A capacitor is charged by a battery. The battery is removed and another identical uncharged capacitor is connected in parallel. The total electrostatic energy of resulting system:

Options

A. increases by a factor of 4
B. decreases by a factor of 2
C. remains the same
D. increases by a factor of 2

Answer

To solve the problem, we need to analyze the situation step by step.

### Step 1: Understand the Initial Conditions
We have one capacitor (let's call it C1) that is charged by a battery to a voltage V. The charge on this capacitor can be calculated using the formula:
\[ Q = C \cdot V \]
where C is the capacitance of the capacitor.

**Hint:** Remember that the charge on a capacitor is directly proportional to the capacitance and the voltage across it.

### Step 2: Calculate the Initial Energy
The electrostatic energy (Ui) stored in the charged capacitor can be calculated using the formula:
\[ U_i = \frac{1}{2} C V^2 \]

**Hint:** The energy stored in a capacitor is given by the formula involving capacitance and the square of the voltage.

### Step 3: Connect the Second Capacitor
After the battery is removed, we connect an identical uncharged capacitor (C2) in parallel with C1. The total capacitance of the system becomes:
\[ C_{total} = C1 + C2 = C + C = 2C \]

**Hint:** When capacitors are connected in parallel, their capacitances add up.

### Step 4: Determine the New Voltage
When the uncharged capacitor is connected, the charge (Q) will redistribute between the two capacitors. The total charge remains the same, but the voltage across both capacitors will change. The new voltage (Vc) across the capacitors can be calculated as:
\[ V_c = \frac{Q_{total}}{C_{total}} = \frac{Q + 0}{2C} = \frac{Q}{2C} \]
Since \( Q = CV \), we can substitute:
\[ V_c = \frac{CV}{2C} = \frac{V}{2} \]

**Hint:** The voltage across capacitors in parallel is the same, and the total charge is conserved.

### Step 5: Calculate the Final Energy
The final energy (Uf) stored in the system after connecting the second capacitor can be calculated as:
\[ U_f = \frac{1}{2} C_{total} V_c^2 = \frac{1}{2} (2C) \left(\frac{V}{2}\right)^2 \]
Substituting the values:
\[ U_f = \frac{1}{2} (2C) \left(\frac{V^2}{4}\right) = \frac{1}{2} C \cdot \frac{V^2}{2} = \frac{1}{4} C V^2 \]

**Hint:** The energy is calculated using the new total capacitance and the new voltage after connecting the second capacitor.

### Step 6: Compare Initial and Final Energies
Now, we can compare the initial and final energies:
- Initial energy: \( U_i = \frac{1}{2} C V^2 \)
- Final energy: \( U_f = \frac{1}{4} C V^2 \)

From this, we can see that:
\[ U_f = \frac{1}{2} U_i \]

This means the final energy is half of the initial energy.

### Conclusion
Thus, the total electrostatic energy of the resulting system decreases by half when the identical uncharged capacitor is connected in parallel.

**Final Answer:** The total electrostatic energy of the resulting system is \( \frac{1}{2} U_i \).