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 address this, we must determine how to augment the capacitance of a parallel plate capacitor. The capacitance \( C \) is governed by the equation:v

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

where:

• \( C \) represents capacitance.
• \( \varepsilon_0 \) denotes the permittivity of free space (a constant).
• \( A \) signifies the area of a single plate, calculated as \( A = l \times b \).
• \( d \) is the separation distance between the plates.

The initial parameters are established as:

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

The initial plate area \( A \) is computed as:

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

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

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

A tenfold increase in capacitance, yielding \( C_{\text{new}} \), necessitates:

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

We must identify which configurations achieve this target capacitance by modifying either plate area or inter-plate distance:

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:
   • \( l = 5 \, \text{cm} = 5 \times 10^{-2} \, \text{m} \)
   • \( b = 2 \, \text{cm} = 2 \times 10^{-2} \, \text{m} \)
   • \( d = 1 \, \mu\text{m} = 1 \times 10^{-6} \, \text{m} \)

The correct selections are C and E exclusively.