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

To find the new capacitance when a dielectric material of dielectric constant \( K \) is introduced to cover half the area of a capacitor, we begin by considering the configuration of the capacitor system. Let the original capacitance of the whole capacitor be \( C \). The original capacitance can be expressed with its physical dimensions and permittivity as:

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

where \( \varepsilon_0 \) is the permittivity of free space, \( A \) is the area of the plates, and \( d \) is the separation between the plates.

When the dielectric is introduced covering half the area, the capacitor can effectively be considered as two capacitors connected in parallel:

• Capacitor 1 with dielectric, covering area \( \frac{A}{2} \) and dielectric constant \( K \).
• Capacitor 2 without dielectric, also covering area \( \frac{A}{2} \).

For these sub-capacitors, their individual capacitances are:

1. \( C_1 = \frac{K\varepsilon_0 \cdot \frac{A}{2}}{d} = \frac{KA\varepsilon_0}{2d} \)

2. \( C_2 = \frac{\varepsilon_0 \cdot \frac{A}{2}}{d} = \frac{A\varepsilon_0}{2d} \)

The total capacitance \( C' \) for these capacitors in parallel is the sum of their capacitances:

\( C' = C_1 + C_2 = \frac{KA\varepsilon_0}{2d} + \frac{A\varepsilon_0}{2d} \)

\( C' = \frac{A\varepsilon_0}{2d}(K + 1) \)

We substitute back the expression for the original capacitance \( C = \frac{\varepsilon_0 A}{d} \):

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

Then half of that would be \( \frac{C}{2} = \frac{A\varepsilon_0}{2d} \), and thus we have:

\( C' = \frac{C}{2}(K + 1) \times 2 = C(K + 1) \)

Therefore, the new capacitance with the dielectric partially inserted is \( C' = (1 + K)C \).

Answer 2

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: \( \text{(3) } (1 + K)C \)