Electrostatics Ques 183
- 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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Answer:
Correct Answer: 183.$\frac{5}{6}\sigma$
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. $(5 / 6) \sigma$.
BETA
