$$ due to electric field of $$(A)$$ and $$\phi_{2}$$ be the flux through $$(A)$$ due to electric field of $$(B)$$ then:
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$$\phi_{1} = \phi_{2}$$
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$$\phi_{1} > \phi_{2}$$
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$$\phi_{1} < \phi_{2}$$
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It depend and on $$x$$ and $$R$$
Explanation
From the defination of flux,
$$\phi=E\cdot dA$$
or, $$\phi= E(A)$$
where, $$A$$ is the area of surface.
Now, according to question,
$$\phi _1:$$ Flux through $$B$$ due to $$A$$
So, $$E$$ due to $$A=\cfrac{\sigma}{2\varepsilon_o}$$ (Non- conducting charge sheet)
$$\Rightarrow \phi_1=E$$ (Area of $$B$$)
$$\Rightarrow \phi_1=\cfrac{\sigma}{\varepsilon_o}(\pi 4R^2)=\cfrac{2\sigma \pi R^2}{\varepsilon_o}$$
Similarly for $$\phi_2$$: Flux through $$A$$ due to $$B$$,
$$E$$ due to $$B=\cfrac{\sigma}{2\varepsilon_o}$$
$$\Rightarrow \phi_2=E$$ (Area of $$A$$)
$$\Rightarrow \phi_2=\cfrac{\sigma}{2\varepsilon_o}\left[16R^2\pi\right]=\cfrac{8\sigma \pi R^2}{\varepsilon_2}$$
So, $$\phi_2>\phi_1$$
The electric flux emerging through the closed surface $${S_1}$$ shown in figure which intersects the spherical conductor $$S,$$ due to the presence of a positive charge very near to conductor is:
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Positive
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Negative
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Zero
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None of the above
Explanation
As you see, due to $$+q$$ charge a $$-ve$$ charge will induce on the surface of conductor which is near to the charge.
But there will be a positive charge also induced on the other side and our gaussion surface will thus have positive charge enclosed in it.
$$\Rightarrow \quad \phi =\dfrac { +q }{ { \varepsilon }_{ 0 } } $$ {positive flux}
$$\therefore $$ Option A is the correct answer.
A point charge $$Q(C)$$ is placed at the origin. Find the electric flux of which an area $$4\pi\ m^2$$ on a concentric spherical shell of radius $$R$$
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$$\large{\frac{Q}{R^2 \epsilon_0}}$$
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$$\large{\frac{Q}{ \epsilon_0}}$$
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$$\large{\frac{Q}{4R^2 \epsilon_0}}$$
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none of above
Explanation
From Gauss law,
$$\phi =\dfrac { { Q }_{ inc } }{ { \varepsilon }_{ 0 } } $$
For a spherical shell
$$\rightarrow $$ flux through an area of $${ 4\pi r }^{ 2 }$$
$${ m }^{ 2 }=\dfrac { Q }{ { \varepsilon }_{ 0 } } $$
$$\rightarrow $$ Flux through 1
$${ m }^{ 2 }$$ area
$$=\dfrac { Q }{ 4\pi { R }^{ 2 }{ \varepsilon }_{ 0 } } $$
Also flux through
$$4\pi { m }^{ 2 }=\dfrac { 4\pi Q }{ 4\pi { R }^{ 2 }{ \varepsilon }_{ 0 } } =\dfrac { Q }{ { R }^{ 2 }{ \varepsilon }_{ 0 } } $$
$$\therefore $$ we can conclude that flux through an area of $$4\pi { m }^{ 2 }$$ on a shell is given by $$=\dfrac { Q }{ { R }^{ 2 }{ \varepsilon }_{ 0 } } $$
$$\therefore $$ Option (A) is the correct answer.
A point charge $$+Q$$ is placed just outside an imaginary hemispherical surface of radius $$R$$ as shown in the figure.
Which of the following statements is/are correct?
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The electric flux passing through the curved surface of the hemisphere is $$-\dfrac { Q }{ { 2 }_{ { \varepsilon }_{ 0 } } } \left( 1-\dfrac { 1 }{ \sqrt { 2 } } \right)$$
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Total flux through the curved and the flat surface is $$\dfrac { Q }{ _{ { \varepsilon }_{ 0 } } } $$
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The component of the electric field normal to the flat surface is constant over the surface
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The circumference of the flat surface is an equipotential
Explanation
The correct answers are options (A) and (D).
