A capacitor is charged with a battery and energy stored is U. After disconnecting battery another capacitor of same capacity is connected in parallel with it. Then energy stored in each capacitor is:

Options

A. U/2
B. U/4
C. 4 U
D. 2 U

Explanation

To solve the problem step by step, let's break it down:

### Step 1: Understand the initial conditions
A capacitor of capacitance \( C \) is charged by a battery to a voltage \( V \). The energy stored in the capacitor is given by the formula:
\[ U = \frac{1}{2} C V^2 \]
Since the charge \( Q \) on the capacitor can also be expressed as:
\[ Q = C V \]
we can substitute \( Q \) into the energy formula:
\[ U = \frac{1}{2} \frac{Q^2}{C} \]

### Step 2: Disconnect the battery
After charging, the battery is disconnected. The charge \( Q \) remains on the capacitor since it is isolated.

### Step 3: Connect a second identical capacitor in parallel
Now, a second capacitor of the same capacitance \( C \) is connected in parallel with the first capacitor. Initially, the second capacitor has no charge.

### Step 4: Determine the new charge distribution
When the two capacitors are connected in parallel, the total charge \( Q \) will redistribute between the two capacitors. Since they are identical, the charge will split equally:
\[ Q_1 = Q_2 = \frac{Q}{2} \]
where \( Q_1 \) and \( Q_2 \) are the charges on the first and second capacitors, respectively.

### Step 5: Calculate the new voltage across the capacitors
The voltage across both capacitors after the redistribution of charge can be calculated using:
\[ V' = \frac{Q_1}{C} = \frac{Q/2}{C} = \frac{Q}{2C} \]

### Step 6: Calculate the energy stored in each capacitor
The energy stored in each capacitor after connecting them in parallel can be calculated using the energy formula:
\[ U' = \frac{1}{2} C (V')^2 \]
Substituting \( V' \):
\[ U' = \frac{1}{2} C \left(\frac{Q}{2C}\right)^2 = \frac{1}{2} C \cdot \frac{Q^2}{4C^2} = \frac{Q^2}{8C} \]

### Step 7: Relate the new energy to the original energy
From the original energy formula:
\[ U = \frac{1}{2} \frac{Q^2}{C} \]
We can see that:
\[ U' = \frac{Q^2}{8C} = \frac{1}{4} \left(\frac{1}{2} \frac{Q^2}{C}\right) = \frac{U}{4} \]

### Conclusion
The energy stored in each capacitor after connecting them in parallel is:
\[ U' = \frac{U}{4} \]

### Final Answer
The energy stored in each capacitor is \( \frac{U}{4} \).