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

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
$C=\frac{\varepsilon_{0} A}{d}$
First capacitance of capacitor,
$C_{1}=\frac{\varepsilon_{0} A}{3 d}$
second capacitance of capacitor,
$C_{2}=\frac{c_{0} A}{3(2 d)}=\frac{\varepsilon_{0} A}{6 d}$
and third capacitance of capacitor,
$C_{3}=\frac{\varepsilon_{0} A}{3(3 d)}=\frac{\varepsilon_{0} A}{9d}$
Equivalent capacitance of capacitor,
$C_{eq} =C_{1}+C_{2}+C_{3}$
$C_{eq} =\frac{\varepsilon_{0} A}{3 d}+\frac{\varepsilon_{0} A}{6 d}+\frac{\varepsilon_{0} A}{9 d} $
$C_{eq}= \frac{\left(6 \varepsilon_{0}+3 \varepsilon_{0}+2 \varepsilon_{0}\right) A}{18 d}$
$=\frac{11 \varepsilon_{0} A}{18 d}$
Hence, the capacitance of the arrangement is
$\frac{11 \varepsilon_{0} A}{18 d}$