Hint: Calculate the electric flux passing through the curved surface.
Step 1: Calculate the electric flux passing through the hemisphere.
As shown in the figure, the component of electric field normal to the flat surface is given by $$Ecos\theta$$
The solid angle subtended by the flat surface at the point $$P$$ is:
$$2\pi (1-\dfrac{1}{\sqrt2})$$
So, the electric flux passing through the curved surface will be:
$$-\dfrac{Q}{\epsilon_0}\dfrac{2\pi(1-\dfrac{1}{\sqrt{2}})}{4\pi}$$
$$=-\dfrac{Q}{2\epsilon_0}{(1-\dfrac{1}{\sqrt{2}})}$$
Step 2: Consider the other options.
Here, both $$E$$ and $4\theta$$ changes continuously for different points on the curved surface. So, accordingly option (C) is incorrect.
From the above calculation, we find that the flux through the curved surface is less than $$\dfrac{Q}{\epsilon _0}$$. So, option (B) is also incorrect.
However, the circumference of the flat surface is equipotential because the distance of each point on the circumference is equal from the charge $$+Q$$. So, option (D) is correct.
Hence, the correct answers are options (A) and (D).
Two large metal sheets having surface charge density $$+\sigma$$ and $$-\sigma$$ are kept parallel to each other at a small separation distance $$d$$. The electric field at any point in the region between the plates is
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$$\large{\frac{\sigma}{\epsilon_0}}$$
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$$\large{\frac{\sigma}{2\epsilon_0}}$$
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$$\large{\frac{2\sigma}{\epsilon_0}}$$
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$$\large{\frac{\sigma}{4\epsilon_0}}$$
Explanation
Electric field due to sheet $$A$$, is $${ \vec { E } }_{ 1 }=\dfrac { \sigma }{ 2{ \varepsilon }_{ 0 } } i$$ {in right side of the sheet}
Also, electric field due to sheet $$B$$, is $${ \vec { E } }_{ 2 }=\dfrac { \sigma }{ 2{ \varepsilon }_{ 0 } } i$$ {in left side of the sheet}
$$\therefore $$ Resultant electric field by parallelogram law of vector addition,
$$\vec { E } ={ \vec { E } }_{ 1 }+{ \vec { E } }_{ 2 }$$ {$$\because \theta ={ 0 }^{ 0 }$$ between $${ \vec { E } }_{ 1 }\& { \vec { E } }_{ 2 }$$}
$$=\dfrac { \sigma }{ 2{ \varepsilon }_{ 0 } } +\dfrac { \sigma }{ 2{ \varepsilon }_{ 0 } } $$
$$\boxed { \vec { E } =\dfrac { \sigma }{ { \varepsilon }_{ 0 } } } $$
$$\therefore $$ Option (A) is the correct answer.
Three infinitely charged sheets are kept parallel to $$x-y$$ plane having charge densities as shown. Then the value of electric field at $$'P'$$ is:
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$$\dfrac { -4\sigma }{ { \in }_{ 0 } } \hat { k }$$
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$$\dfrac { 4\sigma }{ { \in }_{ 0 } } \hat { k }$$
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$$\dfrac { -2\sigma }{ { \in }_{ 0 } } \hat { k }$$
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$$\dfrac { 2\sigma }{ { \in }_{ 0 } } \hat { k }$$
Explanation
$$E$$ at $$P$$
$$=E_{dw\quad to Z=0}+E_{Z=2a}+E_{Z=3a}$$
$$=-\cfrac{\sigma}{2\epsilon_o}-\cfrac{2 \sigma}{2\epsilon_o}-\cfrac{1 \sigma}{2\epsilon_o}$$
$$=-\cfrac{4\sigma}{2\epsilon_o}\hat K$$
$$=-\cfrac{-2\sigma}{\epsilon_o}\hat K$$
Two infinite uniform sheet of charge each with surface charge density $$\sigma$$ are located at $$x\ =\ a$$ and $$x\ =\ -a$$. Which of the following is correct?
