Electrostatics Ques 41
- 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 radius $R_{1}, R_{2}, R_{3}$, and $R_{4}$ respectively. Then,
(2015 Main)
(a) $R_{1} \neq 0$ and $\left(R_{2}-R_{1}\right)>\left(R_{4}-R_{3}\right)$
(b) $R_{1}=0$ and $R_{2}>\left(R_{4}-R_{3}\right)$
(c) $2 R<R_{4}$
(d) $R_{1}=0$ and $R_{2}<\left(R_{4}-R_{3}\right)$
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Answer:
Correct Answer: 41.(b,c)
Solution:
- $V_{0}=$ potential on the surface $=\frac{K q}{R}$
where, $K=\frac{1}{4 \pi \varepsilon_{0}}$ and $q$ is total charge on sphere.
Potential at centre $=\frac{3}{2} \frac{K q}{R}=\frac{3}{2} V_{0}$
Hence, $R_{1}=0$
From centre to surface potential varies between $\frac{3}{2} V_{0}$ and $V_{0}$
From surface to infinity, it varies between $V_{0}$ and $0, \frac{5 V_{0}}{4}$ will
be potential at a point between centre and surface. At any point, at a distance $r(r \leq R)$ from centre potential is given by
$$ \begin{aligned} V & =\frac{K q}{R^{3}} \frac{3}{2} R^{2}-\frac{1}{2} r^{2} \\ & =\frac{V_{0}}{R^{2}} \frac{3}{2} R^{2}-\frac{1}{2} r^{2} \end{aligned} $$
Putting $V=\frac{5}{4} V_{0}$ and $r=R_{2}$ in this equation, we get
$$ R_{2}=\frac{R}{\sqrt{2}} $$
$\frac{3 V_{0}}{4}$ and $\frac{V_{0}}{4}$ are the potentials lying between $V_{0}$ and zero hence these potentials lie outside the sphere. At a distance $r(\geq R)$ from centre potential is given by $V=\frac{K q}{r}=\frac{V_{0} R}{r}$
Putting $V=\frac{3}{4} V_{0}$ and $r=R_{3}$ in this equation we get, $R_{3}=\frac{4}{3} R$
Further putting $V=\frac{V_{0}}{4}$ and $r=R_{4}$ in above equation,
we get $\quad R_{4}=4 R$
Thus, $R_{1}=0, R_{2}=\frac{R}{\sqrt{2}}, R_{3}=\frac{4 R}{3}$ and $R_{4}=4 R$ with these values, option (b) and (c) are correct.
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