The energy of a charged capacitor is U. Another identical capacitor is connected parallel to the first capacitor, after disconnecting the battery. The total energy of the system of these capacitors will be

Options

A. `U/4`
B. `U/2`
C. `(3U)/(2)`
D. `(2U)/(4)`

Explanation

To solve the problem step by step, we need to analyze the situation with the capacitors before and after the second capacitor is connected in parallel.

### Step 1: Understand the initial conditions
Initially, we have a charged capacitor with energy \( U \). The energy stored in a capacitor is given by the formula:
\[ U = \frac{1}{2} C V^2 \]
where \( C \) is the capacitance and \( V \) is the voltage across the capacitor.

### Step 2: Determine the charge on the initial capacitor
The charge \( Q \) on the initial capacitor can be expressed as:
\[ Q = C V \]

### Step 3: Connect the second identical capacitor in parallel
When we connect another identical capacitor (also with capacitance \( C \)) in parallel to the first capacitor after disconnecting the battery, the total capacitance of the system becomes:
\[ C_{\text{total}} = C + C = 2C \]

### Step 4: Find the new voltage across the capacitors
Since the capacitors are now in parallel, they will share the same voltage \( V' \). The total charge in the system remains the same as the charge on the first capacitor, which is \( Q = C V \). Therefore, we can write:
\[ Q = C V = C_{\text{total}} V' = 2C V' \]
From this, we can solve for \( V' \):
\[ C V = 2C V' \implies V' = \frac{V}{2} \]

### Step 5: Calculate the new energy in the system
Now, we can calculate the total energy \( U' \) stored in the two capacitors after they are connected in parallel. The energy stored in the capacitors is given by:
\[ U' = \frac{1}{2} C_{\text{total}} (V')^2 \]
Substituting \( C_{\text{total}} \) and \( V' \):
\[ U' = \frac{1}{2} (2C) \left(\frac{V}{2}\right)^2 \]
\[ U' = \frac{1}{2} (2C) \left(\frac{V^2}{4}\right) \]
\[ U' = \frac{2C V^2}{8} = \frac{C V^2}{4} \]

### Step 6: Relate the new energy to the initial energy
Recall that the initial energy \( U \) was:
\[ U = \frac{1}{2} C V^2 \]
Thus, we can express \( U' \) in terms of \( U \):
\[ U' = \frac{C V^2}{4} = \frac{1}{2} \left(\frac{C V^2}{2}\right) = \frac{1}{2} U \]

### Final Answer
The total energy of the system of these capacitors after connecting the second capacitor in parallel is:
\[ U' = \frac{U}{2} \]