Question

A dielectric of dielectric constant $ K $ is introduced such that half of its area of a capacitor of capacitance $ C $ is occupied by it. The new capacity is

• A. \( 2C \)
• B. \( C \)
• C. \( (1+K)C \)
• D. \( 2C(1+K) \)

Hint:

When a dielectric is introduced into a capacitor, the capacitance increases by a factor of \( 1 + K \), where \( K \) is the dielectric constant of the material.

Answer 1

The Correct Option is C

The capacitance of a parallel plate capacitor with a dielectric inserted is given by: \[ C = \frac{\epsilon_0 A}{d} \] where: 
- \( \epsilon_0 \) is the permittivity of free space, 
- \( A \) is the area of the plates, 
- \( d \) is the distance between the plates. When a dielectric material of dielectric constant \( K \) is inserted into a portion of the capacitor, the capacitance increases. 
If half of the area is occupied by the dielectric, the effective capacitance can be calculated by considering the contribution of the dielectric. Let the original capacitance without the dielectric be \( C_0 = \frac{\epsilon_0 A}{d} \). 
After inserting the dielectric, the capacitance becomes: \[ C_{\text{new}} = C_0 \left( 1 + K \right) \] where \( K \) is the dielectric constant. 
Since half of the area is occupied by the dielectric, the total effective capacitance will be: \[ C_{\text{new}} = (1 + K)C \] 
Thus, the new capacitance is \( (1 + K)C \). 
Therefore, the correct answer is: \( \text{(3) } (1 + K)C \)