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(A) The potential $V$ at a distance $r$ from the center of a conducting spherical shell of radius $R$ with charge $Q$ is given by:$V = \frac{kQ}{R}$ f

Vedclass · Accepted Answer

$2$

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(A) The potential $V$ at a distance $r$ from the center of a conducting spherical shell of radius $R$ with charge $Q$ is given by:$V = \frac{kQ}{R}$ for $r \le R$ and $V = \frac{kQ}{r}$ for $r > R$,where $k = \frac{1}{4\pi\epsilon_0}$.For the point at distance $r = \frac{R}{2}$ (inside both shells):The potential $V_2$ is the sum of potentials due to both shells:$V_2 = \frac{kq}{R} + \frac{k(2q)}{2R} = \frac{kq}{R} + \frac{kq}{R} = \frac{2kq}{R}$.For the point at distance $r = 3R$ (outside both shells):The potential $V_1$ is the sum of potentials due to both shells treated as point charges at the center:$V_1 = \frac{kq}{3R} + \frac{k(2q)}{3R} = \frac{3kq}{3R} = \frac{kq}{R}$.Therefore,the ratio $\frac{V_2}{V_1} = \frac{2kq/R}{kq/R} = 2$.

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