Question

In an isolated parallel capacitor of plate are $A$, plate separation $d$ and charge $q$, the force of attraction between the plates is $F$. Then :

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

Answer 1

Answer: (A), (D)

The correct options are
A $F\alpha q^2$
D $F\alpha \frac{1}{A}$

Electric field due to plate of charge q$=E=\frac{\sigma }{2{ϵ}_0}=\frac{q}{2A{ϵ}_0}$

The force of attraction between the plates,$F=qE=\frac{q^2}{2A{ϵ}_0}$

So, $F\propto q^2,F\propto \frac{1}{A}$

Ans: $\left(A\right),\left(D\right)$