Question

The separation between the plates of a charged parallel plate capacitor is increased. The force between the plates :

• A. increases
• B. decreases
• C. remains same
• D. first increases then decreases

Answer 1

The correct option is C remains same

The force between the plates is $F=\frac{q^2}{2A{ϵ}_0}$ where $q$ is the magnitude of charge of each plate and $A$ is the area of each plate. So $F$ is independent of separation d between the plates and increase of $d$ does no effect on charge $q$. Thus F will remain same.