Three concentric spherical shells have radii $a$, $b$, and $c$ ($a < b < c$) and have surface charge…

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
  1. $\mathrm{V}_{\mathrm{C}}=\mathrm{V}_{\mathrm{A}} \neq \mathrm{V}_{\mathrm{B}}$
  2. $\mathrm{V}_{\mathrm{C}}=\mathrm{V}_{\mathrm{B}} \neq \mathrm{V}_{\mathrm{A}}$
  3. $\mathrm{V}_{\mathrm{C}} \neq \mathrm{V}_{\mathrm{B}} \neq \mathrm{V}_{\mathrm{A}}$
  4. $\mathrm{V}_{\mathrm{C}}=\mathrm{V}_{\mathrm{B}}=\mathrm{V}_{\mathrm{A}}$

Solution

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, $\mathrm{V}_{\mathrm{A}}=\mathrm{V}_{\mathrm{C}} \neq \mathrm{V}_{\mathrm{B}}$

Asked in: NEET 2009 (Screening)

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