Electrostatics Question 12
Question 12 - 30 January - Shift 1
Two isolated metallic solid spheres of radii $R$ and $2 R$ are charged such that both have same charge density $\sigma$. The spheres are then connected by a thin conducting wire. If the new charge density of the bigger sphere is $\sigma^{\prime}$. The ratio $\frac{\sigma^{\prime}}{\sigma}$ is :
(1) $\frac{9}{4}$
(2) $\frac{4}{3}$
(3) $\frac{5}{3}$
(4) $\frac{5}{6}$
Show Answer
Answer: (4)
Solution:
$ \begin{aligned} Q_1 & =\sigma(4 \pi R^{2}) \\ & =4 \pi R^{2} \sigma \end{aligned} $
$ \begin{aligned} Q_2 & =\sigma(4 \pi(2 R^{2})) \\ & =16 \pi R^{2} \sigma \end{aligned} $
$ \frac{Q_1}{4 \pi \varepsilon_0 R}=\frac{Q_2}{4 \pi \varepsilon_0(2 R)} $
$\therefore Q_2^{\prime}=2 Q_1^{\prime}$
$Q_1^{\prime}+Q_2^{\prime}=Q_1+Q_2$
$\therefore \frac{Q_2^{\prime}}{2}+Q_2^{\prime}=20 \pi R^{2} \sigma$
$\frac{3}{2} Q_2^{\prime}=20 \pi R^{2} \sigma$
$\therefore \frac{Q_2^{\prime}}{4 \pi(2 R)^{2}}=\frac{2}{3} \cdot \frac{20 \pi R^{2} \sigma}{16 \pi R^{2}}$
$\therefore \frac{\sigma^{\prime}}{\sigma}=\frac{5}{6}$
