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 step by step, we need to analyze the situation involving a capacitor with a dielectric and how its energy changes when the length is increased.

### Step 1: Understanding the Initial Configuration
The initial capacitor has a dielectric constant \( k = 4 \), length \( L \), and width \( b \). The capacitance \( C_1 \) of the capacitor filled with dielectric is given by the formula:
\[ C_1 = k \cdot \frac{\epsilon_0 A}{d} \]
where \( A = L \cdot b \) is the area of the plates, \( \epsilon_0 \) is the permittivity of free space, and \( d \) is the separation between the plates.

### Step 2: Calculate Initial Capacitance
Substituting the area into the capacitance formula:
\[ C_1 = 4 \cdot \frac{\epsilon_0 (L \cdot b)}{d} = \frac{4 \epsilon_0 L b}{d} \]

### Step 3: Calculate Initial Energy Stored
The energy \( U_1 \) stored in the capacitor is given by:
\[ U_1 = \frac{1}{2} C_1 V^2 \]
where \( V \) is the voltage across the capacitor.

### Step 4: New Configuration After Length Increase
When the length of the capacitor is increased by \( l_1 \), the new length becomes \( L + l_1 \). The new capacitance \( C_2 \) of the capacitor (with the dielectric still in place) is:
\[ C_2 = k \cdot \frac{\epsilon_0 A'}{d} \]
where \( A' = (L + l_1) \cdot b \). Thus,
\[ C_2 = 4 \cdot \frac{\epsilon_0 ((L + l_1) \cdot b)}{d} = \frac{4 \epsilon_0 (L + l_1) b}{d} \]

### Step 5: Calculate New Energy Stored
The new energy \( U_2 \) stored in the capacitor is:
\[ U_2 = \frac{1}{2} C_2 V^2 \]
According to the problem, this new energy is half of the initial energy:
\[ U_2 = \frac{1}{2} U_1 \]

### Step 6: Setting Up the Equation
Substituting the expressions for \( U_1 \) and \( U_2 \):
\[ \frac{1}{2} C_2 V^2 = \frac{1}{2} \left( \frac{1}{2} C_1 V^2 \right) \]
This simplifies to:
\[ C_2 = \frac{1}{2} C_1 \]

### Step 7: Substitute Capacitance Values
Substituting the capacitance values we derived:
\[ \frac{4 \epsilon_0 (L + l_1) b}{d} = \frac{1}{2} \left( \frac{4 \epsilon_0 L b}{d} \right) \]
Cancelling common terms:
\[ 4(L + l_1) = 2L \]

### Step 8: Solve for \( l_1 \)
Rearranging the equation gives:
\[ 4L + 4l_1 = 2L \]
\[ 4l_1 = 2L - 4L \]
\[ 4l_1 = -2L \]
\[ l_1 = -\frac{L}{2} \]

### Conclusion
The length \( l_1 \) by which the capacitor's length should be increased to make the energy stored half of the initial value is:
\[ l_1 = -\frac{L}{2} \]
This indicates that the length cannot be increased positively to achieve the condition; instead, it suggests a conceptual misunderstanding in the problem setup.