If on the concentric hollow spheres of radii | vedclass

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Question

${{\rm{q}}_1} + {{\rm{q}}_2} = {\rm{Q}}$ and $ = \frac{{{{\rm{q}}_1}}}{{4\pi {{\rm{r}}^2}}} = \frac{{{{\rm{q}}_2}}}{{4\pi {{\rm{R}}^2}}}$

$\Ri

Vedclass · Accepted Answer

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

Vedclass · Answer

${{m{q}}_1} + {{m{q}}_2} = {m{Q}}$ and $ = \frac{{{{m{q}}_1}}}{{4\pi {{m{r}}^2}}} = \frac{{{{m{q}}_2}}}{{4\pi {{m{R}}^2}}}$

$\Rightarrow \frac{q_{1}}{q_{2}}=\frac{r^{2}}{R^{2}}$

$\Rightarrow \mathrm{q}_{1}=\frac{\mathrm{Qr}^{2}}{\mathrm{R}^{2}+\mathrm{r}^{2}} $ and $\mathrm{q}_{2}=\frac{\mathrm{QR}^{2}}{\mathrm{R}^{2}+\mathrm{r}^{2}}$

$\mathrm{V}_{	ext {Common }}=\frac{1}{4 \pi \varepsilon_{0}}\left[\frac{\mathrm{q}_{1}}{\mathrm{r}}+\frac{\mathrm{q}_{2}}{\mathrm{R}}ight]$

$\mathrm{V}_{	ext {Common }}=\frac{1}{4 \pi \varepsilon_{0}}\left[\frac{\mathrm{Qr}}{\mathrm{R}^{2}+\mathrm{r}^{2}}+\frac{\mathrm{QR}}{\mathrm{R}^{2}+\mathrm{r}^{2}}ight]=\frac{\mathrm{Q}(\mathrm{R}+\mathrm{r})}{4 \pi \varepsilon_{0}\left(\mathrm{R}^{2}+\mathrm{r}^{2}ight)}$

If on the concentric hollow spheres of radii $r$ and $R( > r)$ the charge $Q$ is distributed such that their surface densities are same then the potential at their common centre is

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