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 of the capacitors and their energy storage.

### Step 1: Understanding the Initial Setup
Initially, we have one capacitor (let's call it C1) that is charged by a battery. This capacitor stores energy U when charged to a voltage V.

**Hint:** Recall the formula for energy stored in a capacitor: \( U = \frac{1}{2} C V^2 \).

### Step 2: Energy Stored in the First Capacitor
The energy stored in the first capacitor can be expressed in terms of charge (Q) and capacitance (C):
\[ U = \frac{Q^2}{2C} \]
Here, Q is the charge on the capacitor when it is charged by the battery.

**Hint:** Remember that the charge Q on a capacitor is related to voltage and capacitance by \( Q = C \cdot V \).

### Step 3: Disconnecting the Battery
Once the battery is disconnected, the charge Q remains on the capacitor since there is no path for it to leave. The energy stored in the capacitor remains U.

**Hint:** Think about how the disconnection of the battery affects the charge on the capacitor.

### Step 4: Connecting an Identical Capacitor
Now, we connect another identical capacitor (C2) in parallel with the first capacitor (C1). Since both capacitors are identical, they both have capacitance C.

**Hint:** Consider how capacitors behave when connected in parallel.

### Step 5: Finding the Equivalent Capacitance
For capacitors in parallel, the equivalent capacitance (C_eq) is the sum of their capacitances:
\[ C_{eq} = C_1 + C_2 = C + C = 2C \]

**Hint:** Use the formula for equivalent capacitance in parallel connections.

### Step 6: Calculating the New Energy Stored
Now, we need to calculate the new energy stored in the system with the two capacitors. The charge Q remains the same, but now the capacitance has changed to C_eq:
\[ U' = \frac{Q^2}{2C_{eq}} = \frac{Q^2}{2(2C)} = \frac{Q^2}{4C} \]

**Hint:** Substitute the equivalent capacitance into the energy formula.

### Step 7: Relating the New Energy to the Original Energy
We know from our earlier expression that:
\[ U = \frac{Q^2}{2C} \]
Thus, we can relate the new energy U' to the original energy U:
\[ U' = \frac{1}{2} \cdot \frac{Q^2}{2C} = \frac{U}{2} \]

**Hint:** This shows that the new energy stored is half of the original energy stored.

### Conclusion
The energy stored by both capacitors in the system after connecting the second capacitor is:
\[ U' = \frac{U}{2} \]

**Final Answer:** The energy stored by both capacitors of the system will be \( \frac{U}{2} \).