Question

Three concentric spherical shells have radii $a$ , $b$ and $c$ $(a<b<c)$ 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. $V_C=V_B\ne V_A$
• B. $V_C\ne V_B\ne V_A$
• C. $V_C=V_B=V_A$
• D. $V_A=V_C>V_B$

Answer 1

Answer: (D)

$V_A=\frac{1}{4\pi {ϵ}_0}\left\{\frac{q_A}{a}+\frac{q_B}{b}+\frac{q_C}{c}\right\}$
$=\frac{4\pi }{4\pi {ϵ}_0}\left\{\frac{a^2\sigma }{a}-\frac{b^2\sigma }{b}+\frac{c^2\sigma }{c}\right\}$
$V_A=\frac{1}{{ϵ}_0}\left\{\frac{a^2\sigma }{a}-\frac{b^2\sigma }{b}+\frac{c^2\sigma }{c}\right\}\Rightarrow V_A=\frac{1}{{ϵ}_0}\left\{a\sigma -b\sigma +c\sigma \right\}=\frac{2a\sigma }{{ϵ}_0}$
$V_B=\frac{1}{{ϵ}_0}\left\{\frac{a^2\sigma }{b}-\frac{b^2\sigma }{b}+\frac{c^2\sigma }{c}\right\}{V}_B=\frac{1}{{ϵ}_0}\left\{\frac{a^2\sigma }{b}-b\sigma +c\sigma \right\}{V}_B=\frac{a\sigma }{{ϵ}_0}\left\{\frac{a}{b}-1\right\}$
$V_C=\frac{1}{{ϵ}_0}\left\{\frac{a^2\sigma }{c}-\frac{b^2\sigma }{c}+\frac{c^2\sigma }{c}\right\}{V}_C=\frac{\sigma }{{ϵ}_0}\left\{\frac{a^2-b^2}{a+b}+c\right\}{V}_C=\frac{\sigma }{{ϵ}_0}\left\{a-b+c\right\}{V}_C=\frac{2\sigma a}{{ϵ}_0}$
$V_A=V_C>V_B$