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Two concentric hollow spherical shells have radii $r$ and $R(R \gg r)$. A charge $Q$ is distributed on them such that the surface charge densities are equal. The electric potential at the centre is
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Verified Answer
The correct answer is:
$\frac{Q(R+r)}{4 \pi \varepsilon_0\left(R^2+r^2\right)}$
We know that, the surface charge densities is
$$
\sigma=\frac{Q}{4 \pi\left(r^2+R^2\right)}
$$
The electric potential at the centre
$$
\begin{aligned}
V & =\frac{1}{4 \pi \varepsilon_0}\left[\frac{\sigma \times 4 \pi r^2}{r}+\frac{\sigma \times 4 \pi R^2}{R}\right] \\
& =\frac{1}{4 \pi \varepsilon_0} \times \sigma \times 4 \pi\left[\frac{r^2}{r}+\frac{R^2}{R}\right] \\
& =\frac{\sigma}{\varepsilon_0}(r+R)
\end{aligned}
$$
and $\quad V=\frac{Q}{\varepsilon_0 \cdot 4 \pi\left(r^2+R^2\right)} \cdot(r+R)$
$$
\Rightarrow \quad V=\frac{Q}{4 \pi \varepsilon_0} \cdot \frac{(r+R)}{\left(r^2+R^2\right)}
$$
$$
\sigma=\frac{Q}{4 \pi\left(r^2+R^2\right)}
$$
The electric potential at the centre
$$
\begin{aligned}
V & =\frac{1}{4 \pi \varepsilon_0}\left[\frac{\sigma \times 4 \pi r^2}{r}+\frac{\sigma \times 4 \pi R^2}{R}\right] \\
& =\frac{1}{4 \pi \varepsilon_0} \times \sigma \times 4 \pi\left[\frac{r^2}{r}+\frac{R^2}{R}\right] \\
& =\frac{\sigma}{\varepsilon_0}(r+R)
\end{aligned}
$$
and $\quad V=\frac{Q}{\varepsilon_0 \cdot 4 \pi\left(r^2+R^2\right)} \cdot(r+R)$
$$
\Rightarrow \quad V=\frac{Q}{4 \pi \varepsilon_0} \cdot \frac{(r+R)}{\left(r^2+R^2\right)}
$$
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