JEE Main 2015 · 4 Apr · Shift 1 · Physics Q15

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A uniformly charged solid sphere of radius R has potential $V_0$ (measured with respect to ∞) on its surface. For this sphere the equipotential surfaces with potentials $\dfrac{3V_0}2$ , $\dfrac{5V_0}4$ , $\dfrac{3V_0}4$ and $\dfrac{V_0}4$ have radius $R_1$ , $R_2$ , $R_3$ and $R_4$ respectively. Then

R_1=0

and

R_2>(R_4-R_3)

R_1\ne0

and

(R_2-R_1)>(R_4-R_3)

R_1=0

and

R_2<(R_4-R_3)

2R<R_4

Answer and worked solution

Official answer: C

Inside: $V=\dfrac{V_0}2\left(3-\dfrac{r^2}{R^2}\right)$ . Outside: $V=\dfrac{V_0R}r$ .

$\frac32V_0$ is the centre: $R_1=0$ .
$\frac54V_0$ : $3-\frac{r^2}{R^2}=\frac52$ , so $R_2=\frac R{\sqrt2}$ .
$\frac34V_0$ : $R_3=\frac43R$ .
$\frac14V_0$ : $R_4=4R$ .

$R_4-R_3=\frac83R>R_2$ , so (3) holds. Also $R_4=4R>2R$ , so (4) holds. The key accepts either (3) or (4).

Topics:Electric potential Equipotential surfaces

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