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$$E\ =\ \large{\frac{-\sigma}{\epsilon_0}}\vec{i}$$ ; For $$x<-a$$
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$$E\ =\ 0$$ ; For $$-a \lt x \lt a$$
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$$E\ =\ \large{\frac{\sigma}{\epsilon_0}}\vec{i}$$ ; For $$x>a$$
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$$E\ =\ \large{\frac{2\sigma}{\epsilon_0}}\vec{i}$$ ; For $$-a \lt x \lt a$$
Explanation
At (1) $$\vec { E } =\dfrac { \sigma }{ 2{ \varepsilon }_{ 0 } } +\dfrac { \sigma }{ 2{ \varepsilon }_{ 0 } } $$ {due to sheet A & B}
$$=\dfrac { \sigma }{ { \varepsilon }_{ 0 } } =\dfrac { \sigma }{ { \varepsilon }_{ 0 } } \{ -i\} $$ (directed toward -ve x-axis)
Similarly, at (3) $$\vec { E } =\dfrac { \sigma }{ 2{ \varepsilon }_{ 0 } } +\dfrac { \sigma }{ 2{ \varepsilon }_{ 0 } } $$ {due to sheet A & B}
$$=\dfrac { \sigma }{ { \varepsilon }_{ 0 } } (i)$$ (directed towards +ve x-axis)
But at (2), $$\vec { E } =\dfrac { \sigma }{ 2{ \varepsilon }_{ 0 } } -\dfrac { \sigma }{ 2{ \varepsilon }_{ 0 } } $$ {due to opposite direction at electric field}
$$=0$$
$$\therefore $$ Option (A), (B) & (C) satisfy the equ. (1), (2) & (3) respectively.
That is why (A), (B) & (C) are correct answer.
Consider a triangular surface whose vertices are three points having co-ordinate A ( 2a, 0, 0 ), B(0, a, 0), C(0, 0, a). If there is a uniform electric field $$E_0\hat{i} + 2 E_0\hat {j} + 3 E_0 \hat {k} $$ then flux linled to triangular surface ABC is:
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$$\dfrac{7E_0a^2}{2}$$
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$$3 E_0a^2$$
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$$\dfrac{11 E_0a^2}{2}$$
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Zero
Explanation
$$\phi = E_0\times \dfrac{1}{2}a^2 + 2E_0\times \dfrac{1}{2}\times 3E_0\times 3a^2$$
$$= \dfrac{11}{2} E_0a^2$$
A uniformly charged rod of length 4 cm and linear charge density $$\lambda = 30 \mu C/m$$ is placed as shown in figure. Calculate the x-component of electric field at point P
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$$36 \times 10 ^5 N/C$$
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$$9 \times 10 ^5 N/C$$
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$$1.8 \times 10 ^5 N/C$$
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$$27 \times 10 ^5 N/C$$
Two non-conducting spheres of radii $${R}_{1}$$ and $${R}_{2}$$ and carrying uniform volume charge densities $$+\rho$$ and $$-\rho$$, respectively, are placed such that they partially overlap, as shown in the figure. At all points in the overlapping region,
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the electrostatic field is zero
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the electrostatic potential is constant
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the electrostatic field is constant in magnitude
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the electrostatic field has different direction
Explanation
In triangle $$PC_1 C_2$$
$$ \vec{r_2} = \vec{d} = \vec{r_1}$$
The electrostatic field at point $$P$$ is
$$\vec{E} = \dfrac{ K \left( \rho \dfrac{4}{3} \pi {R_1}^3 \right) \vec{r_2} }{ {R_1}^3} + \dfrac{ K \left( \rho \dfrac{4}{3} \pi {R_2}^3 \right) \left( \vec{-r_1} \right) }{ {R_2}^3} $$
$$\vec{E} = K \rho \dfrac{4}{3} \pi \left( \vec{r_2} - \vec{r_1} \right)$$
$$\vec{E} = \dfrac{\rho}{3 \epsilon_0} \vec{d}$$
Hence, option (C) is correct.
