Question

A uniformly charged solid sphere of radius $R$ has potential $V _{0}\, 9$ 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 radius $R _{1}, R _{2}, R _{3}$ and $R _{4}$ respectively. Then

• A. $R_1 \ne 0$ and $(R_2-R_1) >(R_4-R_3)$
• B. $R_1 = 0$ and $R_2 >(R_4-R_3)$
• C. $2R < R_4$
• D. $R_1 = 0$ and $R_2 < (R_4-R_3)$

Answer 1

Answer: (C)

$2R < R_4$

Answer 2

$R_{1}=\frac{3 V_{0}}{2} ; R_{2}=\frac{5 V_{0}}{4} ; R_{3} \frac{3 V_{0}}{4} ; R_{4}=\frac{V_{0}}{4}$
$\therefore r< R \,\,\,\,\,\,V=\frac{K Q}{2 R^{3}}\left(3 R^{2}-r^{2}\right)$
$v =\frac{3 V _{0}}{2}, \,R _{1}=0$
$\frac{5 V_{0}}{4}=\frac{K Q}{2 R^{3}}\left(3 R^{2}-R_{2}^{2}\right)$
$\therefore R _{2}=\frac{ R }{\sqrt{2}}$
$r>R$
$\frac{3 V_{0}}{4}=\frac{K Q}{R_{3}}$
$R_{3}=\frac{4 K Q}{3 V_{0}}=\frac{K Q \times R}{3 \times K Q}=\frac{R}{3}$
$\frac{ V _{0}}{4}=\frac{ KQ }{ R _{4}}$
$\therefore R _{4}=\frac{4 KQ }{ V _{0}}=\frac{4 KQ }{ KQ } \times R =4 R$
On comparing we get
$(1) \&(2)$