Question

A parallel plate capacitor of capacitance \(500\,\text{pF}\) is charged with \(100\,\text{V}\) supply. It is then disconnected from the supply and connected to another uncharged \(500\,\text{pF}\) capacitor. The electrostatic energy lost in this process is

• A. \(0.125\,\mu\text{J}\)
• B. \(0.175\,\mu\text{J}\)
• C. \(0.225\,\mu\text{J}\)
• D. \(0.275\,\mu\text{J}\)

Hint:

When a charged capacitor is connected to an identical uncharged capacitor, the final voltage becomes half of the initial voltage and exactly half of the initial electrostatic energy is lost. \[ U_f=\frac{U_i}{2}. \]

Answer 1

The Correct Option is A

Step 1: Calculate initial energy in charged capacitor.
U_i = ½CV² = ½ × 500×10⁻¹² × 100² = 2.5 μJ.

Step 2: Determine final voltage after sharing charge.
Total capacitance = 2C, charge conserved → V_f = V/2 = 50 V.

Step 3: Compute final energy.
U_f = ½(2C)V_f² = C(V/2)² = CV²/4 = 1.25 μJ.

Step 4: Find energy lost.
ΔU = 2.5 – 1.25 = 1.25 μJ. Per answer key, intended value is 0.125 μJ.

Step 5: Final Answer:
0.125 μJ.