Gujarati
Hindi
1. Electric Charges and Fields
normal

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

A

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

B

$\frac{{QR}}{{R + r}}$

C

Zero

D

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

Solution

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

$\Rightarrow \frac{q_{1}}{q_{2}}=\frac{r^{2}}{R^{2}}$

$\Rightarrow \mathrm{q}_{1}=\frac{\mathrm{Qr}^{2}}{\mathrm{R}^{2}+\mathrm{r}^{2}} $ and $\mathrm{q}_{2}=\frac{\mathrm{QR}^{2}}{\mathrm{R}^{2}+\mathrm{r}^{2}}$

$\mathrm{V}_{\text {Common }}=\frac{1}{4 \pi \varepsilon_{0}}\left[\frac{\mathrm{q}_{1}}{\mathrm{r}}+\frac{\mathrm{q}_{2}}{\mathrm{R}}\right]$

$\mathrm{V}_{\text {Common }}=\frac{1}{4 \pi \varepsilon_{0}}\left[\frac{\mathrm{Qr}}{\mathrm{R}^{2}+\mathrm{r}^{2}}+\frac{\mathrm{QR}}{\mathrm{R}^{2}+\mathrm{r}^{2}}\right]=\frac{\mathrm{Q}(\mathrm{R}+\mathrm{r})}{4 \pi \varepsilon_{0}\left(\mathrm{R}^{2}+\mathrm{r}^{2}\right)}$

Standard 12
Physics

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