Question

Three identical, parallel conducting plates $A,B$ and $C$ are placed as shown. Switches $S_1$ and $S_2$ are open and can connect $A$ and $C$ to earth when closed. $+Q$ charge is given to $B$. Find the magnitude of charge flown through $S_1$ and $S_2$ when both switches are closed.

• A. $Q_{S_1}=\frac{Q}{3},Q_{S_2}=\frac{2Q}{3}$
  
• B. $Q_{S_1}=\frac{Q}{2},Q_{S_2}=\frac{2Q}{3}$
  
• C. $Q_{S_1}=\frac{2Q}{3},Q_{S_2}=\frac{Q}{3}$
  
• D. $Q_{S_1}=\frac{Q}{3},Q_{S_2}=\frac{Q}{3}$
  

Answer 1

The correct option is A $Q_{S_1}=\frac{Q}{3},Q_{S_2}=\frac{2Q}{3}$


When both $S_1$ and $S_2$ are closed, then the charge distribution on the surfaces of the plates are as shown in figure.


Plates $A$ and $B$ form parallel plate capacitor of capacitance $C_1$ and plates $B$ and $C$ form parallel plate capacitor of capacitance $C_2$.

So,
$C_1=\frac{{\epsilon }_{0}A}{2d}=\frac{C}{2}$ (Say $C=\frac{{\epsilon }_{0}A}{d})$

$C_2=\frac{{\epsilon }_{0}A}{d}=C$

$\therefore \frac{C_1}{C_2}=\frac{1}{2}....\left(1\right)$

Also, we know that
$\Rightarrow Q\propto C$ [When $V=$ Constant]

As plates $C_1$ and $C_2$ are at same voltage,

$\frac{C_1}{C_2}=\frac{Q_1}{Q_2}$

$\Rightarrow \frac{Q_1}{Q_2}=\frac{1}{2}$ [from (1)]

$\Rightarrow \frac{Q_1}{Q-Q_1}=\frac{1}{2}$ [$\because Q_1+Q_2=Q$ ]

$\Rightarrow 2Q_1=Q-Q_1$

$\Rightarrow Q_1=\frac{Q}{3}$

$\Rightarrow Q_2=\frac{2Q}{3}$

Hence, option (a) is the correct answer.