Question

The configuration of the metal plate can be shown in figure. The area of each plate is A. The distance between each of the plates is d. Then the capacitance between A and B is

• A. $\frac{3}{2}\frac{A{ϵ}_0}{d}$
• B. $\frac{A{ϵ}_0}{d}$
• C. $\frac{5}{2}\frac{A{ϵ}_0}{d}$
• D. None of the above

Answer 1

The correct option is A $\frac{3}{2}\frac{A{ϵ}_0}{d}$

The equivalent diagram of these three capacitors can be shown asThe more simplified diagram can be shown as

Then, the equivalent capacitance of $C_1$ and $C_3$ can be given as,
$C_{eq}=\frac{C_1{C}_3}{C_1+C_3}$
Now the equivalent diagram will be,
$C_{net}=C_2+\frac{C_1{C}_3}{C_1+C_3}$
$C_{net}=C+\frac{CC}{C+C}=\frac{3C}{2}=\frac{3{ϵ}_{0}A}{2d}$
For detailed solution watch the next video.