Question

Two concentric hollow spherical shells have radii $r$ and $R \left(R\,>>\,r\right)$ . A charge $Q$ is distributed on them such that the surface charge densities are equal. The electric potential at the centre is

• A. $\frac{Q(R+r)}{4 \pi \varepsilon_{0}\left(R^{2}+r^{2}\right)}$
• B. $\frac{Q\left(R^{2}+r^{2}\right)}{4 \pi \varepsilon_{0}(R+r)}$
• C. $\frac{Q}{\left(R+r\right)}$
• D. zero

Answer 1

Answer: (A)

$\frac{Q(R+r)}{4 \pi \varepsilon_{0}\left(R^{2}+r^{2}\right)}$

Answer 2

We know that, the surface charge densities is $\sigma=\frac{Q}{4 \pi\left(r^{2}+R^{2}\right)}\,\,\,...(i)$ The electric potential at the centre $V =\frac{1}{4 \pi \varepsilon_{0}}\left[\frac{\sigma \times 4 \pi r^{2}}{r}+\frac{\sigma \times 4 \pi R^{2}}{R}\right]$ $=\frac{1}{4 \pi \varepsilon_{0}} \times \sigma \times 4 \pi\left[\frac{r^{2}}{r}+\frac{R^{2}}{R}\right]$ $=\frac{\sigma}{\varepsilon_{0}}(r+R)$ and $V=\frac{Q}{\varepsilon_{0} \cdot 4 \pi\left(r^{2}+R^{2}\right)} \cdot(r+R)$ $\Rightarrow V =\frac{Q}{4 \pi \varepsilon_{0}} \cdot \frac{(r+R)}{\left(r^{2}+R^{2}\right)}$