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Q.

Concentric metallic hollow spheres of radii R and 6R hold charges $Q_{1}$ and $Q_{2}$ respectively. Given that surface charge densities of the concentric spheres are equal, the potential difference $V (R) - V (4 R)$ is

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a

$\frac{3 Q_{1}}{4 π ε_{0} R}$

b

$\frac{3 Q_{1}}{16 π ε_{0} R}$

c

$\frac{5 Q_{1}}{24 π ε_{0} R}$

d

$\frac{3 Q_{2}}{4 π ε_{0} R}$

answer is C.

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Detailed Solution

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$σ = \frac{Q_{1}}{4 π R^{2}} = \frac{Q_{2}}{4 π 16 R^{2}} V_{R} - V_{4 R} = \frac{K Q}{R} + \frac{K Q_{2}}{6 R} - \frac{K Q}{6 R} - \frac{K Q_{2}}{6 R} 16 Q_{1} = Q_{2} V_{R} - V_{4 R} = \frac{K Q_{1}}{R} [1 - \frac{1}{6}] = \frac{5}{6} \frac{K Q_{1}}{R} = \frac{5 Q_{1}}{24 π \in_{0} R}$

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