A uniformly charged solid sphere of radius R has potential V₀… | JEE Main 2015 ((one Or More Correct))

Q105JEE Main 2015 · (One or more correct)Difficult

Electric Potential & Equipotentials

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 $\frac{3V_0}{2}, \frac{5V_0}{4}, \frac{3V_0}{4}$ and $\frac{V_0}{4}$ have radius $R_1, R_2, R_3$ and $R_4$ respectively. Then

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Answer: (a)

Inside: $V = \dfrac{V_0(3R^2 - r^2)}{2R^2}$ ; outside: $V = \dfrac{V_0R}{r}$ .
$\dfrac{3V_0}{2}$ is the centre value, so $R_1 = 0$ .
$\dfrac{5V_0}{4}$ : $3 - \dfrac{r^2}{R^2} = \dfrac{5}{2} \Rightarrow R_2 = \dfrac{R}{\sqrt2} \approx 0.71R$ .
$\dfrac{3V_0}{4}$ : $R_3 = \dfrac{4R}{3}$ ; $\dfrac{V_0}{4}$ : $R_4 = 4R$ , so $R_4 - R_3 = \dfrac{8R}{3} > R_2$ .
So the answer is (a) $R_1 = 0$ and $R_2 < (R_4 - R_3)$ .

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