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Three concentric spherical shells have radii a, b, and $\mathrm{c}(\mathrm{a} < \mathrm{b} < \mathrm{c})$ 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 :
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Verified Answer
The correct answer is:
$\mathrm{V}_{\mathrm{C}}=\mathrm{V}_{\mathrm{A}} \neq \mathrm{V}_{\mathrm{B}}$
$$
\begin{aligned}
\mathrm{q}_{\mathrm{A}} & =4 \pi \mathrm{a}^2 \sigma, \\
\mathrm{q}_{\mathrm{B}} & =-4 \pi \mathrm{b}^2 \sigma, \\
\mathrm{q}_{\mathrm{C}} & =4 \pi \mathrm{c}^2 \sigma, \mathrm{c}=\mathbf{a}+\mathrm{b} \\
\mathrm{V}_{\mathrm{A}} & =\frac{1}{4 \pi \in_0}\left(\frac{\mathrm{q}_{\mathrm{A}}}{\mathrm{a}}+\frac{\mathrm{q}_{\mathrm{B}}}{\mathrm{b}}+\frac{\mathrm{q}_{\mathrm{C}}}{\mathrm{c}}\right) \\
& =\frac{2 \sigma \mathrm{a}}{\epsilon_0} \\
\mathrm{~V}_{\mathrm{B}} & =\frac{1}{4 \pi \epsilon_0}\left(\frac{\mathrm{q}_{\mathrm{A}}}{\mathrm{a}}+\frac{\mathrm{q}_{\mathrm{B}}}{\mathrm{b}}+\frac{\mathrm{q}_{\mathrm{C}}}{\mathrm{c}}\right)
\end{aligned}
$$
$$
\begin{aligned}
& =\frac{\sigma}{\epsilon_0}\left(\frac{\mathbf{a}^2}{\mathbf{b}}-\mathbf{b}+\mathrm{c}\right) \\
& =\frac{\sigma}{\epsilon_0}\left(\mathbf{a}+\frac{\mathbf{a}^2}{\mathbf{b}}\right) \\
\mathrm{V}_{\mathrm{C}} & =\frac{1}{4 \pi \in_0}\left(\frac{\mathrm{q}_{\mathrm{A}}}{\mathbf{a}}+\frac{\mathrm{q}_{\mathrm{B}}}{\mathrm{b}}+\frac{\mathrm{q}_{\mathrm{C}}}{\mathrm{c}}\right) \\
& =\frac{\sigma}{\epsilon_0}\left(\frac{\mathrm{a}^2-\mathbf{b}}{\mathrm{c}}+\mathrm{c}\right)=\frac{2 \sigma \mathbf{a}}{\epsilon_0}
\end{aligned}
$$
So, $V_C=V_A \neq V_B$
\begin{aligned}
\mathrm{q}_{\mathrm{A}} & =4 \pi \mathrm{a}^2 \sigma, \\
\mathrm{q}_{\mathrm{B}} & =-4 \pi \mathrm{b}^2 \sigma, \\
\mathrm{q}_{\mathrm{C}} & =4 \pi \mathrm{c}^2 \sigma, \mathrm{c}=\mathbf{a}+\mathrm{b} \\
\mathrm{V}_{\mathrm{A}} & =\frac{1}{4 \pi \in_0}\left(\frac{\mathrm{q}_{\mathrm{A}}}{\mathrm{a}}+\frac{\mathrm{q}_{\mathrm{B}}}{\mathrm{b}}+\frac{\mathrm{q}_{\mathrm{C}}}{\mathrm{c}}\right) \\
& =\frac{2 \sigma \mathrm{a}}{\epsilon_0} \\
\mathrm{~V}_{\mathrm{B}} & =\frac{1}{4 \pi \epsilon_0}\left(\frac{\mathrm{q}_{\mathrm{A}}}{\mathrm{a}}+\frac{\mathrm{q}_{\mathrm{B}}}{\mathrm{b}}+\frac{\mathrm{q}_{\mathrm{C}}}{\mathrm{c}}\right)
\end{aligned}
$$
$$
\begin{aligned}
& =\frac{\sigma}{\epsilon_0}\left(\frac{\mathbf{a}^2}{\mathbf{b}}-\mathbf{b}+\mathrm{c}\right) \\
& =\frac{\sigma}{\epsilon_0}\left(\mathbf{a}+\frac{\mathbf{a}^2}{\mathbf{b}}\right) \\
\mathrm{V}_{\mathrm{C}} & =\frac{1}{4 \pi \in_0}\left(\frac{\mathrm{q}_{\mathrm{A}}}{\mathbf{a}}+\frac{\mathrm{q}_{\mathrm{B}}}{\mathrm{b}}+\frac{\mathrm{q}_{\mathrm{C}}}{\mathrm{c}}\right) \\
& =\frac{\sigma}{\epsilon_0}\left(\frac{\mathrm{a}^2-\mathbf{b}}{\mathrm{c}}+\mathrm{c}\right)=\frac{2 \sigma \mathbf{a}}{\epsilon_0}
\end{aligned}
$$
So, $V_C=V_A \neq V_B$
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