Question

From given figure find charge flown through $A$ to $B$ when switch $S$ is closed.

• A. $\frac{-\epsilon C}{2}$
• B. $\frac{\epsilon C}{2}$
• C. $-\epsilon C$
• D. $\epsilon C$

Answer 1

The correct option is A $\frac{-\epsilon C}{2}$


Initially net charge on plate 2 and 3 was zero.
Let charge on $C$ be $q_1$ and on $2C$ be $q_2$. So net charge on plate 2 and 3 after closing switch is $q_2-q_1$
$C_{eq}=\frac{3C}{2}$
So $q_1+q_2=\frac{3C}{2}\epsilon$
Where $\frac{q_1}{C}=\frac{q_2}{2C}\Rightarrow q_1=\frac{q_2}{2}$
$q_2=\epsilon C$ & $q_1=\frac{\epsilon C}{2}$
$\Rightarrow$ So $q_2-q_1=\frac{\epsilon C}{2}$
Net charge $q_2-q_1$ is positive so charge flown from $A$ to $B$ is negative.