A charge Q is distributed over two concentric | vedclass

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Question

Since, the surface densities are equal, hence

${\frac{{{{\rm{q}}_1}}}{{4\pi {{\rm{r}}^2}}} = \frac{{{{\rm{q}}_2}}}{{4\pi {{\rm{R}}^2}}}}$  &nb

Vedclass · Accepted Answer

$\frac{{Q\left( {R + r} \right)}}{{4\pi {\varepsilon _0}\left( {{R^2} + {r^2}} \right)}}$

Vedclass · Answer

Since, the surface densities are equal, hence

${\frac{{{{m{q}}_1}}}{{4\pi {{m{r}}^2}}} = \frac{{{{m{q}}_2}}}{{4\pi {{m{R}}^2}}}}$            (where ${{m{ }}{{m{q}}_1} + {{m{q}}_2} = {m{Q}}}$)

or  $\frac{\mathrm{q}_{1}}{\mathrm{r}^{2}}=\frac{\mathrm{q}_{2}}{\mathrm{R}^{2}}=\frac{\mathrm{q}_{1}+\mathrm{q}_{2}}{\mathrm{r}^{2}+\mathrm{R}^{2}}=\frac{\mathrm{Q}}{\mathrm{r}^{2}+\mathrm{R}^{2}}$

$	herefore $  $\mathrm{q}_{1}=\frac{\mathrm{Q}}{\mathrm{r}^{2}+\mathrm{R}^{2}} 	imes \mathrm{r}^{2}$

and $\mathrm{q}_{2}=\frac{\mathrm{Q}}{\mathrm{r}^{2}+\mathrm{R}^{2}} 	imes \mathrm{R}^{2}$

So, potential at the common centre,

${\mathrm{V}=\frac{\mathrm{q}_{1}}{4 \pi \varepsilon_{0} \mathrm{r}}+\frac{\mathrm{q}_{2}}{4 \pi \varepsilon_{0} \mathrm{R}}=\frac{1}{4 \pi \varepsilon_{0}}\left(\frac{\mathrm{q}_{1}}{\mathrm{r}}+\frac{\mathrm{q}_{2}}{\mathrm{R}}ight)}$

${=\frac{1}{4 \pi \varepsilon_{0}}\left[\frac{\mathrm{Q}}{\mathrm{R}^{2}+\mathrm{r}^{2}} 	imes \frac{\mathrm{r}^{2}}{\mathrm{r}}+\frac{\mathrm{Q}}{\mathrm{R}^{2}+\mathrm{r}^{2}} 	imes \frac{\mathrm{R}^{2}}{\mathrm{R}}ight]} $

${=\frac{1}{4 \pi \varepsilon_{0}} \frac{\mathrm{Q}(\mathrm{R}+\mathrm{r})}{\left(\mathrm{R}^{2}+\mathrm{r}^{2}ight)}}$

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

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