A dielectric having dielectric constant k =4 is filled in a capacitor having length l and width b now length of capacitor is increased by `l_1` for which energy stored become half of initial value `l_1` should be

Options

A. `2l`
B. `6l`
C. `8l`
D. `4l`

Answer

To solve the problem, we need to analyze the situation step by step. We have a capacitor with a dielectric material that has a dielectric constant \( k = 4 \). The initial length of the capacitor is \( L \) and the width is \( B \). When the length is increased by \( L_1 \), the energy stored in the capacitor becomes half of its initial value. We need to find the value of \( L_1 \).

### Step 1: Write down the expression for the initial capacitance

The capacitance \( C \) of a parallel plate capacitor filled with a dielectric is given by:

\[ C = k \frac{A \epsilon_0}{d} \]

where:
- \( k \) is the dielectric constant,
- \( A \) is the area of the plates,
- \( \epsilon_0 \) is the permittivity of free space,
- \( d \) is the separation between the plates.

For our capacitor, the area \( A \) can be expressed as:

\[ A = L \times B \]

Thus, the initial capacitance becomes:

\[ C = k \frac{L \cdot B \cdot \epsilon_0}{d} = 4 \frac{L \cdot B \cdot \epsilon_0}{d} \]

### Step 2: Write down the expression for the energy stored in the capacitor

The energy \( U \) stored in a capacitor is given by:

\[ U = \frac{1}{2} C V^2 \]

where \( V \) is the voltage across the capacitor. Since the charge \( Q \) remains constant when the length is increased, we can express \( V \) in terms of \( Q \) and \( C \):

\[ V = \frac{Q}{C} \]

Thus, the energy can also be expressed as:

\[ U = \frac{Q^2}{2C} \]

### Step 3: Analyze the new configuration after increasing the length

When the length of the capacitor is increased by \( L_1 \), the new length becomes \( L + L_1 \). The new capacitance \( C' \) can be expressed as the sum of two capacitors in parallel: one with the dielectric and one without.

1. The capacitance of the part with the dielectric (length \( L \)):
\[ C_d = k \frac{(L \cdot B) \epsilon_0}{d} = 4 \frac{L \cdot B \epsilon_0}{d} \]

2. The capacitance of the part without the dielectric (length \( L_1 \)):
\[ C_0 = \frac{(L_1 \cdot B) \epsilon_0}{d} \]

Thus, the total capacitance \( C' \) becomes:

\[ C' = C_d + C_0 = 4 \frac{L \cdot B \epsilon_0}{d} + \frac{(L_1 \cdot B) \epsilon_0}{d} \]

### Step 4: Set up the equation for the energy condition

According to the problem, the new energy stored is half of the initial energy:

\[ U' = \frac{U}{2} \]

Substituting the expressions for energy:

\[ \frac{Q^2}{2C'} = \frac{1}{2} \left( \frac{Q^2}{2C} \right) \]

This simplifies to:

\[ C' = 2C \]

### Step 5: Substitute the expressions for capacitance

Substituting \( C' \) and \( C \) into the equation:

\[ 4 \frac{L \cdot B \epsilon_0}{d} + \frac{(L_1 \cdot B) \epsilon_0}{d} = 2 \left( 4 \frac{L \cdot B \epsilon_0}{d} \right) \]

This simplifies to:

\[ 4L + L_1 = 8L \]

### Step 6: Solve for \( L_1 \)

Rearranging gives:

\[ L_1 = 8L - 4L = 4L \]

Thus, the required length \( L_1 \) is:

\[ L_1 = 4L \]

### Final Answer
\[ L_1 = 4L \]