A spherical shell with an inner radius a and | vedclass

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(D) $1$. Since the shell is made of conducting material,the electric field inside the material of the shell $(a < R < b)$ must be zero.$2$. To ensure

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$K\frac{Q-q}{b}$

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(D) $1$. Since the shell is made of conducting material,the electric field inside the material of the shell $(a $2$. To ensure the electric field is zero inside the conductor,an induced charge of $-Q$ must appear on the inner surface of the shell (at radius $a$).$3$. The total charge on the shell is $-q$. Since the inner surface has a charge of $-Q$,the charge on the outer surface (at radius $b$) must be $q_{outer} = (-q) - (-Q) = Q - q$.$4$. The potential at a point $R$ inside the material of the shell $(a $5$. Potential $V(R) = V_{point} + V_{inner} + V_{outer} = \frac{KQ}{R} + \frac{K(-Q)}{R} + \frac{K(Q-q)}{b}$.$6$. Simplifying this,$V(R) = 0 + \frac{K(Q-q)}{b} = \frac{K(Q-q)}{b}$.

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