A capacitor is charged by connecting a battery across its plates. It stores energy U. Now the battery is disconnected and another identical capacior is connected across it, then the energy stores by both capacitors of the system will be

Options

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

Answer

To solve the problem step by step, we will analyze the situation involving the capacitors and the energy stored in them.

### Step 1: Understand the Initial Condition
Initially, we have a single capacitor \( C \) connected to a battery with voltage \( V \). The energy \( U \) stored in this capacitor can be expressed using the formula:
\[ U = \frac{1}{2} C V^2 \]

### Step 2: Disconnect the Battery
After charging, the battery is disconnected. The capacitor retains the charge \( Q \) which can be calculated as:
\[ Q = C V \]
At this point, the voltage across the capacitor is still \( V \).

### Step 3: Connect an Identical Capacitor
Now, we connect another identical capacitor \( C \) which is initially uncharged. When two identical capacitors are connected in parallel, the total capacitance becomes:
\[ C_{\text{total}} = C + C = 2C \]

### Step 4: Calculate the New Voltage
Since the charge \( Q \) is conserved, the total charge \( Q \) will now be distributed across the two capacitors. The new charge on each capacitor will be:
\[ Q' = \frac{Q}{2} = \frac{C V}{2} \]
The new voltage \( V' \) across the combined capacitors can be found using the formula:
\[ V' = \frac{Q'}{C_{\text{total}}} = \frac{Q}{2C} = \frac{C V}{2 \cdot 2C} = \frac{V}{2} \]

### Step 5: Calculate the New Energy Stored
Now, we need to calculate the energy stored in the system of two capacitors. The energy \( U' \) stored in the two capacitors can be calculated using the new voltage:
\[ U' = \frac{1}{2} C_{\text{total}} (V')^2 = \frac{1}{2} (2C) \left(\frac{V}{2}\right)^2 \]
Substituting the values:
\[ U' = \frac{1}{2} (2C) \left(\frac{V^2}{4}\right) = \frac{1}{2} \cdot 2C \cdot \frac{V^2}{4} = \frac{C V^2}{4} \]

### Step 6: Relate New Energy to Initial Energy
Now, we can relate the new energy \( U' \) to the initial energy \( U \):
\[ U' = \frac{1}{2} C V^2 = \frac{U}{2} \]

### Conclusion
Thus, the energy stored by both capacitors in the system after connecting the second identical capacitor is:
\[ \text{Energy stored by both capacitors} = \frac{U}{2} \]