Question

The capacitance of the capacitor of plate areas $A_1$ and $A_2(A_1<A_2)$ at a distance d, as shown in figure is :

• A. $\frac{{ϵ}_0(A_1+A_2)}{2d}$
• B. $\frac{{ϵ}_0{A}_2}{d}$
• C. $\frac{{ϵ}_0\sqrt{A_1{A}_2}}{d}$
• D. $\frac{{ϵ}_0{A}_1}{d}$

Answer 1

The correct option is D $\frac{{ϵ}_0{A}_1}{d}$

As $A_1<A_2$ so the common area of plates is $A=A_1$.
Thus, the capacitance of parallel plate capacitor is $C=\frac{A{ϵ}_0}{d}=\frac{A_1{ϵ}_0}{d}$