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. dereases two times.
D. increases four times.

Answer

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

### Step 1: Understand the formula for energy stored in a capacitor
The energy (U) stored in a capacitor is given by the formula:
\[ U = \frac{1}{2} \frac{Q^2}{C} \]
where \( Q \) is the charge and \( C \) is the capacitance.

### Step 2: Determine the capacitance of a cylindrical capacitor
The capacitance \( C \) of a cylindrical capacitor can be expressed as:
\[ C = \frac{2 \pi \epsilon_0 L}{\ln\left(\frac{B}{A}\right)} \]
where:
- \( \epsilon_0 \) is the permittivity of free space,
- \( L \) is the length of the capacitor,
- \( A \) and \( B \) are the inner and outer radii of the cylindrical capacitor.

### Step 3: Substitute the capacitance into the energy formula
Substituting the expression for capacitance into the energy formula, we get:
\[ U = \frac{1}{2} \frac{Q^2}{\frac{2 \pi \epsilon_0 L}{\ln\left(\frac{B}{A}\right)}} \]
This simplifies to:
\[ U = \frac{Q^2 \ln\left(\frac{B}{A}\right)}{4 \pi \epsilon_0 L} \]

### Step 4: Analyze the changes when charge and length are doubled
Now, if we double the charge and length:
- New charge \( Q' = 2Q \)
- New length \( L' = 2L \)

### Step 5: Substitute the new values into the energy formula
The new energy \( U' \) can be calculated as follows:
\[ U' = \frac{(2Q)^2 \ln\left(\frac{B}{A}\right)}{4 \pi \epsilon_0 (2L)} \]
This simplifies to:
\[ U' = \frac{4Q^2 \ln\left(\frac{B}{A}\right)}{8 \pi \epsilon_0 L} \]
\[ U' = \frac{Q^2 \ln\left(\frac{B}{A}\right)}{2 \pi \epsilon_0 L} \]

### Step 6: Relate the new energy to the original energy
Now, we can relate \( U' \) to the original energy \( U \):
\[ U' = 2 \cdot \frac{Q^2 \ln\left(\frac{B}{A}\right)}{4 \pi \epsilon_0 L} \]
\[ U' = 2U \]

### Conclusion
Thus, the energy stored in the capacitor doubles when both the charge and length of the capacitor are doubled.

### Final Answer
The energy stored in the capacitor becomes \( 2U \).