Question

A parallel plate capacitor is connected to a cell. It's positive plate A and the 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 the plates A and B parallel to them What is the charge on the inner surface of A ?

• A. $Q$
• B. $\frac{Q}{2}$
• C. $\frac{3Q}{2}$
• D. $\frac{Q}{4}$

Answer 1

The correct option is B $\frac{Q}{2}$

Let the capacitance of the A+B system be given by C = $\frac{ϵA}{d}$
Now when the plate is introduced in middle, the distance of A+C and C+B system is changed to $\frac{d}{2}$ So the capacitance will be $\frac{2ϵA}{d}$ = 2C
Now, Total charge in the system is Q assume A+C has q charge and C+B has Q+q charge
So, $\frac{q}{2C}$ + $\frac{Q+q}{2C}$ = $\frac{Q}{C}$
So, q = $\frac{Q}{2}$