Three concentric spherical shells have radii $a$, $b$, and $c$ ($a < b < c$) and have surface charge…
- $\mathrm{V}_{\mathrm{C}}=\mathrm{V}_{\mathrm{A}} \neq \mathrm{V}_{\mathrm{B}}$
- $\mathrm{V}_{\mathrm{C}}=\mathrm{V}_{\mathrm{B}} \neq \mathrm{V}_{\mathrm{A}}$
- $\mathrm{V}_{\mathrm{C}} \neq \mathrm{V}_{\mathrm{B}} \neq \mathrm{V}_{\mathrm{A}}$
- $\mathrm{V}_{\mathrm{C}}=\mathrm{V}_{\mathrm{B}}=\mathrm{V}_{\mathrm{A}}$
Solution
$\begin{aligned}
& \mathrm{V}_{\mathrm{B}}=\frac{1}{4 \pi \varepsilon_0} \cdot \frac{\sigma 4 \pi \mathrm{a}^2}{\mathrm{a}}-\frac{1}{4 \pi \varepsilon_0} \frac{\sigma 4 \pi \mathrm{b}^2}{\mathrm{~b}} \\
&+\frac{1}{4 \pi \varepsilon_0} \cdot \frac{\sigma 4 \pi \mathrm{c}^2}{\mathrm{c}}
\end{aligned}$
$\begin{aligned}
& =\frac{\sigma}{\varepsilon_0}\left(\frac{a^2}{c}-b+c\right)=\frac{\sigma}{\varepsilon_0}(2 a)(\because c=a+b) \\
& \text { and } V_C=\frac{1}{4 \pi \varepsilon_0} \cdot \frac{\sigma 4 \pi a^2}{c}-\frac{1}{4 \pi \varepsilon_0} \frac{\sigma 4 \pi b^2}{c} \\
& +\frac{1}{4 \pi \varepsilon_0} \cdot \frac{\sigma 4 \pi c^2}{c} \\
& =\frac{\sigma}{\varepsilon_0}\left(\frac{a^2}{c}-\frac{b^2}{c}+c\right)=\frac{\sigma}{\varepsilon_0}(2 a)(\because c=a+b) \\
&
\end{aligned}$
Hence, $\mathrm{V}_{\mathrm{A}}=\mathrm{V}_{\mathrm{C}} \neq \mathrm{V}_{\mathrm{B}}$Asked in: NEET 2009 (Screening)