Question

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

• A. $\frac{11 \varepsilon_0 A}{18 d}$
• B. $\frac{13 \varepsilon_0 A}{17 d}$
• C. $\frac{11 \varepsilon_0 A}{20 d}$
• D. $\frac{18 \varepsilon_0 A}{11 d}$

Answer 1

Answer: (A)

$\frac{11 \varepsilon_0 A}{18 d}$

Answer 2

Step 1: Capacitor arrangement The given system consists of three capacitors connected in parallel, each having:

Area of overlap = $\frac{A}{3}$.

The distances between the plates are $d$, $2d$, and $3d$, respectively.

Step 2: Capacitance of each section The capacitance of a parallel plate capacitor is given by:

$C = \frac{\epsilon_0 A}{d}.$

For the three sections:
1. $C_1 = \frac{\epsilon_0 A}{3d} = \frac{\epsilon_0 A}{3d}$,
2. $C_2 = \frac{\epsilon_0 A}{6d}$,
3. $C_3 = \frac{\epsilon_0 A}{9d}$.

Step 3: Total capacitance Since the capacitors are in parallel, the equivalent capacitance is:

$C_{\text{eq}} = C_1 + C_2 + C_3.$

Substitute the values:

$C_{\text{eq}} = \frac{\epsilon_0 A}{3d} + \frac{\epsilon_0 A}{6d} + \frac{\epsilon_0 A}{9d}.$

Take the common denominator:

$C_{\text{eq}} = \frac{\epsilon_0 A}{d} \left( \frac{1}{3} + \frac{1}{6} + \frac{1}{9} \right).$

Simplify the fraction:

$\frac{1}{3} + \frac{1}{6} + \frac{1}{9} = \frac{6 + 3 + 2}{18} = \frac{11}{18}.$

Thus:

$C_{\text{eq}} = \frac{\epsilon_0 A}{d} \cdot \frac{11}{18}.$

Final Answer: $C_{\text{eq}} = \frac{11 \epsilon_0 A}{18d}.$