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EEJ MAIN Physics QUESTION #6995
Question 1

A parallel plate capacitor of capacitance 12 pF is charged to 10 V and then disconnected from the battery. A porcelain slab of dielectric constant 6.5 is slipped between the plates. What is the work done by the capacitor on the slab?

  • 692 pJ
  • 508 pJ✔️
  • 560 pJ
  • 600 pJ
Correct Answer Explanation

Since the battery is disconnected, charge $Q$ is constant:

$Q = C_0 V_0 = 12 \times 10^{-12} \times 10 = 120\ \text{pC}$

Initial energy: $U_i = \frac{Q^2}{2C_0} = \frac{(120\times10^{-12})^2}{2\times12\times10^{-12}} = 600\ \text{pJ}$

New capacitance: $C = KC_0 = 6.5 \times 12 = 78\ \text{pF}$

Final energy: $U_f = \frac{Q^2}{2C} = \frac{(120)^2 \times 10^{-24}}{2\times78\times10^{-12}} \approx 92.3\ \text{pJ}$

Work done by capacitor on slab $= U_i - U_f = 600 - 92.3 \approx 508\ \text{pJ}$