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A point charge +Q is placed at point B at a distance 2R from the center O of an uncharged thin conducting shell of radius R as shown in the figure. lf $V_{A}$ be the potential at point A, which is at a radial distance of R/2 from the center of the shell, then

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$V_{A} > \frac{Q}{8 π ε_{0} R}$

$V_{A} = \frac{Q}{8 π ε_{0} R}$

$V_{A} < \frac{Q}{8 π ε_{0} R}$

$V_{A} = 0$

answer is C.

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Detailed Solution

Since electric field inside the conducting shell is zero,

$V_{0} = V_{A} a n d V_{0} = \frac{Q}{8 π ε_{0} R}$

$∴ V_{A} = \frac{Q}{8 π ε_{0} R}$

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