Question

A parallel plate capacitor has two square plates of side 5 cm separated by 15 mm. A Pyrex glass slab of dielectric constant \(K = 4.5\) and thickness 10 mm is placed between the plates. The capacitance of the capacitor is:

• A. \(3.11 \times 10^{-12} \, {F}\)
• B. \(311 \, \mu{F}\)
• C. \(311 \, {pF}\)
• D. \(3.11 \, {nF}\)

Hint:

Remember that the presence of a dielectric reduces the effective separation between capacitor plates, thereby increasing the capacitance.

Answer 1

The Correct Option is A

The capacitance of a parallel plate capacitor with a dielectric material is given by the formula:

\[ C = \frac{K \epsilon_0 A}{d} \]

where:

• \( C \) is the capacitance,
• \( K \) is the dielectric constant of the material (for Pyrex, \( K = 4.5 \)),
• \( \epsilon_0 \) is the permittivity of free space \( = 8.854 \times 10^{-12} \, \text{C}^2 / \text{N m}^2 \),
• \( A \) is the area of one of the plates,
• \( d \) is the separation between the plates.

Given:

• Side of the plates \( = 5 \, \text{cm} = 0.05 \, \text{m} \),
• Separation between the plates \( d = 15 \, \text{mm} = 0.015 \, \text{m} \),
• Thickness of the dielectric slab \( = 10 \, \text{mm} = 0.01 \, \text{m} \),
• Dielectric constant \( K = 4.5 \).

The area \( A \) of the square plates is:

\[ A = \text{side}^2 = (0.05)^2 = 0.0025 \, \text{m}^2 \]

Now, substitute these values into the formula for capacitance:

\[ C = \frac{4.5 \times (8.854 \times 10^{-12}) \times 0.0025}{0.015} \] \[ C = \frac{4.5 \times 8.854 \times 10^{-12} \times 0.0025}{0.015} \] \[ C = 3.11 \times 10^{-12} \, \text{F} \]

Thus, the correct answer is Option (1), \( 3.11 \times 10^{-12} \, \text{F} \).

Answer 2

To solve the problem, we need to calculate the capacitance of a parallel plate capacitor with a dielectric slab inserted between the plates.

1. Capacitance of a Parallel Plate Capacitor:
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 (\( 8.85 \times 10^{-12} \, \text{F/m} \)),
\( A \) is the area of each plate,
\( d \) is the distance between the plates.

For a capacitor with a dielectric material between the plates, the formula becomes: \[ C = \frac{\varepsilon_0 K A}{d} \] where \( K \) is the dielectric constant of the material. In this case, \( K = 4.5 \), and the given dimensions are: - The side of each square plate is 5 cm (\( 0.05 \, \text{m} \)), - The distance between the plates is 15 mm (\( 0.015 \, \text{m} \)), - The thickness of the dielectric material is 10 mm (\( 0.01 \, \text{m} \)).

2. Calculating the Area of the Plates:
The area \( A \) of each square plate is given by: \[ A = (\text{side})^2 = (0.05 \, \text{m})^2 = 0.0025 \, \text{m}^2 \]

3. Calculating the Capacitance:
Now, substitute the values into the formula for capacitance: \[ C = \frac{(8.85 \times 10^{-12}) \times 4.5 \times 0.0025}{0.01} \] \[ C = \frac{9.9375 \times 10^{-14}}{0.01} = 3.11 \times 10^{-12} \, \text{F} \]

4. Identifying the Correct Answer:
The capacitance of the capacitor is \( 3.11 \times 10^{-12} \, \text{F} \).

Final Answer:
The correct answer is Option A: \( 3.11 \times 10^{-12} \, \text{F} \).