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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 Vc denote the potentials of the three shells, then for c = a + b, we have:
Option: 1
$\text{V}_{\text{C}}=\text{V}_{\text{B}}\neq\text{V}_{\text{A}}$

Option: 2
$\text{V}_{\text{C}}\neq\text{V}_{\text{B}}\neq\text{V}_{\text{A}}$

Option: 3
$\text{V}_{\text{C}}=\text{V}_{\text{B}}=\text{V}_{\text{A}}$

Option: 4
$\text{V}_{\text{C}}=\text{V}_{\text{A}}\neq\text{V}_{\text{B}}$

Answers (1)

best_answer

Outside the sphere (P lies outside the sphere) -

$dpi{100} E_{out}=frac{1}{4pi epsilon _{0}}frac{Q}{r^{2}}=frac{sigma R^{2}}{epsilon _{0}r^{2}}$

$V_{out}=frac{1}{4pi epsilon _{0}}frac{Q}{r}=frac{sigma R^{2}}{epsilon _{0}r}$

where

$sigma$ - surface charge density.

At the surface of the Sphere -

$V=R$

$E_{s}=frac{1}{4pi epsilon _{0}}frac{Q}{R^{2}}=frac{sigma }{epsilon _{0}}$

$V_{s}=frac{1}{4pi epsilon _{0}}frac{Q}{R}=frac{sigma R}{epsilon _{0}}$

$V_{A} = \frac{kq_{A}}{a}+\frac{kq_{B}}{b}+\frac{kq_{C}}{c}$

$= \frac{k.\sigma .4\pi a^{2}}{a}-\frac{k.\sigma .4\pi b^{2}}{b}+\frac{k.\sigma .4\pi c^{2}}{c}$

$V_{A}=k.\sigma .4\pi \left ( a-b+c \right )$

$V_{B}= \frac{kq_{A}}{b}+\frac{kq_{B}}{b}+\frac{kq_{C}}{c}$

$=\frac{k.\sigma .4\pi a^{2}}{b}+\frac{k\left ( -\sigma .4\pi b^{2} \right )}{b}+\frac{k.\sigma .4\pi c^{2}}{c}$

$= 4\pi k\sigma \left [ \frac{a^{2}}{b}-b+c\right ]$

$= 4\pi k\sigma \left ( \frac{a^{2}-b^{2}+bc}{b} \right )$

$V_{c} =\frac{k.q_{A}}{a}+\frac{k.q_{B}}{b}+\frac{k.q_{C}}{c}$

$= \frac{k.\sigma 4\pi a^{2}}{c}-\frac{k.\sigma 4\pi b^{2}}{c}+\frac{k.\sigma 4\pi c^{2}}{c}$

$= 4\pi k\sigma \frac{\left [ a^{2}-b^{2}+c^{2}\right ]}{c}$

$4\pi k\sigma \left [ \frac{\left ( a-b \right ).\left ( a+b \right )}{c} +c\right ]$

$=4\pi k\sigma \left [ a-b+c \right ]\left [ \therefore a+b=c \right ]$

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

Posted by

HARSH KANKARIA

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