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, we need to analyze the situation step by step.

### Step 1: Initial Energy Stored in the Capacitor
When the capacitor (C) is charged by a battery with voltage (V), the charge (Q) on the capacitor is given by:
\[ Q = C \cdot V \]
The energy (U) stored in the capacitor is given by the formula:
\[ U = \frac{1}{2} \frac{Q^2}{C} \]
Substituting \( Q \) into the energy formula:
\[ U = \frac{1}{2} \frac{(C \cdot V)^2}{C} = \frac{1}{2} C V^2 \]

### Step 2: Disconnecting the Battery
After charging, the battery is disconnected. The capacitor now holds charge Q and energy U.

### Step 3: Connecting Another Capacitor in Parallel
Now, we connect another capacitor of the same capacitance (C) in parallel with the charged capacitor. When capacitors are connected in parallel, the voltage across both capacitors remains the same, but the total charge gets redistributed.

### Step 4: Charge Distribution
The total charge before connecting the second capacitor is Q. After connecting the second capacitor, the total capacitance becomes:
\[ C_{total} = C + C = 2C \]
Since the voltage (V') across both capacitors must be the same, we can use the relationship:
\[ Q = C \cdot V \]
The new voltage across the combined capacitors can be expressed as:
\[ Q = 2C \cdot V' \]
Setting the two equations for charge equal gives:
\[ C \cdot V = 2C \cdot V' \]
From this, we can solve for the new voltage:
\[ V' = \frac{V}{2} \]

### Step 5: New Charge on Each Capacitor
The charge on each capacitor after connecting them in parallel will be:
\[ Q' = C \cdot V' = C \cdot \frac{V}{2} = \frac{Q}{2} \]
Thus, each capacitor now has a charge of \( \frac{Q}{2} \).

### Step 6: New Energy Stored in Each Capacitor
The energy stored in each capacitor can be calculated using the formula:
\[ U' = \frac{1}{2} \frac{(Q')^2}{C} \]
Substituting \( Q' = \frac{Q}{2} \):
\[ U' = \frac{1}{2} \frac{\left(\frac{Q}{2}\right)^2}{C} = \frac{1}{2} \frac{\frac{Q^2}{4}}{C} = \frac{Q^2}{8C} \]

### Step 7: Relating New Energy to Original Energy
Recall that the original energy \( U \) was:
\[ U = \frac{1}{2} \frac{Q^2}{C} \]
Now, we can express the new energy stored in each capacitor:
\[ U' = \frac{1}{4} \cdot \frac{1}{2} \frac{Q^2}{C} = \frac{U}{4} \]

### Conclusion
Thus, the energy stored in each capacitor after connecting the second capacitor in parallel is:
\[ U' = \frac{U}{4} \]