Question

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. $V_C = V_B \ne V_A$
• B. $V_C \ne V_B \ne V_A$
• C. $V_C = V_B = V_A$
• D. $V_C = V_A \ne V_B$

Answer 1

Answer: (D)

$V_C = V_A \ne V_B$

Answer 2

$q _{ A } =4 \pi a ^{2} \sigma$ $q _{ B } =-4 \pi b ^{2} \sigma$ $q _{ C }=4 \pi c ^{2} \sigma,\, c = a + b$ $V _{ A } =\frac{1}{4 \pi \in_{0}}\left(\frac{ q _{ A }}{ a }+\frac{ q _{ B }}{ b }+\frac{ q _{ C }}{ c }\right)$ $=\frac{2 \sigma a }{\epsilon_{0}}$ $V _{ B } =\frac{1}{4 \pi \in_{0}}\left(\frac{ q _{ A }}{ a }+\frac{ q _{ B }}{ b }+\frac{ q _{ C }}{ c }\right)$ $=\frac{\sigma}{\epsilon_{0}}\left(\frac{ a ^{2}}{ b }- b + c \right)$ $=\frac{\sigma}{\epsilon_{0}}\left( a +\frac{ a ^{2}}{ b }\right)$ $V _{ C } =\frac{1}{4 \pi \in_{0}}\left(\frac{ q _{ A }}{ a }+\frac{ q _{ B }}{ b }+\frac{ q _{ C }}{ c }\right)$ $=\frac{\sigma}{\epsilon_{0}}\left(\frac{ a ^{2}- b }{ c }+ c \right)=\frac{2 \sigma a }{\epsilon_{0}}$ So, $V _{ C }= V _{ A } \neq V _{ B }$