Question

A parallel plate capacitor has a capacitance of \( 4 \, \mu\text{F} \). If the dielectric constant of the material between the plates is \( 5 \), what will be the new capacitance?

• A. \( 20 \, \mu\text{F} \)
• B. \( 15 \, \mu\text{F} \)
• C. \( 8 \, \mu\text{F} \)
• D. \( 10 \, \mu\text{F} \)

Hint:

Remember: The capacitance of a parallel plate capacitor increases by a factor of the dielectric constant when a dielectric is inserted between the plates.

Answer 1

The Correct Option is A

Step 1: Use the formula for capacitance of a parallel plate capacitor
The capacitance \( C \) of a parallel plate capacitor with a dielectric material is given by: \[ C = C_0 \times K \] where:
- \( C_0 \) is the capacitance without the dielectric (initial capacitance),
- \( K \) is the dielectric constant.
Step 2: Substitute the given values
We are given:
- Initial capacitance \( C_0 = 4 \, \mu\text{F} \),
- Dielectric constant \( K = 5 \).
Substitute these values into the formula: \[ C = 4 \, \mu\text{F} \times 5 = 20 \, \mu\text{F} \] Answer: Therefore, the new capacitance is \( 20 \, \mu\text{F} \). So, the correct answer is option (1).