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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
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Verified Answer
The correct answer is:
$\mathrm{V}_{\mathrm{C}}=\mathrm{V}_{\mathrm{A}} \neq \mathrm{V}_{\mathrm{B}}$
Here,
$$
\begin{aligned}
& \mathrm{V}_{\mathrm{A}}=\frac{1}{4 \pi \varepsilon_0} \frac{\sigma 4 \pi \mathrm{a}^2}{\mathrm{a}}-\frac{1}{4 \pi \varepsilon_0} \frac{\sigma 4 \pi \mathrm{b}^2}{\mathrm{~b}} \\
& +\frac{1}{4 \pi \varepsilon_0} \cdot \frac{\sigma 4 \pi c^2}{c} \\
& =\frac{\sigma}{\varepsilon_0}(a-b+c)=\frac{\sigma}{\varepsilon_0}(2 a) \quad(\because c=a+b) \\
&
\end{aligned}
$$
$$
\begin{aligned}
& \mathrm{V}_{\mathrm{B}}=\frac{1}{4 \pi \varepsilon_0} \cdot \frac{\sigma 4 \pi \mathrm{a}^2}{\mathrm{a}}-\frac{1}{4 \pi \varepsilon_0} \frac{\sigma 4 \pi \mathrm{b}^2}{\mathrm{~b}} \\
&+\frac{1}{4 \pi \varepsilon_0} \cdot \frac{\sigma 4 \pi \mathrm{c}^2}{\mathrm{c}}
\end{aligned}
$$
$$
\begin{aligned}
& =\frac{\sigma}{\varepsilon_0}\left(\frac{a^2}{c}-b+c\right)=\frac{\sigma}{\varepsilon_0}(2 a)(\because c=a+b) \\
& \text { and } V_C=\frac{1}{4 \pi \varepsilon_0} \cdot \frac{\sigma 4 \pi a^2}{c}-\frac{1}{4 \pi \varepsilon_0} \frac{\sigma 4 \pi b^2}{c} \\
& +\frac{1}{4 \pi \varepsilon_0} \cdot \frac{\sigma 4 \pi c^2}{c} \\
& =\frac{\sigma}{\varepsilon_0}\left(\frac{a^2}{c}-\frac{b^2}{c}+c\right)=\frac{\sigma}{\varepsilon_0}(2 a)(\because c=a+b) \\
&
\end{aligned}
$$
Hence, $\mathrm{V}_{\mathrm{A}}=\mathrm{V}_{\mathrm{C}} \neq \mathrm{V}_{\mathrm{B}}$
$$
\begin{aligned}
& \mathrm{V}_{\mathrm{A}}=\frac{1}{4 \pi \varepsilon_0} \frac{\sigma 4 \pi \mathrm{a}^2}{\mathrm{a}}-\frac{1}{4 \pi \varepsilon_0} \frac{\sigma 4 \pi \mathrm{b}^2}{\mathrm{~b}} \\
& +\frac{1}{4 \pi \varepsilon_0} \cdot \frac{\sigma 4 \pi c^2}{c} \\
& =\frac{\sigma}{\varepsilon_0}(a-b+c)=\frac{\sigma}{\varepsilon_0}(2 a) \quad(\because c=a+b) \\
&
\end{aligned}
$$
$$
\begin{aligned}
& \mathrm{V}_{\mathrm{B}}=\frac{1}{4 \pi \varepsilon_0} \cdot \frac{\sigma 4 \pi \mathrm{a}^2}{\mathrm{a}}-\frac{1}{4 \pi \varepsilon_0} \frac{\sigma 4 \pi \mathrm{b}^2}{\mathrm{~b}} \\
&+\frac{1}{4 \pi \varepsilon_0} \cdot \frac{\sigma 4 \pi \mathrm{c}^2}{\mathrm{c}}
\end{aligned}
$$
$$
\begin{aligned}
& =\frac{\sigma}{\varepsilon_0}\left(\frac{a^2}{c}-b+c\right)=\frac{\sigma}{\varepsilon_0}(2 a)(\because c=a+b) \\
& \text { and } V_C=\frac{1}{4 \pi \varepsilon_0} \cdot \frac{\sigma 4 \pi a^2}{c}-\frac{1}{4 \pi \varepsilon_0} \frac{\sigma 4 \pi b^2}{c} \\
& +\frac{1}{4 \pi \varepsilon_0} \cdot \frac{\sigma 4 \pi c^2}{c} \\
& =\frac{\sigma}{\varepsilon_0}\left(\frac{a^2}{c}-\frac{b^2}{c}+c\right)=\frac{\sigma}{\varepsilon_0}(2 a)(\because c=a+b) \\
&
\end{aligned}
$$
Hence, $\mathrm{V}_{\mathrm{A}}=\mathrm{V}_{\mathrm{C}} \neq \mathrm{V}_{\mathrm{B}}$
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