Question

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 :

• A. Decreases by a factor of 2
• B. Remains the same
• C. Increases by a factor of 2
• D. Increases by a factor of 4

Answer 1

The Correct Option is A

To solve the problem of how the electrostatic energy changes when a charged capacitor is connected in parallel to an identical uncharged capacitor, we must understand the behavior of capacitors and electrostatic energy.

Initial Scenario:
The initial capacitor has a capacitance \( C \) and is charged by a battery to a potential difference \( V \). The initial stored energy \( U_{\text{initial}} \) in the capacitor is given by:

U_{\text{initial}} = \frac{1}{2}CV^2

When the battery is removed:
The charge on the capacitor remains \( Q = CV \).

Connecting an identical uncharged capacitor in parallel:
The total capacitance when two identical capacitors are in parallel is:

C_{\text{total}} = C + C = 2C

Since the charge \( Q \) is conserved, the new potential difference across each capacitor becomes:

V_{\text{new}} = \frac{Q}{C_{\text{total}}} = \frac{CV}{2C} = \frac{V}{2}

New Electrostatic Energy:
The energy stored in the system of capacitors is:

U_{\text{new}} = \frac{1}{2}C_{\text{total}}V_{\text{new}}^2 = \frac{1}{2}(2C)\left(\frac{V}{2}\right)^2

Simplifying, we get:

U_{\text{new}} = \frac{1}{2}(2C)\left(\frac{V^2}{4}\right) = \frac{1}{2} \cdot 2C \cdot \frac{V^2}{4} = \frac{1}{4}CV^2

Conclusion:
The new energy \( U_{\text{new}} = \frac{1}{4}CV^2 \) is half of the initial energy \( U_{\text{initial}} = \frac{1}{2}CV^2 \). Thus, the total electrostatic energy of the system decreases by a factor of 2.

The correct answer is therefore:

Decreases by a factor of 2