A cylindrical capacitor has charge Q and length L. If both the charge and length of the capacitor are doubled by keeping other parameters fixed, the energy stored in the capacitor

Options

A. remains same
B. increases two times
C. decreases two times
D. increases four times

Answer

To solve the problem, we need to analyze how the energy stored in a cylindrical capacitor changes when both the charge and the length of the capacitor are doubled.

### Step-by-Step Solution:

1. **Understanding the Energy Stored in a Capacitor**:
The energy (E) stored in a capacitor is given by the formula:
\[ E = \frac{1}{2} \frac{Q^2}{C} \]
where \( Q \) is the charge and \( C \) is the capacitance.

2. **Capacitance of a Cylindrical Capacitor**:
The capacitance \( C \) of a cylindrical capacitor is given by:
\[ C = \frac{2 \pi \epsilon_0 L}{\ln(B/A)} \]
where \( L \) is the length of the capacitor, \( B \) is the outer radius, \( A \) is the inner radius, and \( \epsilon_0 \) is the permittivity of free space.

3. **Initial Energy Expression**:
Substituting the expression for capacitance into the energy formula, we have:
\[ E = \frac{1}{2} \frac{Q^2 \ln(B/A)}{2 \pi \epsilon_0 L} \]

4. **Doubling Charge and Length**:
If both the charge \( Q \) and the length \( L \) are doubled, we denote the new charge as \( Q' = 2Q \) and the new length as \( L' = 2L \).

5. **New Capacitance**:
The new capacitance \( C' \) after doubling the length becomes:
\[ C' = \frac{2 \pi \epsilon_0 (2L)}{\ln(B/A)} = 2 \cdot \frac{2 \pi \epsilon_0 L}{\ln(B/A)} = 2C \]

6. **New Energy Expression**:
Now substituting \( Q' \) and \( C' \) into the energy formula:
\[ E' = \frac{1}{2} \frac{(2Q)^2}{C'} = \frac{1}{2} \frac{4Q^2}{2C} = \frac{2Q^2}{C} \]

7. **Relating New Energy to Initial Energy**:
From the initial energy expression \( E = \frac{1}{2} \frac{Q^2}{C} \), we can see that:
\[ E' = 2 \cdot \frac{1}{2} \frac{Q^2}{C} = 2E \]

8. **Conclusion**:
The energy stored in the capacitor after doubling both the charge and the length is:
\[ E' = 2E \]
Therefore, the energy stored in the capacitor increases by a factor of 2.

### Final Answer:
The energy stored in the capacitor increases two times.