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In the following diagram the conducting shells are concentric. The amount of charge that flows through $\small s$ after closing it.

Option: 1
$\frac{Q}{2}$

Option: 2
$\frac{Q}{4}$

Option: 3
$\frac{Q}{3}$

Option: 4
$\frac{Q}{5}$

Answers (1)

best_answer

The charges on the shells are as shown in figures,

$\begin{aligned} & q_2-q_1=-2 Q&-(1) \\ & q_1+\left(-q_2\right)+q_3=Q+3 Q&-(2) \end{aligned}$

$\text { Also } \quad V_A=V_C$

$\text { or } \begin{aligned} & \frac{1}{4 \pi \epsilon_0}\left[\frac{q_1}{R}-\frac{q_1}{2 R}+\frac{q_2}{2 R}-\frac{q_2}{4 R}+\frac{q_3}{4 R}\right] \\ = & \frac{1}{4 \pi \epsilon_0_11}\left[\frac{q_1-q_2+q_2-q_2+q_3}{4 R}\right] \end{aligned}$

$q_1-\frac{q_2}{2}+\frac{q_2}{2}-\frac{q_2}{4}+\frac{q_3}{4}=\frac{q_3}{4}$ $-(3)$

$Solving\ above\ equations\ we\ get - \\ q_1=\frac{2 Q}{3}$

So that change flows through $s$ is $Q-\frac{2 Q}{3}$

$=\frac{Q}{3}$

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Pankaj

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