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:
- \( \frac{11 \varepsilon_0 A}{18 d} \)
- \( \frac{13 \varepsilon_0 A}{17 d} \)
- \( \frac{11 \varepsilon_0 A}{20 d} \)
- \( \frac{18 \varepsilon_0 A}{11 d} \)
The Correct Option is A
Approach Solution - 1
The given problem involves finding the capacitance of a capacitor with a stair-like structure. Let's break it down step-by-step.
Understanding the Structure:
We have a flat plate of area \( A \) and a second plate with a stair-like structure. The stair structure consists of three parts, each with an area \( \frac{A}{3} \) and height \( d \).
Concept of Capacitance:
The capacitance \( C \) of a parallel-plate capacitor is given by the formula:
\(C = \frac{\varepsilon_0 A}{d}\)
where \( \varepsilon_0 \) is the permittivity of free space, \( A \) is the area of the plate, and \( d \) is the separation between the plates.
Solution:
Each step of the stair functions as an individual capacitor in parallel with the others. The areas and distances are as follows:
- First stair: Area = \( \frac{A}{3} \), Distance = \( d \)
- Second stair: Area = \( \frac{A}{3} \), Distance = \( 2d \)
- Third stair: Area = \( \frac{A}{3} \), Distance = \( 3d \)
Each of these capacitors contributes to the total capacitance inversely proportional to their distance:
Capacitance for each stair:
- \(C_1 = \frac{\varepsilon_0 A/3}{d}\)
- \(C_2 = \frac{\varepsilon_0 A/3}{2d}\)
- \(C_3 = \frac{\varepsilon_0 A/3}{3d}\)
Since these capacitors are in parallel, their total capacitance \( C_{\text{total}} \) is the sum:
\(C_{\text{total}} = C_1 + C_2 + C_3\\)
Substituting the values:
\(C_{\text{total}} = \frac{\varepsilon_0 A}{3d} + \frac{\varepsilon_0 A}{6d} + \frac{\varepsilon_0 A}{9d}\)
Taking a common denominator:
\(C_{\text{total}} = \frac{6 \varepsilon_0 A}{18d} + \frac{3 \varepsilon_0 A}{18d} + \frac{2 \varepsilon_0 A}{18d}\)
\(C_{\text{total}} = \frac{(6+3+2) \varepsilon_0 A}{18d} = \frac{11 \varepsilon_0 A}{18d}\)
Conclusion:
The capacitance of the given arrangement is \(\frac{11 \varepsilon_0 A}{18 d}\), matching the correct answer.
Approach Solution -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: \( C_{\text{eq}} = \frac{11 \epsilon_0 A}{18d}. \)
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