Electrostatics 7 Question 31
33. Two isolated metallic solid spheres of radii $R$ and $2 R$ are charged such that both of these have same charge density $\sigma$. The spheres are located far away from each other and connected by a thin conducting wire. Find the new charge density on the bigger sphere.
(1996, 3M)
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Solution:
- Let $q _1$ and $q _2$ be the charges on the two spheres before connecting them.
Then, $\quad q _1=\sigma\left(4 \pi R^{2}\right)$, and $q _2=\sigma(4 \pi)(2 R)^{2}=16 \sigma \pi R^{2}$
Therefore, total charge $(q)$ on both the spheres is
$$ q=q _1+q _2=20 \sigma \pi R^{2} $$
Now, after connecting, the charge is distributed in the ratio of their capacities, which in turn depends on the ratio of their $\operatorname{radii}\left(C=4 \pi \varepsilon _0 R\right)$
$$ \begin{array}{lll} \therefore & & \frac{q _1^{\prime}}{q _2^{\prime}}=\frac{R}{2 R}=\frac{1}{2} \\ & \therefore & q _1{ }^{\prime}=\frac{q}{3}=\frac{20}{3} \sigma \pi R^{2} \\ & \text { and } & q _2{ }^{\prime}=\frac{2 q}{3}=\frac{40}{3} \sigma \pi R^{2} \end{array} $$
Therefore, surface charge densities on the spheres are
$$ \begin{aligned} & \sigma _1=\frac{q _1{ }^{\prime}}{4 \pi R^{2}}=\frac{(20 / 3) \sigma \pi R^{2}}{4 \pi R^{2}}=\frac{5}{3} \sigma \\ & \sigma _2=\frac{q _2^{\prime}}{4 \pi(2 R)^{2}}=\frac{(40 / 3) \sigma \pi R^{2}}{16 \pi R^{2}}=\frac{5}{6} \sigma \end{aligned} $$
and
Hence, surface charge density on the bigger sphere is $\sigma _2$ i.e. $(5 / 6) \sigma$.
