A capacitor of capacitance 900 μ F is charged by a 100 V battery. The capacitor is disconnected from the battery and connected to another uncharged identical capacitor such that one plate of uncharged capacitor connected to positive plate and another plate of uncharged capacitor connected to negative plate of the charged capacitor. The loss of energy in this process is measured as x × 10⁻ ² J. The value of x is _____________.

Solution

Physics sir · 4 steps

Charge is conserved when the capacitors share it, energy is not: compare the energy before and after.

1. Charge before. The charged capacitor holds Q = CV = 900 × 10⁻⁶ × 100 = 0.09 C. Its energy is 1/2CV² = 1/2 × 900 × 10⁻⁶ × 100² = 4.5 J.
2. Share the charge. Joining plus to plus and minus to minus puts the two capacitors in parallel, so the total capacitance is 2C. The charge 0.09 C is conserved, so the common voltage is Q/2C = 50 V.

   Δ U = 1/2CV² - 1/2(2C)(V/2)² = 1/4CV²

3. Energy after. The final energy is 1/2 × 1800 × 10⁻⁶ × 50² = 2.25 J. The loss is 4.5 - 2.25 = 2.25 J, lost as heat in the connecting wires and as radiation.
4. So. The loss is 2.25 J = 225 × 10⁻² J, so x = 225.

The common slip: 450. Taking the whole initial energy 1/2CV² = 4.5 J as the loss gives 450. Only half of it is lost, because the two capacitors still store 2.25 J at the end.

The idea to keep. When a charged capacitor shares its charge with an identical uncharged one, the voltage halves and exactly half of the stored energy is lost. Charge is conserved, energy is not.

Answer from the NTA final answer key. Solution by PrepWiser.