Question

A capacitor is made of a flat plate of area $A$ and a second plate of stair-like structure as shown in the figure. The area of each stair is $\frac{A}{3}$ and the height is $d$. The capacitance of the arrangement is

• A. $\frac{{\epsilon }_{0}A}{3d}$
• B. $\frac{6{\epsilon }_{0}A}{11d}$
• C. $\frac{3{\epsilon }_{0}A}{d}$
• D. $\frac{11{\epsilon }_{0}A}{18d}$

Answer 1

Answer: (D)

According to the question,

Given, area of each stair $=\frac{A}{3}$
and height of each stair $=d$
$\therefore$ Capacitance of parallel plate capacitor is given as
$C=\frac{{\epsilon }_{0}A}{d}$
First capacitance of capacitor,
$C_1=\frac{{\epsilon }_{0}A}{3d}$
second capacitance of capacitor,
$C_2=\frac{c_{0}A}{3(2d)}=\frac{{\epsilon }_{0}A}{6d}$
and third capacitance of capacitor,
$C_3=\frac{{\epsilon }_{0}A}{3(3d)}=\frac{{\epsilon }_{0}A}{9d}$
Equivalent capacitance of capacitor,
$C_{eq}=C_1+C_2+C_3$
$C_{eq}=\frac{{\epsilon }_{0}A}{3d}+\frac{{\epsilon }_{0}A}{6d}+\frac{{\epsilon }_{0}A}{9d}$
$C_{eq}=\frac{\left(6{\epsilon }_0+3{\epsilon }_0+2{\epsilon }_0\right)A}{18d}$
$=\frac{11{\epsilon }_{0}A}{18d}$
Hence, the capacitance of the arrangement is
$\frac{11{\epsilon }_{0}A}{18d}$