The flux of the electric field due to charges distributed in a sphere of radius $$5$$ cm is $$10$$ Vm. What will be the electric flux, through a concentric sphere of radius $$10$$ cm ?
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$$20\ Vm$$
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$$30 \ Vm$$
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$$5\ Vm$$
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$$10 \ Vm$$
Explanation
Given :- $${ \phi }_{ A }=10Vm$$
$$r=5cm=5\times { 10 }^{ -2 }m$$
Now, from Gauss law,
$$\phi =\dfrac { { Q }_{ inc } }{ { \varepsilon }_{ 0 } } $$
$$\therefore $$ we can conclude from Gauss law that flux is a quantity which depend on the total charge inside the gaussion surface.
Now, Initially charge contained is same as finally,
$$\therefore $$ flux will also be some,
$$\Rightarrow { \phi }_{ B }=10Vm$$ through a surface of $$10cm$$
$$\therefore $$ Option (D) is the correct answer.
Which of the following statement(s) is/are correct?
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If the electric field due to a point charge varies as $${r}^{2.5}$$ instead $${r}^{-2}$$, then the Gauss law will still be valid.
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The Gauss law can be used to calculate the field distribution around an electric dipole.
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If the electric field between two point charges is zero somewhere, then the sign of the two charges is the same.
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The work done by the external force in moving a unit positive charge from point $$A$$ at potential $${V}_{A}$$ to point $$B$$ at potential $${V}_{B}$$ is $$({V}_{B}-{V}_{A})$$.
A positively charged thin metal ring of radius $$R$$ is fixed in the $$xy-$$plane with its centre at the origin $$O$$. A negatively charge particle $$P$$ is released from rest from the point $$(0, 0, {Z}_{0})$$ where $${ Z }_{ 0 }>0$$. Then the motion of $$P$$ is
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Periodic, for all values of $${Z}_{0}$$ satisfying $$0<{ z }_{ 0 }<\infty $$
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simple harmonic, for all values of $$0<{ z }_{ 0 }\le R$$
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approximately simple harmonic, provided $${ z }_{ 0 }\ll R$$
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such that $$P$$ crosses $$O$$ and continues to move along negative $$z-$$axis towards $$z=-\infty$$
Explanation
Ring will attract the (-ve)ly charged particle so, it will come down and when it will go below the ring then its velocity will decrease as force will be in opposite direction with the velocity and then finally it will stop and then will go up and a so, it will have a periodic motion.
Now, if $$z_0<<R,$$ then $$\left| F \right| \simeq \dfrac { kQq\left( { z }_{ 0 } \right) }{ { \left( { R }^{ 2 } \right) }^{ 3/2 } } $$
$$\left| F \right| \simeq \dfrac { kQq\left( { z }_{ 0 } \right) }{ { R }^{ 3 } } $$
$$\left| F \right| \alpha\;z$$ and it is opposite in direction of displacement .
So, it will be nearly a simple harmonic motion.
Hence, this question has multiple answers.
A number of spherical conductors of different radii are changed to same potential. The surface charge density of each conductor is related with its radius as
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$$\sigma \propto \dfrac{1}{R^2}$$
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$$\sigma \propto \dfrac{1}{R}$$
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$$\sigma \propto R$$
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None of these
Explanation
$$\sigma=\cfrac{q}{A}\,\,\,\,\sigma=\cfrac{q}{4\pi r^{2}}$$(as sphere).
$$v=\cfrac{q}{4\pi\epsilon_{0}r}=\cfrac{\sigma r}{\epsilon_{0}}$$
$$\therefore \sigma=\cfrac{v\epsilon_{0}}{r}$$(As $$v$$ is constant).
$$\therefore \sigma\propto\cfrac{1}{r}$$
An insulated conductor initially free from charge is charged by repeated contacts with a plate which after each contact is replenished to a charge $$Q$$ from an electrophorus. If $$q$$ is the charge on the conductor after the first operation, The maximum charge which can be given to the conductor in this way is $$Qq/(Q - q)$$.
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0%
True
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False
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Practice Class 12 Medical Physics Quiz Questions and Answers
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