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Three concentric spherical shells have radii a, b and c (a < b < c) and have surface charge densities $\sigma$, $-\sigma$ and $\sigma$ respectively. If $V_A$, $V_B$ and $V_C$ denote the potentials of the three shells, then, for c = a + b, we have
A
$V_C = V_B = V_A$
B
$V_C = V_A \neq V_B$
C
$V_C = V_B \neq V_A$
D
$V_C \neq V_B \neq V_A$
Detailed Solution
Charges on the shells: $q_A = 4\pi a^2\sigma$, $q_B = -4\pi b^2\sigma$, $q_C = 4\pi c^2\sigma$
Potential of a shell of radius R and charge q: $\frac{q}{4\pi\varepsilon_0R}$ at any point on or inside it, and $\frac{q}{4\pi\varepsilon_0r}$ at an outside point at distance r.
$V_A = \frac{1}{4\pi\varepsilon_0}\left(\frac{q_A}{a} + \frac{q_B}{b} + \frac{q_C}{c}\right) = \frac{\sigma}{\varepsilon_0}(a - b + c)$
$V_B = \frac{1}{4\pi\varepsilon_0}\left(\frac{q_A}{b} + \frac{q_B}{b} + \frac{q_C}{c}\right) = \frac{\sigma}{\varepsilon_0}\left(\frac{a^2}{b} - b + c\right)$
$V_C = \frac{1}{4\pi\varepsilon_0}\left(\frac{q_A + q_B + q_C}{c}\right) = \frac{\sigma}{\varepsilon_0}\left(\frac{a^2 - b^2 + c^2}{c}\right)$
Put c = a + b: $V_A = \frac{\sigma}{\varepsilon_0}(2a)$
$V_C = \frac{\sigma}{\varepsilon_0}\left[\frac{(a - b)(a + b)}{a + b} + (a + b)\right] = \frac{\sigma}{\varepsilon_0}(a - b + a + b) = \frac{\sigma}{\varepsilon_0}(2a)$
$V_B = \frac{\sigma}{\varepsilon_0}\left(\frac{a^2}{b} + a\right)$, which is not equal to $\frac{\sigma}{\varepsilon_0}(2a)$ since $a \neq b$
Hence $V_C = V_A \neq V_B$.
Potential of a shell of radius R and charge q: $\frac{q}{4\pi\varepsilon_0R}$ at any point on or inside it, and $\frac{q}{4\pi\varepsilon_0r}$ at an outside point at distance r.
$V_A = \frac{1}{4\pi\varepsilon_0}\left(\frac{q_A}{a} + \frac{q_B}{b} + \frac{q_C}{c}\right) = \frac{\sigma}{\varepsilon_0}(a - b + c)$
$V_B = \frac{1}{4\pi\varepsilon_0}\left(\frac{q_A}{b} + \frac{q_B}{b} + \frac{q_C}{c}\right) = \frac{\sigma}{\varepsilon_0}\left(\frac{a^2}{b} - b + c\right)$
$V_C = \frac{1}{4\pi\varepsilon_0}\left(\frac{q_A + q_B + q_C}{c}\right) = \frac{\sigma}{\varepsilon_0}\left(\frac{a^2 - b^2 + c^2}{c}\right)$
Put c = a + b: $V_A = \frac{\sigma}{\varepsilon_0}(2a)$
$V_C = \frac{\sigma}{\varepsilon_0}\left[\frac{(a - b)(a + b)}{a + b} + (a + b)\right] = \frac{\sigma}{\varepsilon_0}(a - b + a + b) = \frac{\sigma}{\varepsilon_0}(2a)$
$V_B = \frac{\sigma}{\varepsilon_0}\left(\frac{a^2}{b} + a\right)$, which is not equal to $\frac{\sigma}{\varepsilon_0}(2a)$ since $a \neq b$
Hence $V_C = V_A \neq V_B$.