A uniformly charged solid sphere of radius R | vedclass

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Question

We know, $\mathrm{V}_{0}=\frac{\mathrm{Kq}}{\mathrm{R}}=\mathrm{V\,surface}$

Now, $\mathrm{V}_{\mathrm{i}}=\frac{\mathrm{Kq}}{2 \mathrm{R}^{3}}\lef

Vedclass · Accepted Answer

$2R < R_4$

Vedclass · Answer

We know, $\mathrm{V}_{0}=\frac{\mathrm{Kq}}{\mathrm{R}}=\mathrm{V\,surface}$

Now, $\mathrm{V}_{\mathrm{i}}=\frac{\mathrm{Kq}}{2 \mathrm{R}^{3}}\left(3 \mathrm{R}^{2}-\mathrm{r}^{2}\right) \quad[\mathrm{For}\, \mathrm{r}<\mathrm{R}]$

At the centre of sphare $r=0 .$

Here $\mathrm{V}=\frac{3}{2} \mathrm{V}_{0}$

Now, $\frac{5}{4} \frac{\mathrm{Kq}}{\mathrm{R}}=\frac{\mathrm{Kq}}{2 \mathrm{R}^{3}}\left(3 \mathrm{R}^{2}-\mathrm{r}^{2}\right)$

$\mathrm{R}_{2}=\frac{\mathrm{R}}{\sqrt{2}}$

$\frac{3}{4} \frac{\mathrm{Kq}}{\mathrm{R}}=\frac{\mathrm{Kq}}{\mathrm{R}^{3}}$

$\frac{1}{4} \frac{\mathrm{Kq}}{\mathrm{R}}=\frac{\mathrm{Kq}}{\mathrm{R}_{4}}$

$\mathrm{R}_{4}=4 \mathrm{R}$

Also, $\mathrm{R}_{1}=0$ and $\mathrm{R}_{2}<\left(\mathrm{R}_{4}-\mathrm{R}_{3}\right)$

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