Question

A parallel-plate capacitor is connected to a cell. Its positive plate $A$ and its negative plate $B$ have charges $+Q$ and $-Q$ respectively. A third plate $C$, identical to $A$ and $B$, with charge $+Q$, is now introduced midway between $A$ and $B$, parallel to them. Which of the following are correct?

• A. The charge on the inner face of $B$ is now $-\frac{3Q}{2}$
• B. There is no change in the potential difference between $A$ and $B$
• C. The potential difference between $A$ and $C$ is one-third of the potential difference between $B$ and $C$
• D. The charge on the inner face of $A$ is now $Q/2$

Answer 1

Answer: (A), (B), (C), (D)

The correct options are
A The charge on the inner face of $B$ is now $-\frac{3Q}{2}$
B There is no change in the potential difference between $A$ and $B$
C The potential difference between $A$ and $C$ is one-third of the potential difference between $B$ and $C$
D The charge on the inner face of $A$ is now $Q/2$

Let the charges will be distributed among the surfaces of plates according to figure below.
As the field inside the every plate is zero, so
for plate A: $\frac{1}{2A{ϵ}_0}(Q-q_1-q_1-Q+q_2-q_2+Q+q_3-q_3)=0$
$\Rightarrow q_1=\frac{Q}{2}$
for C: $\frac{1}{2A{ϵ}_0}(Q-q_1+q_1+Q-q_2-q_2+Q+q_3-q_3)=0$
$\Rightarrow q_2=\frac{3Q}{2}$
for B: $\frac{1}{2A{ϵ}_0}(q_3+Q+q_3-q_2-Q+q_2-q_1-Q+q_1)=0$
$\Rightarrow q_3=\frac{Q}{2}$
thus, the charge on inner face of plate A $=q_1=\frac{Q}{2}$.
The charge on inner face of B $=-Q-q_3=-Q-\frac{Q}{2}=-\frac{3Q}{2}$
The potential between A and B before introduced plate C is $V_{AB}=\frac{Q}{A{ϵ}_0}d$
Now $V_{AC}=\frac{Q/2}{A{ϵ}_0}d/2$ and $V_{BC}=\frac{3Q/2}{A{ϵ}_0}d/2$
thus, $V_{BC}=3V_{AC}\Rightarrow V_{AC}=\frac{1}{3}{V}_{BC}$
After introduced C , $V_{AB}^{\prime }=V_{AC}+V_{BC}=V_{AC}+3V_{AC}=4V_{AC}$
or $V_{AB}^{\prime }=4\frac{Q/2}{A{ϵ}_0}d/2=\frac{Q}{A{ϵ}_0}d=V_{AB}$