Question

A parallel plate capacitor is made up of stair like structure with a plate area A of each stair and that is connected with a wire of length b, as shown in the figure. The capacitance of the arrangement is \(\frac{x}{15} \frac{ε_0A}{b}\), the value of \(x\) is ______?

Answer 1

Correct Answer: 23

To solve for the value of x in the given expression for the capacitance, let's examine the structure of the capacitor.

The staircase structure consists of multiple parallel plate capacitors in series. Each 'stair' has a plate area A and a distance b between them. The formula for the capacitance C of a single parallel plate capacitor is:

\(C = \frac{ε_0 A}{d}\)

For our case, the distance d is b. Therefore, the capacitance of one stair is:

\(C_{\text{stair}} = \frac{ε_0 A}{b}\)

In the given structure, there are multiple such capacitors in series (5 in this diagram). For capacitors in series, the total capacitance C_{\text{total}} is given by:

\(\frac{1}{C_{\text{total}}} = \frac{1}{C_1} + \frac{1}{C_2} + \cdots + \frac{1}{C_n}\)

Since all capacitors have the same capacitance, the formula becomes:

\(\frac{1}{C_{\text{total}}} = \frac{n}{C_{\text{stair}}}\)

Solving for C_{\text{total}}:

\(C_{\text{total}} = \frac{C_{\text{stair}}}{n}\)

Given that the capacitance of the arrangement is:

\(C_{\text{arrangement}} = \frac{x}{15} \frac{ε_0 A}{b}\)

We equate this to our calculated total capacitance:

\(\frac{ε_0 A}{nb} = \frac{x}{15} \frac{ε_0 A}{b}\)

Canceling terms and solving for x, we get:

\(x = 15n\)

Here, n is 5, so:

\(x = 15 \times 5 = 75\)

The value of x is 75, which is not within the expected range of 23,23, indicating a need to review the problem context or given range.