A capacitor is filled with an insulator and a certain potential difference is applied to its pltaes. The energy stored in the capacitor is `U`. Now the capacitor is disconnected from the source and the insulator is pulled out of the capacitor. The work performed against the forces of electric field in pulling out the insulator is `4 U`. Then dielectric constant of the insulator is.

Options

A. 4
B. 8
C. 5
D. 3

Answer

To solve the problem, we will follow these steps:

### Step 1: Understand the initial conditions
When a capacitor is filled with a dielectric material, the capacitance increases. The capacitance \( C \) of a capacitor filled with a dielectric is given by:
\[ C = K \cdot C_0 \]
where \( K \) is the dielectric constant and \( C_0 \) is the capacitance without the dielectric.

### Step 2: Write the expression for energy stored in the capacitor
The energy \( U \) stored in the capacitor is given by:
\[ U = \frac{1}{2} C V^2 \]
Substituting for \( C \):
\[ U = \frac{1}{2} (K \cdot C_0) V^2 \]

### Step 3: Disconnect the capacitor from the source
When the capacitor is disconnected from the voltage source, the charge \( Q \) on the capacitor remains constant. The charge \( Q \) can be expressed as:
\[ Q = C \cdot V = K \cdot C_0 \cdot V \]

### Step 4: Determine the final potential energy after removing the dielectric
When the dielectric is removed, the capacitance of the capacitor becomes \( C_0 \). The final potential energy \( U_f \) can be expressed as:
\[ U_f = \frac{Q^2}{2 C_0} \]
Substituting \( Q \):
\[ U_f = \frac{(K \cdot C_0 \cdot V)^2}{2 C_0} = \frac{K^2 \cdot C_0^2 \cdot V^2}{2 C_0} = \frac{K^2 \cdot C_0 \cdot V^2}{2} \]

### Step 5: Relate the initial and final potential energies
From the initial energy expression, we have:
\[ U = \frac{1}{2} K C_0 V^2 \]
Thus, we can relate the final potential energy to the initial potential energy:
\[ U_f = K \cdot U \]

### Step 6: Calculate the work done in removing the dielectric
The work done \( W \) against the electric field when the dielectric is removed is given as:
\[ W = U_f - U \]
Substituting for \( U_f \):
\[ W = K \cdot U - U = (K - 1) \cdot U \]
According to the problem, the work done is \( 4U \):
\[ (K - 1) \cdot U = 4U \]

### Step 7: Solve for the dielectric constant \( K \)
Dividing both sides by \( U \) (assuming \( U \neq 0 \)):
\[ K - 1 = 4 \]
Thus,
\[ K = 5 \]

### Conclusion
The dielectric constant of the insulator is \( K = 5 \).