Three concentric spherical shells have radii a, band c (a < b < c) and have surface charge densities $σ, - σ and σ$ respectively. If $V_{A}, V_{B} and V_{C}$ denote the potentials of the three shells, then for c = a + b, we have :

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$V_{C} = V_{B} = V_{A}$

$V_{C} = V_{A} \neq V_{B}$

$V_{C} = V_{B} \neq V_{A}$

$V_{C} \neq V_{B} = V_{A}$

answer is B.

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Detailed Solution

$V_{A} = \frac{k (σ x 4 {πa}^{2})}{a} + \frac{k (- σ x 4 {πb}^{2})}{b} + \frac{k (σ x 4 {πc}^{2})}{c} = 4 πkσ (a - b + c)$

$V_{B} = \frac{k (σ x 4 {πa}^{2})}{b} + \frac{k (- σ x 4 {πb}^{2})}{b} + \frac{k (σ x 4 {πc}^{2})}{c} = 4 πkσ (\frac{a^{2}}{b} - b + c)$

$V_{C} = \frac{k (σ x 4 {πa}^{2})}{c} + \frac{k (- σ x 4 {πb}^{2})}{c} + \frac{k (σ x 4 {πc}^{2})}{c} = 4 πkσ (\frac{a^{2}}{c} - \frac{b^{2}}{c} + c)$

$\Rightarrow V_{A} = V_{C} \neq V_{B}$

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