Electrostatics 2 Question 18
19. 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,
(a) $R _1 \neq 0$ and $\left(R _2-R _1\right)>\left(R _4-R _3\right)$
(2015 Main)
(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: 19. (a) $V _A=\frac{\sigma}{\varepsilon _0}(a-b+c), V _B=\frac{\sigma}{\varepsilon _0} \frac{a^{2}}{b}-b+c$,
$$ V _C=\frac{\sigma}{\varepsilon _0} \frac{a^{2}}{c}-\frac{b^{2}}{c}+c \quad \text { (b) } a+b=c \quad \text { 20. } \frac{Q(R+r)}{4 \pi \varepsilon _0\left(R^{2}+r^{2}\right)} $$
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.
