Electrostatic Potential and Capacitance - Result Question 9
10. 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 $V_C$ denotes the potentials of the three shells, then for $c=a+b$, we have
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(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) $V_C=V_A \neq V_B$
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Answer:
Correct Answer: 10. (d)
Solution:
- (d) $c=a+b$.
$ \begin{aligned} V_A & =\frac{1}{4 \pi E_o}[\frac{qA}{a}+\frac{qB}{b}+\frac{qC}{a}] \\ & =\frac{1}{4 \pi E_o}[\frac{4 \pi a^{2} \sigma}{a}-\frac{4 \pi b^{2} \sigma}{b}+\frac{4 \pi c^{2} \sigma}{c}] \\ & =\frac{4 \pi}{4 \pi E_o}[\frac{a^{2} \sigma}{a}-\frac{b^{2} \sigma}{b}+\frac{c^{2} \sigma}{c}] \\ \Rightarrow V_A & =\frac{\sigma a}{\varepsilon_0}-\frac{\sigma b}{\varepsilon_0}+\frac{\sigma c}{\varepsilon_0}=\frac{\sigma}{\varepsilon_0}[c-(b-a)] \end{aligned} $
Similarly $V_B=\frac{1}{4 \pi E_O}[\frac{q A}{a}+\frac{q B}{b}+\frac{q C}{c}]$
$\Rightarrow V_B=\frac{1}{4 \pi E_O}[\frac{4 \pi \sigma a^{2}}{b}-\frac{4 \pi \sigma b^{2}}{b}+\frac{4 \pi \sigma c^{2}}{c}]$
$V_B=\frac{-\sigma b}{\varepsilon_0}+\frac{1}{4 \pi \varepsilon_0} \cdot \frac{\sigma \times 4 \pi a^{2}}{b}+\frac{\sigma c}{\varepsilon_0}$ $=\frac{\sigma}{\varepsilon_0}[c-\frac{(b^{2}-a^{2})}{b}]$
$V_C=\frac{4 \pi}{4 \pi E_o}[\frac{a^{2} \sigma}{c}-\frac{b^{2} \sigma}{c}+\frac{\sigma c^{2}}{c}]$
$V_C=\frac{\sigma c}{\varepsilon_0}-\frac{1}{4 \pi \varepsilon_0} \cdot \frac{\sigma \times 4 \pi b^{2}}{c}+\frac{1}{4 \pi \varepsilon_0} \cdot \frac{\sigma \times 4 \pi a^{2}}{c}$
$=\frac{\sigma}{\varepsilon_0}[c-\frac{(b^{2}-a^{2})}{c}]$
$=\frac{\sigma}{\varepsilon_0}[c-(b-a)]$
$V_A=V_C \neq V_B$
