# Solve this following

Question:

A charge $Q$ is distributed over two concentric conducting thin spherical shells radii $r$ and $R$ $(\mathrm{R}>\mathrm{r})$. If the surface charge densities on the two shells are equal, the electric potential at the common centre is :

1. $\frac{1}{4 \pi \varepsilon_{0}} \frac{(R+2 r) Q}{2\left(R^{2}+r^{2}\right)}$

2. $\frac{1}{4 \pi \varepsilon_{0}} \frac{(\mathrm{R}+\mathrm{r})}{2\left(\mathrm{R}^{2}+\mathrm{r}^{2}\right)} \mathrm{Q}$

3. $\frac{1}{4 \pi \varepsilon_{0}} \frac{(\mathrm{R}+\mathrm{r})}{\left(\mathrm{R}^{2}+\mathrm{r}^{2}\right)} \mathrm{Q}$

4. $\frac{1}{4 \pi \varepsilon_{0}} \frac{(2 \mathrm{R}+\mathrm{r})}{\left(\mathrm{R}^{2}+\mathrm{r}^{2}\right)} \mathrm{Q}$

Correct Option: , 3

Solution:

Let the charges on inner and outer spheres are $Q_{1}$ and $Q_{2}$.

Since charge density ' $\sigma$ ' is same for both spheres, so

$\sigma=\frac{Q_{1}}{4 \pi r^{2}}=\frac{Q_{2}}{4 \pi R^{2}} \Rightarrow \frac{Q_{1}}{Q_{2}}=\frac{r^{2}}{R^{2}}$

$\mathrm{Q}_{1}+\mathrm{Q}_{2}=\mathrm{Q} \Rightarrow \frac{\mathrm{Q}_{2} \mathrm{r}^{2}}{\mathrm{R}^{2}}+\mathrm{Q}_{2}=\mathrm{Q}$

$\Rightarrow Q_{2}=\frac{Q R^{2}}{\left(r^{2}+R^{2}\right)}$

$Q_{1}=\frac{r^{2}}{R^{2}} \cdot \frac{Q R^{2}}{\left(R^{2}+r^{2}\right)}=\frac{Q r^{2}}{\left(R^{2}+r^{2}\right)}$

Potential at centre ' $\mathrm{O}^{\prime}=\frac{\mathrm{kQ}_{1}}{\mathrm{r}}+\frac{\mathrm{kQ}_{2}}{\mathrm{R}}$

$=\mathrm{k}\left[\frac{\mathrm{Qr}^{2}}{\mathrm{r}\left(\mathrm{R}^{2}+\mathrm{r}^{2}\right)}+\frac{\mathrm{Q} \mathrm{R}^{2}}{\mathrm{R}\left(\mathrm{R}^{2}+\mathrm{r}^{2}\right)}\right]$

$=\frac{\mathrm{kQ}(\mathrm{r}+\mathrm{R})}{\left(\mathrm{R}^{2}+\mathrm{r}^{2}\right)}=\frac{1}{4 \pi \epsilon_{0}} \frac{(\mathrm{R}+\mathrm{r})}{\left(\mathrm{R}^{2}+\mathrm{r}^{2}\right)} \mathrm{Q}$