Question

A capacitor is made of a flat plate of area A and second plate having a stair-like structure as shown in figure. The width of each plate is a and the height is b. The capacitance of the capacitor is :

• A. $\frac{2A{ϵ}_0}{3(d+b)}$
• B. $\frac{A{ϵ}_0(3d^2+6bd+2b^2)}{3d(b+d)(d+2b))}$
• C. $\frac{A{ϵ}_0(d^2+2bd+b^2)}{3d(d+b)(d+2b))}$
• D. $\frac{2A{ϵ}_0(d^2+2bd+b^2)}{3d(d+b)(d+2b))}$

Answer 1

The correct option is B $\frac{A{ϵ}_0(3d^2+6bd+2b^2)}{3d(b+d)(d+2b))}$

Capacitance of a system between two ends...is nothing but the charge needed for every unit potential developed between the two ends.
Note that all capacitors have same surface area = a = $\frac{A}{3}$
So..
First capacitance $C_1=\frac{{\epsilon }_{0}A}{3d}$
Sec capacitance $C_2=\frac{{\epsilon }_{0}A}{3(d+b)}$
Third capacitance $C_3=\frac{{\epsilon }_{0}A}{3(d+2b)}$
Adding up all as all the above specified capacitances are parallel.
The net Capacitance is $C_n=C_1+C_2+C_3$
Hence the answer.