For the situation shown in the figure,mark th | vedclass

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Question

(D) conductor is an equipotential body. The potential at any point inside or on the surface of the conductor is the same as the potential at its cente

Vedclass · Accepted Answer

Potential at point $B$ due to induced charges is $\frac{-qR}{4\pi \varepsilon_0(d+R)d}$

Vedclass · Answer

(D) conductor is an equipotential body. The potential at any point inside or on the surface of the conductor is the same as the potential at its center.Let $V_c$ be the potential at the center of the conductor. The potential at the center is due to the external charge $q$ and the induced charges on the surface of the conductor.The distance of the center from charge $q$ is $(d+R)$.Potential due to charge $q$ at the center is $V_q = \frac{1}{4\pi \varepsilon_0} \frac{q}{(d+R)}$.Since the conductor is neutral,the net charge on it is zero. Therefore,the potential at the center due to induced charges $(V_{ind})$ must be such that the total potential at the center is the potential due to the external charge at the center if the conductor were not present,but this is incorrect. Actually,for a neutral conductor,the total potential at the center is simply the potential due to the external charge $q$ at the center,because the net potential contribution from induced charges at the center is zero if the conductor were isolated and neutral. However,the potential of the conductor is constant and equal to the potential at the center due to the external charge $q$ at distance $(d+R)$.Thus,the potential of the conductor is $V = \frac{1}{4\pi \varepsilon_0} \frac{q}{(d+R)}$.Now,consider the potential at point $B$ (on the surface). $V_B = V_{q,B} + V_{ind,B} = V_{conductor}$.$V_{q,B} = \frac{1}{4\pi \varepsilon_0} \frac{q}{d}$.So,$V_{ind,B} = V_{conductor} - V_{q,B} = \frac{1}{4\pi \varepsilon_0} \left( \frac{q}{d+R} - \frac{q}{d} \right) = \frac{q}{4\pi \varepsilon_0} \left( \frac{d - (d+R)}{d(d+R)} \right) = \frac{-qR}{4\pi \varepsilon_0 d(d+R)}$.Therefore,option $D$ is correct.

For the situation shown in the figure,mark the correct statement. $A$ point charge $q$ is placed at a distance $d$ from the surface of a neutral hollow conductor of radius $R$.

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