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(D) The potential at any point on a shell is the sum of potentials due to all shells. The charge on a shell of radius $r$ and surface charge density $

Vedclass · Accepted Answer

$V_C = V_A 
eq V_B$

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(D) The potential at any point on a shell is the sum of potentials due to all shells. The charge on a shell of radius $r$ and surface charge density $\sigma$ is $q = 4\pi r^2 \sigma$.Thus,$q_A = 4\pi a^2 \sigma$,$q_B = -4\pi b^2 \sigma$,and $q_C = 4\pi c^2 \sigma$.The potentials are:$V_A = \frac{1}{4\pi\epsilon_0} [\frac{q_A}{a} + \frac{q_B}{b} + \frac{q_C}{c}] = \frac{1}{\epsilon_0} [a\sigma - b\sigma + \frac{c^2\sigma}{c}] = \frac{\sigma}{\epsilon_0} [a - b + c]$$V_B = \frac{1}{4\pi\epsilon_0} [\frac{q_A}{b} + \frac{q_B}{b} + \frac{q_C}{c}] = \frac{1}{\epsilon_0} [\frac{a^2\sigma}{b} - b\sigma + \frac{c^2\sigma}{c}] = \frac{\sigma}{\epsilon_0} [\frac{a^2}{b} - b + c]$$V_C = \frac{1}{4\pi\epsilon_0} [\frac{q_A}{c} + \frac{q_B}{c} + \frac{q_C}{c}] = \frac{1}{\epsilon_0} [\frac{a^2\sigma}{c} - \frac{b^2\sigma}{c} + c\sigma] = \frac{\sigma}{\epsilon_0} [\frac{a^2 - b^2}{c} + c]$Given $c = a + b$,then $c - b = a$ and $c - a = b$. Also $a^2 - b^2 = (a - b)(a + b) = (a - b)c$.Substituting $c = a + b$ into $V_A$: $V_A = \frac{\sigma}{\epsilon_0} [a - b + (a + b)] = \frac{2a\sigma}{\epsilon_0}$.Substituting $c = a + b$ into $V_C$: $V_C = \frac{\sigma}{\epsilon_0} [\frac{(a - b)(a + b)}{c} + c] = \frac{\sigma}{\epsilon_0} [\frac{(a - b)c}{c} + c] = \frac{\sigma}{\epsilon_0} [a - b + a + b] = \frac{2a\sigma}{\epsilon_0}$.Since $V_A = V_C = \frac{2a\sigma}{\epsilon_0}$ and $V_B = \frac{\sigma}{\epsilon_0} [\frac{a^2}{b} - b + a + b] = \frac{\sigma}{\epsilon_0} [\frac{a^2}{b} + a]$,clearly $V_A = V_C 
eq V_B$.

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

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