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

The Correct Option is A

This problem concerns calculating the capacitance of a capacitor with a stepped geometry. The solution proceeds through a systematic analysis.

Structure Definition:

A flat plate of area \( A \) is paired with a stepped plate. The stepped structure comprises three sections, each possessing an area of \( \frac{A}{3} \) and a height of \( d \).

Capacitance Principle:

The fundamental formula for the capacitance \( C \) of a parallel-plate capacitor is:

\(C = \frac{\varepsilon_0 A}{d}\)

Here, \( \varepsilon_0 \) represents the permittivity of free space, \( A \) is the area of one plate, and \( d \) denotes the separation distance between the plates.

Solution Methodology:

Each step of the structure acts as an individual capacitor connected in parallel with the others. The respective areas and separation distances are as follows:

1. Step 1: Area = \( \frac{A}{3} \), Distance = \( d \)
2. Step 2: Area = \( \frac{A}{3} \), Distance = \( 2d \)
3. Step 3: Area = \( \frac{A}{3} \), Distance = \( 3d \)

The capacitance contributed by each step is inversely proportional to its separation distance:

Individual capacitances for each step:

1. \(C_1 = \frac{\varepsilon_0 A/3}{d}\)
2. \(C_2 = \frac{\varepsilon_0 A/3}{2d}\)
3. \(C_3 = \frac{\varepsilon_0 A/3}{3d}\)

As these capacitors are in parallel, the total capacitance \( C_{\text{total}} \) is the arithmetic sum of their individual capacitances:

\(C_{\text{total}} = C_1 + C_2 + C_3\\)

Substituting the calculated values:

\(C_{\text{total}} = \frac{\varepsilon_0 A}{3d} + \frac{\varepsilon_0 A}{6d} + \frac{\varepsilon_0 A}{9d}\)

To sum these fractions, a common denominator is employed:

\(C_{\text{total}} = \frac{6 \varepsilon_0 A}{18d} + \frac{3 \varepsilon_0 A}{18d} + \frac{2 \varepsilon_0 A}{18d}\)

\(C_{\text{total}} = \frac{(6+3+2) \varepsilon_0 A}{18d} = \frac{11 \varepsilon_0 A}{18d}\)

Final Result:

The calculated capacitance for the described configuration is \(\frac{11 \varepsilon_0 A}{18 d}\).