Question 11 of 11intermediate🔧 ApplyNumerical3 marks

A 600pF capacitor is charged by a 200V supply. It is then disconnected from the supply and is connected to another uncharged 600 pF capacitor. How much electrostatic energy is lost in the process?

Correct Answer

The electrostatic energy lost in the process is 6×106 \times 10⁻⁶ J.

Exercise: EXERCISES | Q: 2.11 | (Page 80)
For More Understanding

Explanation

This question is identical in concept to Example 2.10 in the textbook. When a charged capacitor is connected to an uncharged capacitor of equal capacitance, the charge redistributes equally. The key insight from the textbook is that 'though no charge is lost; the final energy is only half the initial energy.' The remaining energy is lost as heat and electromagnetic radiation during the transient current flow.

Solution Steps

  1. Step 1: Calculate initial charge on the capacitor Q = CV = 600×10600 \times 10⁻¹² F × 200 V = 12×1012 \times 10⁻⁸ C

  2. Step 2: Calculate initial electrostatic energy stored Initial Energy = (1/2) CV² = (1/2) × 600×10600 \times 10⁻¹² × (200)² = (1/2) × 600×10600 \times 10⁻¹² × 40000 = 1.2×101.2 \times 10⁻⁵ J

  3. Step 3: Determine charge distribution after connection When connected to another uncharged 600 pF capacitor, by charge conservation, total charge Q remains same. Charge on each capacitor becomes Q/2 = 6×106 \times 10⁻⁸ C. The common potential V' = V/2 = 100 V.

  4. Step 4: Calculate final electrostatic energy Final Energy = 2 × (1/2) C(V')² = 2 × (1/2) × 600×10600 \times 10⁻¹² × (100)² = 6×106 \times 10⁻⁶ J (Alternatively: Final Energy = (1/2) × Initial Energy = 0.6×100.6 \times 10⁻⁵ J)

  5. Step 5: Calculate energy lost Energy lost = Initial Energy – Final Energy = 1.2×101.2 \times 10⁻⁵ – 0.6×100.6 \times 10⁻⁵ = 6×106 \times 10⁻⁶ J

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