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  •  A uniformly charged solid sphere of radius R has potential V0 (measured with respect to <img alt="

 A uniformly charged solid sphere of radius R has potential V0 (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}  has radius R1 , R2  , R3  .and  R4  respectively. Then

 

 

Option: 1

R_{1}= 0 \: and\: R_{2}> (R_{4}-R_{3})


Option: 2

R_{1}\neq 0 \: and\: (R_{2}-R_{1}) > (R_{4}-R_{3})


Option: 3

R_{1}= 0 \: and\: R_{2}< (R_{4}-R_{3})


Option: 4

2R> R_{4}


Answers (1)

best_answer

V(R)=\frac{k Q}{R}=V_{0}$ \\ We use results derived for potential.\\ $V(r)=\frac{k Q}{r}$\\ for $r>R$ \\ $V(r)=\frac{k Q}{2 R^{3}}\left(3 R^{2}-r^{2}\right)$\\ $V(r=0)=\frac{3 k Q}{2 R}=\frac{3 V_{0}}{2}$\\ Hence, $V\left(R_{1}\right)$ matches with $V$ at center. $\therefore R_{1}=0

At \ \ R_{2}$ \\ $V\left(R_{2}\right)=\frac{5 V_{0}}{4}=\frac{k Q}{2 R^{3}}\left(3 R^{2}-R_{2}^{2}\right)$ \\ $\Rightarrow \frac{5 V_{0}}{4}=\frac{1}{2}\left(\frac{k Q}{R}\right)\left(3-\frac{R_{2}^{2}}{R^{2}}\right)\\ \Rightarrow \frac{1}{2} V_{0}\left(3-\frac{R_{2}^{2}}{R^{2}}\right)$ $\Rightarrow \frac{5}{2}=3-\frac{R_{2}^{2}}{R^{2}}$ $\Rightarrow \frac{1}{2}=\frac{R_{2}^{2}}{R^{2}}$ $\Rightarrow R_{2}=\frac{R}{\sqrt{2}}

For \ \ R_{3}$ \\ $V\left(R_{3}\right)=\frac{3 V_{0}}{4}=\frac{k Q}{R_{3}}$ $\Rightarrow \frac{3 V_{0}}{4}=V_{0} \frac{R}{R_{3}}$ $\Rightarrow R_{3}=\frac{4 R}{3}

For \ \ R_{4}$ \\ $V\left(R_{4}\right)=\frac{V_{0}}{4}=\frac{k Q}{R_{4}}$ $\Rightarrow R_{4}=4 R

\begin{aligned} &\text { Thus, } R_{4}-R_{3}=4 R-\frac{4 R}{3}=\frac{8 R}{3}\\ &R_{2}=\frac{R}{\sqrt{2}}<\frac{8 R}{3}=R_{4}-R_{3} \end{aligned}

\text { Also, } 2 R<4 R=R_{4}

 

Posted by

Devendra Khairwa

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