Electric Potential MCQ Questions & Answers in Electrostatics and Magnetism | Physics

Potential at $$B,{V_B}$$  is maximum
$${V_B} > {V_C} > {V_A}$$
As in the direction of electric field potential decreases.

$$\eqalign{ & {V_A} = \frac{1}{{4\pi {\varepsilon _0}}}\frac{{\sigma 4\pi {a^2}}}{a} - \frac{1}{{4\pi {\varepsilon _0}}}\frac{{\sigma 4\pi {b^2}}}{b} + \frac{1}{{4\pi {\varepsilon _0}}} \cdot \frac{{\sigma 4\pi {c^2}}}{c} \cr & = \frac{\sigma }{{{\varepsilon _0}}}\left( {a - b + c} \right) = \frac{\sigma }{{{\varepsilon _0}}}\left( {2a} \right)\,\,\,\,\left( {\because c = a + b} \right) \cr & {V_B} = \frac{1}{{4\pi {\varepsilon _0}}} \cdot \frac{{\sigma 4\pi {a^2}}}{c} - \frac{1}{{4\pi {\varepsilon _0}}}\frac{{\sigma 4\pi {b^2}}}{b} + \frac{1}{{4\pi {\varepsilon _0}}} \cdot \frac{{\sigma 4\pi {c^2}}}{c} \cr & = \frac{\sigma }{{{\varepsilon _0}}}\left( {\frac{{{a^2}}}{c} - b + c} \right) = \frac{\sigma }{{{\varepsilon _0}}}\left( {2a} \right)\,\,\,\,\left( {\because c = a + b} \right) \cr} $$
$$\eqalign{ & {\text{and}}\,\,{V_C} = \frac{1}{{4\pi {\varepsilon _0}}} \cdot \frac{{\sigma 4\pi {a^2}}}{c} - \frac{1}{{4\pi {\varepsilon _0}}}\frac{{\sigma 4\pi {b^2}}}{c} + \frac{1}{{4\pi {\varepsilon _0}}} \cdot \frac{{\sigma 4\pi {c^2}}}{c} \cr & = \frac{\sigma }{{{\varepsilon _0}}}\left( {\frac{{{a^2}}}{c} - \frac{{{b^2}}}{c} + c} \right) = \frac{\sigma }{{{\varepsilon _0}}}\left( {2a} \right)\,\,\,\,\left( {\because c = a + b} \right) \cr & {\text{Hence,}}\,\,{V_A} = {V_C} = {V_B} \cr} $$

$$\eqalign{ & {\text{Work}}\,{\text{done}}\,\,{W_{21}} = \left( {{V_1} - {V_2}} \right)q \cr & {V_1} = \frac{1}{{4\pi { \in _0}}}\left[ {\frac{{{Q_1}}}{R} + \frac{{{Q_2}}}{{\sqrt 2 R}}} \right] \cr & {\text{and}}\,\,{V_2} = \frac{1}{{4\pi { \in _0}}}\left[ {\frac{{{Q_2}}}{R} + \frac{{{Q_1}}}{{\sqrt 2 R}}} \right] \cr & {\text{Thus,}}\,{W_{21}} = \frac{{q\left( {{Q_1} - {Q_2}} \right)\left( {\sqrt 2 - 1} \right)}}{{\sqrt 2 .4\pi { \in _0}R}}. \cr} $$

An equipotential surface has a constant value of potential at all points on its surface.
The surface and interior of a charged conductor is equipotential. Therefore, the potential is same throughout the charged conductor.

Potential due to a charged shell at an outside point
$$ = \frac{1}{{4\pi {\varepsilon _0}}} \times \frac{Q}{r}$$
while that inside the shell $$ = \frac{{\sigma R}}{{{\varepsilon _0}}},$$  where $$R$$ is the radius of the shell.

Since the given system is closed, the increase in $$KE$$  is equal to decrease in $$P.E.$$
$$\eqalign{ & \Rightarrow \frac{3}{2}m{v^2} = \frac{{2k{q^2}}}{{{r_0}}} - \frac{{3k{q^2}}}{r} \cr & \Rightarrow v = \sqrt {\frac{{2k{q^2}\left( {r - {r_0}} \right)}}{{mr{r_0}}}} ,\,\,v\,{\text{will}}\,{\text{be}}\,{\text{max}}\,{\text{when}}\,r \to \infty \cr & \Rightarrow {v_{\max }} = \sqrt {\frac{{2k{q^2}}}{{m{r_0}}}} \cr} $$
The work performed by the interaction force during the variation of the system’s configuration is equal to the decrease in the potential energy
$$W = {U_1} - {U_2} = \frac{{3k{q^2}}}{{{r_0}}}$$
∴ Work done per particle $$ = \frac{{k{q^2}}}{{{r_0}}}$$

As we know that relation between potential difference and electric field $$E$$ in a particular region is given by, $$E = - \frac{{dV}}{{dr}}$$
As, $$V = 6x - 8xy - 8y + 6yz$$
So, $$E = - \frac{{dV}}{{dr}}$$
$$ = - \left[ {\left( {6 - 8y} \right)\hat i + \left( { - 8x - 8 + 6z} \right)\hat j + 6y\hat k} \right]$$
The value of $$E$$ at coordinate $$\left( {1,1,1} \right)$$
$$\eqalign{ & E = - \left[ { - 2\hat i - 10\hat j + 6\hat k} \right] \cr & {\text{So,}}\,\,{E_{{\text{net}}}} = \sqrt {{{\left( { - 2} \right)}^2} + {{\left( { - 10} \right)}^2} + {6^2}} = 2\sqrt {35} \,N/C \cr} $$
and force on charge $$q$$ due to $${E_{{\text{net}}}}$$ is given by
$$F = q{E_{{\text{net}}}} = 2 \times 2\sqrt {35} = 4\sqrt {35} \,N$$

$$\eqalign{ & E = \frac{V}{d},F = eE = \frac{{eV}}{d} \cr & a = \frac{F}{m} = \frac{{eV}}{{md}} \cr & d = \frac{1}{2}a{t^2}\,{\text{or}}\,t = \sqrt {\frac{{2\;d}}{a}} \cr & {\text{or}}\,\,t = \sqrt {\frac{{2d\,md}}{{eV}}} = \sqrt {\frac{{2m{d^2}}}{{eV}}} \cr} $$

Potential energy decreases whenever there is attraction. A negative charge placed at centroid causes attraction.

$$\eqalign{ & \int_{\ell \, = \,\infty }^{\ell \, = \,0} { - \overrightarrow E .\overrightarrow {d\ell } } = {V_0} - {V_\infty } \cr & = \frac{{kq}}{r} - 0 \cr & = \frac{{9 \times {{10}^9} \times 1.11 \times {{10}^{ - 10}}}}{{0.5}} \cr & \approx 2\,V \cr} $$