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A uniformly charged solid sphere of radius $R$ has potential $V_0$ (measured with respect to $\infty$) on its surface. For this sphere the equipotential surfaces with potentials $\frac{{3{V_0}}}{2},\;\frac{{5{V_0}}}{4},\;\frac{{3{V_0}}}{4}$ and $\frac{{{V_0}}}{4}$ have rasius $R_1,R_2,R_3$ and $R_4$ respectively. Then
$R_1$$ \ne 0$ and $(R_2-R_1) > (R_4-R_3)$
$R_1$ $ = 0$ and $R_2 < (R_4-R_3)$
$2R < R_4$
$R_1$ $ = 0$ and $ R_2 > (R_4-R_3)$
Solution
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 $\mathrm{R}_{2}<\left(\mathrm{R}_{4}-\mathrm{R}_{3}\right)$
