Question

In a parallel-plate capacitor of plate area A, plate separation d and charge Q, the force of attraction between the plates is F.

• A. $F\propto Q^2$
• B. $F\propto \frac{1}{A}$
• C. $F\propto d$
• D. $F\propto \frac{1}{d}$

Answer 1

Answer: (A), (B)

The correct options are
A $F\propto Q^2$
B $F\propto \frac{1}{A}$

The electric field due to the plate of charge Q is $E=\frac{Q}{2{ϵ}_{0}A}$

The force between the plates $F=QE=\frac{Q^2}{2{ϵ}_{0}A}$

thus, $F\propto Q^2,$ and $F\propto \frac{1}{A}$