A parallel plate capacitor filled with a material of dielectric constant K is charged to a certain voltage and is isolated. The dielectric material is removed. Then
(a) The capacitance decreases by a factor K
(b) The electric field reduces by a factor K
(c ) The voltage across the capacitor increases by a factor K
(d) The charge strored in the capacitor increases by a factor K

Options

A. a and b are true
B. a and c are true
C. b and c are true
D. b and d are true

Answer

To solve the problem, we need to analyze the behavior of a parallel plate capacitor when the dielectric material is removed. Here are the steps to derive the correct answers:

### Step 1: Understand the Initial Conditions
- A parallel plate capacitor has a capacitance \( C_1 \) when filled with a dielectric material of dielectric constant \( K \).
- The capacitance with the dielectric is given by:
\[ C_1 = K \frac{\epsilon_0 A}{d} \]
where \( \epsilon_0 \) is the permittivity of free space, \( A \) is the area of the plates, and \( d \) is the separation between the plates.

### Step 2: Charge and Voltage with Dielectric
- When the capacitor is charged to a voltage \( V \), the charge \( Q \) stored in the capacitor is:
\[ Q = C_1 V = K \frac{\epsilon_0 A}{d} V \]

### Step 3: Remove the Dielectric
- When the dielectric is removed, the new capacitance \( C_2 \) becomes:
\[ C_2 = \frac{\epsilon_0 A}{d} \]
- The capacitance decreases by a factor of \( K \):
\[ C_2 = \frac{C_1}{K} \]

### Step 4: Analyze the Effect on Voltage
- Since the capacitor is isolated, the charge \( Q \) remains constant when the dielectric is removed.
- The new voltage \( V_2 \) across the capacitor can be calculated using:
\[ V_2 = \frac{Q}{C_2} \]
- Substituting \( Q \) and \( C_2 \):
\[ V_2 = \frac{K \frac{\epsilon_0 A}{d} V}{\frac{\epsilon_0 A}{d}} = K V \]
- Thus, the voltage increases by a factor of \( K \).

### Step 5: Analyze the Electric Field
- The electric field \( E \) in the capacitor with dielectric is given by:
\[ E_1 = \frac{V}{d} \]
- After removing the dielectric, the electric field becomes:
\[ E_2 = \frac{V_2}{d} = \frac{K V}{d} \]
- Hence, the electric field increases by a factor of \( K \).

### Step 6: Analyze the Charge Stored
- The charge \( Q \) remains constant as it was isolated. Thus, it does not increase or decrease.

### Conclusion
Based on the analysis:
- (a) The capacitance decreases by a factor \( K \) - **True**
- (b) The electric field reduces by a factor \( K \) - **False** (it increases)
- (c) The voltage across the capacitor increases by a factor \( K \) - **True**
- (d) The charge stored in the capacitor increases by a factor \( K \) - **False** (it remains constant)

### Final Answers
- Correct options: (a) and (c)