Electric Potential MCQ Questions & Answers in Electrostatics and Magnetism | Physics
Learn Electric Potential MCQ questions & answers in Electrostatics and Magnetism are available for students perparing for IIT-JEE, NEET, Engineering and Medical Enternace exam.
51.
A charge of 3 coulomb moving in a uniform electric field experiences a force of 3000 newton. The potential difference between the two points situated in a field at a distance of $$1\,cm$$ from each other will be :
A
100
B
5000
C
10
D
50
Answer :
10
Force on a charge in an electric field is
$$\eqalign{
& F = qE \cr
& {\text{But}}\,E = \frac{V}{d} \cr
& \therefore F = \frac{{qV}}{d} \cr
& {\text{or}}\,\,V = \frac{{Fd}}{q} = \frac{{300\,N \times 1 \times {{10}^{ - 2}}\,m}}{{3C}} = 10\,volt \cr} $$
52.
Equipotential surfaces are shown in figure. Then the electric field strength will be
A
$$100\,V{m^{ - 1}}$$ along $$X$$-axis
B
$$100\,V{m^{ - 1}}$$ along $$Y$$-axis
C
$$200\,V{m^{ - 1}}$$ at an angle $${120^ \circ }$$ with $$X$$-axis
D
$$50\,V{m^{ - 1}}$$ at an angle $${120^ \circ }$$ with $$X$$-axis
Answer :
$$200\,V{m^{ - 1}}$$ at an angle $${120^ \circ }$$ with $$X$$-axis
Using $$dV = - \overrightarrow E .d\overrightarrow r $$
$$\eqalign{
& \Rightarrow \Delta V = - E\Delta r\cos \theta \cr
& \Rightarrow E = \frac{{ - \Delta V}}{{\Delta r\cos \theta }} \cr
& \Rightarrow E = \frac{{ - \left( {20 - 10} \right)}}{{10 \times {{10}^{ - 2}}\cos {{120}^ \circ }}} \cr
& = \frac{{ - 10}}{{10 \times {{10}^{ - 2}}\left( { - \sin {{30}^ \circ }} \right)}} \cr
& = 200\,V/m \cr} $$
Direction of $$E$$ be perpendicular to the equipotential surface i.e. at $${{{120}^ \circ }}$$ with $$X$$-axis.
53.
In the electric field of a point charge $$q,$$ a certain charge is carried from point $$A$$ to $$B,C,D$$ and $$E.$$ Then the work done is
A
least along the path $$AB$$
B
least along the path $$AD$$
C
zero along all the paths $$AB,AC,AD$$ and $$AE$$
D
least along $$AE$$
Answer :
zero along all the paths $$AB,AC,AD$$ and $$AE$$
$$ABCDE$$ is an equipotential surface, on equipotential surface no work is done in shifting a charge from one place to another.
54.
If the electrostatic potential were given by $$\phi = {\phi _0}\left( {{x^2} + {y^2} + {z^2}} \right),$$ where $${\phi _0}$$ is constant, then the charge density giving rise to the above potential would be.
A
0
B
$$ - 6{\phi _0}{\varepsilon _0}$$
C
$$ - 2{\phi _0}{\varepsilon _0}$$
D
$$ - \frac{{6{\phi _0}}}{{{\varepsilon _0}}}$$
Answer :
$$ - 6{\phi _0}{\varepsilon _0}$$
$$\eqalign{
& E = - \nabla \phi = - {\phi _0}2\left[ {x\hat i + y\hat i + z\hat x} \right] \cr
& = {\varepsilon _0}\nabla .E = - 2{\varepsilon _0}{\phi _0}\nabla .\left( {x\hat i + y\hat i + z\hat x} \right) \cr
& n = - 6{\phi _0}{\varepsilon _0} \cr} $$
55.
Three identical metallic uncharged spheres $$A,B$$ and $$C$$ each of radius $$a,$$ are kept at the corners of an equilateral triangle of side $$d\left( {d > > a} \right)$$ as shown in Fig. The fourth sphere (of radius $$a$$ ), which has a charge $$q,$$ touches $$A$$ and is then removed to a position far away. $$B$$ is earthed and then the earth connection is removed. $$C$$ is then earthed. The charge on $$C$$ is
A
$$\frac{{qa}}{{2d}}\left( {\frac{{2d - a}}{{2d}}} \right)$$
B
$$\frac{{qa}}{{2d}}\left( {\frac{{2d - a}}{d}} \right)$$
C
$$ - \frac{{qa}}{{2d}}\left( {\frac{{d - a}}{d}} \right)$$
D
$$\frac{{2qa}}{d}\left( {\frac{{d - a}}{{2d}}} \right)$$
56.
A conducting sphere of radius $$R$$ is given a charge $$Q.$$ The electric potential and the electric field at the centre of the sphere respectively are
A
zero and $$\frac{Q}{{4\pi {\varepsilon _0}{R^2}}}$$
B
$$\frac{Q}{{4\pi {\varepsilon _0}R}}$$ and zero
C
$$\frac{Q}{{4\pi {\varepsilon _0}R}}{\text{and}}\frac{Q}{{4\pi {\varepsilon _0}{R^2}}}$$
D
Both and zero
Answer :
$$\frac{Q}{{4\pi {\varepsilon _0}R}}$$ and zero
In a conducting sphere charge is present on the surface of the sphere. So, electric field inside will be zero and potential remains constant from centre to surface of sphere and is equal to $$\frac{1}{{4\pi {\varepsilon _0}}}\frac{Q}{R}$$
57.
