Electrostatics 7 Question 30
32. A conducting sphere $S_{1}$ of radius $r$ is attached to an insulating handle. Another conducting sphere $S_{2}$ of radius $R$ is mounted on an insulating stand. $S_{2}$ is initially uncharged. $S_{1}$ is given a charge $Q$, brought into contact with $S_{2}$ and removed. $S_{1}$ is recharged such that the charge on it is again $Q$ and it is again brought into contact with $S_{2}$ and removed. This procedure is repeated $n$ times.
(1998, 8M)
(a) Find the electrostatic energy of $S_{2}$ after $n$ such contacts with $S_{1}$.
(b) What is the limiting value of this energy as $n \rightarrow \infty$ ?
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Answer:
Correct Answer: 32. (a) $U_n=\frac{q_n^2}{8 \pi \varepsilon_0 R}$
(b) $U_{\infty}=\frac{Q^2 R}{8 \pi \varepsilon_0 r^2} \quad$ Here, $q_n=\frac{Q R}{r}\left[1-\left(\frac{R}{R+r}\right)^n\right]$
Solution:
- Capacities of conducting spheres are in the ratio of their radii. Let $C_{1}$ and $C_{2}$ be the capacities of $S_{1}$ and $S_{2}$, then
$$ \frac{C_{2}}{C_{1}}=\frac{R}{r} $$
(a) Charges are distributed in the ratio of their capacities. Let in the first contact, charge acquired by $S_{2}$ is $q_{1}$. Therefore, charge on $S_{1}$ will be $Q-q_{1}$. Say it is $q_{1}^{\prime}$
$$ \begin{aligned} \therefore \quad \frac{q_{1}}{q_{1}{ }^{\prime}} & =\frac{q_{1}}{Q-q_{1}} \\ & =\frac{C_{2}}{C_{1}}=\frac{R}{r} \\ \therefore \quad q_{1} & =Q \frac{R}{R+r} \end{aligned} $$
In the second contact, $S_{1}$ again acquires the same charge $Q$.
Therefore, total charge in $S_{1}$ and $S_{2}$ will be
$$ Q+q_{1}=Q \quad 1+\frac{R}{R+r} $$
This charge is again distributed in the same ratio. Therefore, charge on $S_{2}$ in second contact,
$$ \begin{aligned} q_{2} & =Q 1+\frac{R}{R+r} \frac{R}{R+r} \\ & =Q \frac{R}{R+r}+\frac{R}{R+r}^{2} \end{aligned} $$
Similarly, $q_{3}=Q \frac{R}{R+r}+\frac{R}{R+r}^{2}+\frac{R}{R+r}^{3}$
and $\quad q_{n}=Q \frac{R}{R+r}+\frac{R}{R+r}^{2}+\ldots+\frac{R}{R+r}^{n}$ or
$$ q_{n}=Q \frac{R}{r} 1-\frac{R}{R+r}^{n} \quad S_{n}=\frac{a\left(1-r^{n}\right)}{(1-r)} $$
Therefore, electrostatic energy of $S_{2}$ after $n$ such contacts
$$ =\frac{q_{n}^{2}}{2\left(4 \pi \varepsilon_{0} R\right)} \text { or } U_{n}=\frac{q_{n}^{2}}{8 \pi \varepsilon_{0} R} $$
where, $q_{n}$ can be written from Eq. (ii).
(b) As $n \rightarrow \infty \quad q_{\infty}=Q \frac{R}{r}$
$$ \therefore U_{\infty}=\frac{q_{\infty}^{2}}{2 C}=\frac{Q^{2} R^{2} / r^{2}}{8 \pi \varepsilon_{0} R} $$
or $\quad U_{\infty}=\frac{Q^{2} R}{8 \pi \varepsilon_{0} r^{2}}$
