Question

A parallel plate capacitor was made with two rectangular plates, each with a length of \( l = 3 \, {cm} \) and breadth of \( b = 1 \, {cm} \). The distance between the plates is \( d = 3 \, \mu{m} \). Out of the following, which are the ways to increase the capacitance by a factor of 10?

• A. l = 30 cm, b = 1 cm, d = 1 μm
• B. l = 3 cm, b = 1 cm, d = 30 μm
• C. l = 6 cm, b = 5 cm, d = 3 μm
• D. l = 1 cm, b = 1 cm, d = 10 μm
• E. l = 5 cm, b = 2 cm, d = 1 μm

• A. C and E only
• B. B and D only
• C. A only
• D. C only

Hint:

In a parallel plate capacitor, the capacitance is proportional to the area of the plates and inversely proportional to the distance between them. To increase the capacitance by a factor of 10, ensure the product of the length and breadth increases while adjusting the distance accordingly.

Answer 1

The Correct Option is A

To solve this problem, we need to understand how the capacitance of a parallel plate capacitor can be increased. The capacitance \( C \) is given by the formula:v

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

where:

• \( C \) is the capacitance.
• \( \varepsilon_0 \) is the permittivity of free space (a constant).
• \( A \) is the area of one of the plates, computed as \( A = l \times b \).
• \( d \) is the distance between the plates.

Initially, the values given are:

• \( l = 3 \, \text{cm} = 3 \times 10^{-2} \, \text{m} \)
• \( b = 1 \, \text{cm} = 1 \times 10^{-2} \, \text{m} \)
• \( d = 3 \, \mu\text{m} = 3 \times 10^{-6} \, \text{m} \)

The area \( A \) is:

\(A = l \times b = 3 \times 10^{-2} \times 1 \times 10^{-2} = 3 \times 10^{-4} \, \text{m}^2\)

Therefore, the initial capacitance \( C_{\text{initial}} \) is:

\(C_{\text{initial}} = \frac{{\varepsilon_0 \cdot 3 \times 10^{-4}}}{{3 \times 10^{-6}}} = \varepsilon_0 \times 100 \, \text{F}\)

To increase the capacitance by a factor of 10, the new capacitance \( C_{\text{new}} \) should be:

\(C_{\text{new}} = 10 \times C_{\text{initial}} = 10 \times \varepsilon_0 \times 100 = \varepsilon_0 \times 1000 \, \text{F}\)

We need to identify which options achieve this new capacitance by either increasing the plate area or reducing the distance between the plates:

1. Option A:
   • \( l = 30 \, \text{cm} = 30 \times 10^{-2} \, \text{m} \)
   • \( b = 1 \, \text{cm} = 1 \times 10^{-2} \, \text{m} \)
   • \( d = 1 \, \mu\text{m} = 1 \times 10^{-6} \, \text{m} \)
2. Option B:
   • \( l = 3 \, \text{cm} \)
   • \( b = 1 \, \text{cm} \)
   • \( d = 30 \, \mu\text{m} \)
3. Option C:
   • \( l = 6 \, \text{cm} = 6 \times 10^{-2} \, \text{m} \)
   • \( b = 5 \, \text{cm} = 5 \times 10^{-2} \, \text{m} \)
   • \( d = 3 \, \mu\text{m} \)
4. Option D:
   • \( l = 1 \, \text{cm} \)
   • \( b = 1 \, \text{cm} \)
   • \( d = 10 \, \mu\text{m} \)
5. Option E:
   • \( l = 5 \, \text{cm} = 5 \times 10^{-2} \, \text{m} \)
   • \( b = 2 \, \text{cm} = 2 \times 10^{-2} \, \text{m} \)
   • \( d = 1 \, \mu\text{m} = 1 \times 10^{-6} \, \text{m} \)

Thus, the correct choices are C and E only.

Answer 2

Given:

- \( A \) is the plate area. - \( d \) is the distance between the plates. - The initial capacitance \( C \) is given by: \[ C = \frac{A \epsilon_0}{d}, \] where \( \epsilon_0 \) is the permittivity of free space.

Step 1: Calculate the initial capacitance

The initial capacitance is given by the equation: \[ C = \frac{\epsilon_0 A}{d}. \] According to the problem, when \( b = 5 \, \text{cm} \) and \( d = 3 \, \text{cm} \), the capacitance is: \[ C = 10 \epsilon_0 \, \text{units}. \]

Step 2: Analyze the second option (Option 'E')

In Option 'E', we have: - \( b = 2 \, \text{cm} \), - \( d = 1 \, \text{cm} \). Substituting these values into the capacitance formula: \[ C = \frac{\epsilon_0 A}{d} = 10 \epsilon_0 \, \text{units}. \]

Final Answer:

The capacitance is given by \( C = 10 \epsilon_0 \, \text{units} \) for both cases.