Question

Three concentric spherical shells have radii $a$ , $b$ and $c$ $\left(\right.a < b < c\left.\right)$ and have surface charge densities $\sigma \text{, } - \sigma \text{ 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. $V_{C}=V_{B}\neq V_{A}$
• B. $V_{C}\neq V_{B}\neq V_{A}$
• C. $V_{C}=V_{B}=V_{A}$
• D. $V_{A}=V_{C}>V_{B}$

Answer 1

Answer: (D)

$V_{A}=\frac{1}{4 \pi \epsilon _{0}}\left\{\frac{q_{A}}{a} + \frac{q_{B}}{b} + \frac{q_{C}}{c}\right\}$
$=\frac{4 \pi }{4 \pi \epsilon _{0}}\left\{\frac{a^{2} \sigma }{a} - \frac{b^{2} \sigma }{b} + \frac{c^{2} \sigma }{c}\right\}$
$V_{A}=\frac{1}{\epsilon _{0}}\left\{\frac{a^{2} \sigma }{a} - \frac{b^{2} \sigma }{b} + \frac{c^{2} \sigma }{c}\right\}\Rightarrow V_{A}=\frac{1}{\epsilon _{0}}\left\{a \sigma - b \sigma + c \sigma \right\}=\frac{2 a \sigma }{\epsilon _{0}}$
$V_{B}=\frac{1}{\epsilon _{0}}\left\{\frac{a^{2} \sigma }{b} - \frac{b^{2} \sigma }{b} + \frac{c^{2} \sigma }{c}\right\}V_{B}=\frac{1}{\epsilon _{0}}\left\{\frac{a^{2} \sigma }{b} - b \sigma + c \sigma \right\}V_{B}=\frac{a \sigma }{\epsilon _{0}}\left\{\frac{a}{b} - 1\right\}$
$V_{C}=\frac{1}{\epsilon _{0}}\left\{\frac{a^{2} \sigma }{c} - \frac{b^{2} \sigma }{c} + \frac{c^{2} \sigma }{c}\right\}V_{C}=\frac{\sigma }{\epsilon _{0}}\left\{\frac{a^{2} - b^{2}}{a + b} + c\right\}V_{C}=\frac{\sigma }{\epsilon _{0}}\left\{a - b + c\right\}V_{C}=\frac{2 \sigma a}{\epsilon _{0}}$
$V_{A}=V_{C}>V_{B}$