Each corner of a cube of side $$l$$ has a negative charge, $$-q.$$ The electrostatic potential energy of a charge $$q$$ at the centre of the cube is
A
$$ - \frac{{4{q^2}}}{{\sqrt 2 \pi {\varepsilon _0}l}}$$
B
$$\frac{{\sqrt 3 {q^2}}}{{4\pi {\varepsilon _0}l}}$$
C
$$\frac{{4{q^2}}}{{\sqrt 2 \pi {\varepsilon _0}l}}$$
D
$$ - \frac{{4{q^2}}}{{\sqrt 3 \pi {\varepsilon _0}l}}$$
Length of body diagonal $$ = \sqrt 3 \ell $$
∴ Distance of centre of cube from each corner $$r = \frac{{\sqrt 3 }}{2}\ell $$
$$P.E.$$ at centre $$ = 8 \times {\text{Potential Energy due to}}$$
$$\eqalign{
& A = 8 \times \frac{{Kq \times \left( { - q} \right)}}{r} \cr
& = 8 \times \frac{1}{{4\pi {\varepsilon _0}\sqrt 3 l}} \times 2 \times q \times \left( { - q} \right) = - \frac{{4{q^2}}}{{\sqrt 3 \pi {\varepsilon _0}l}} \cr} $$
58.
As per this diagram a point charge $$+q$$ is placed at the origin $$O.$$ Work done in taking another point charge $$- Q$$ from the point $$A$$ [coordinates $$\left( {0,a} \right)$$ ] to another point $$B$$ [coordinates $$\left( {a,0} \right)$$ ] along the straight path $$AB$$ is
A
zero
B
$$\left( {\frac{{ - qQ}}{{4\pi {\varepsilon _0}}}\frac{1}{{{a^2}}}} \right)\sqrt 2 a$$
C
$$\left( {\frac{{qQ}}{{4\pi {\varepsilon _0}}}\frac{1}{{{a^2}}}} \right) \cdot \frac{a}{{\sqrt 2 }}$$
D
$$\left( {\frac{{qQ}}{{4\pi {\varepsilon _0}}}\frac{1}{{{a^2}}}} \right)\sqrt 2 a$$
Answer :
zero
Potential at $$A,$$ $${V_A} = \frac{1}{{4\pi {\varepsilon _0}}}\frac{q}{a}$$
Potential at $$B,$$ $${V_B} = \frac{1}{{4\pi {\varepsilon _0}}}\frac{q}{a}$$
Thus, work done in carrying a test charge $$-Q$$ from $$A$$ to $$B$$
$$W = \left( {{V_A} - {V_B}} \right)\left( { - Q} \right) = 0$$
59.
Two concentric spheres of radii $$R$$ and $$r$$ have similar charges with equal surface charge densities $$\left( \sigma \right).$$ What is the electric potential at their common centre ?
A
$$\frac{\sigma }{{{\varepsilon _0}}}$$
B
$$\frac{\sigma }{{{\varepsilon _0}}}\left( {R - r} \right)$$
C
$$\frac{\sigma }{{{\varepsilon _0}}}\left( {R + r} \right)$$
Let $$Q$$ and $$q$$ be the charges on the spheres. The potential at the common centre will be
$$\eqalign{
& V = \frac{1}{{4\pi {\varepsilon _0}}}\left( {\frac{Q}{R}} \right) + \frac{4}{{4\pi {\varepsilon _0}}}\left( {\frac{q}{r}} \right) \cr
& = \frac{1}{{{\varepsilon _0}}}\left[ {\frac{Q}{{4\pi {R^2}}} \times R + \frac{q}{{4\pi {r^2}}} \times r} \right] \cr} $$
$$\eqalign{
& {\text{But,}}\,\,\frac{Q}{{4\pi {R^2}}} = \frac{q}{{4\pi {r^2}}} = \sigma \cr
& \therefore V = \frac{1}{{{\varepsilon _0}}}\left[ {\sigma R + \sigma r} \right] \cr
& = \frac{\sigma }{{{\varepsilon _0}}}\left( {R + r} \right) \cr} $$
60.
An electric dipole, consisting of two opposite charges of $$2 \times {10^{ - 6}}C$$ each separated by a distance $$3\,cm$$ is placed in an electric field of $$2 \times {10^5}N/C.$$ Torque on the dipole is
A
$$12 \times {10^{ - 1}}N - m$$
B
$$12 \times {10^{ - 2}}N - m$$
C
$$12 \times {10^{ - 3}}N - m$$
D
$$12 \times {10^{ - 4}}N - m$$
Answer :
$$12 \times {10^{ - 3}}N - m$$
When a dipole of dipole moment $$p$$ is placed in a uniform external electric field $$E,$$ then torque acting on dipole is
$$\eqalign{
& \tau = p \times E = \left| p \right| \times \left| E \right|\sin \theta \cr
& {\text{where,}}\,\,p = q \times 2l\,\,\left( {2l\,{\text{is length of dipole}}} \right) \cr
& {\text{Given,}}\,\,q = 2 \times {10^{ - 6}}C \cr
& 2l = 3\,cm = 3 \times {10^{ - 2}}m \cr
& E = 2 \times {10^5}\,N/C \cr
& {\text{As,}}\,\,\tau = pE\sin \theta \,\,\left( {\theta = {{90}^ \circ }p \bot E} \right) \cr
& {\text{So,}}\,\,\tau = pE \cr
& \therefore \tau = \left( {2 \times {{10}^{ - 6}}} \right) \times \left( {3 \times {{10}^{ - 2}}} \right) \times \left( {2 \times {{10}^5}} \right) \cr
& = 12 \times {10^{ - 3}}N - m \cr} $$