- Home
- Standard 12
- Physics
A charge $Q$ is distributed over two concentric hollow spheres or radius $r$ and $R(> r)$ such that the surface densities are equal. The potential at the common centre is
$\frac{{Q\left( {{R^2} + {r^2}} \right)}}{{4\pi {\varepsilon _0}\left( {R + r} \right)}}$
$\frac{Q}{{R + r}}$
Zero
$\frac{{Q\left( {R + r} \right)}}{{4\pi {\varepsilon _0}\left( {{R^2} + {r^2}} \right)}}$
Solution
Since, the surface densities are equal, hence
${\frac{{{{\rm{q}}_1}}}{{4\pi {{\rm{r}}^2}}} = \frac{{{{\rm{q}}_2}}}{{4\pi {{\rm{R}}^2}}}}$ (where ${{\rm{ }}{{\rm{q}}_1} + {{\rm{q}}_2} = {\rm{Q}}}$)
or $\frac{\mathrm{q}_{1}}{\mathrm{r}^{2}}=\frac{\mathrm{q}_{2}}{\mathrm{R}^{2}}=\frac{\mathrm{q}_{1}+\mathrm{q}_{2}}{\mathrm{r}^{2}+\mathrm{R}^{2}}=\frac{\mathrm{Q}}{\mathrm{r}^{2}+\mathrm{R}^{2}}$
$\therefore $ $\mathrm{q}_{1}=\frac{\mathrm{Q}}{\mathrm{r}^{2}+\mathrm{R}^{2}} \times \mathrm{r}^{2}$
and $\mathrm{q}_{2}=\frac{\mathrm{Q}}{\mathrm{r}^{2}+\mathrm{R}^{2}} \times \mathrm{R}^{2}$
So, potential at the common centre,
${\mathrm{V}=\frac{\mathrm{q}_{1}}{4 \pi \varepsilon_{0} \mathrm{r}}+\frac{\mathrm{q}_{2}}{4 \pi \varepsilon_{0} \mathrm{R}}=\frac{1}{4 \pi \varepsilon_{0}}\left(\frac{\mathrm{q}_{1}}{\mathrm{r}}+\frac{\mathrm{q}_{2}}{\mathrm{R}}\right)}$
${=\frac{1}{4 \pi \varepsilon_{0}}\left[\frac{\mathrm{Q}}{\mathrm{R}^{2}+\mathrm{r}^{2}} \times \frac{\mathrm{r}^{2}}{\mathrm{r}}+\frac{\mathrm{Q}}{\mathrm{R}^{2}+\mathrm{r}^{2}} \times \frac{\mathrm{R}^{2}}{\mathrm{R}}\right]} $
${=\frac{1}{4 \pi \varepsilon_{0}} \frac{\mathrm{Q}(\mathrm{R}+\mathrm{r})}{\left(\mathrm{R}^{2}+\mathrm{r}^{2}\right)}}$
