Question:
A parallel plate capacitor having capacitance 12 pF is charged by a battery to a potential difference of 10 V between its plates. The charging battery is now disconnected and a porcelain slab of dielectric constant 6.5 is slipped between the plates. The work done by the capacitor on the slab is :
A parallel plate capacitor having capacitance 12 pF is charged by a battery to a potential difference of 10 V between its plates. The charging battery is now disconnected and a porcelain slab of dielectric constant 6.5 is slipped between the plates. The work done by the capacitor on the slab is :
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With the battery disconnected the charge is constant; the energy falls by a factor K, and the loss in energy is the work done on the slab.
Updated On: Nov 5, 2026
- 508 pJ
- 692 pJ
- 560 pJ
- 600 pJ
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The Correct Option is A
Solution and Explanation
Initial energy \(U_i=\frac12CV^2=\frac12\times12\times10^{-12}\times100=600\) pJ.
Charge stays constant, \(C'=KC\), so \(U_f=\frac{Q^2}{2KC}=\frac{U_i}{K}=\frac{600}{6.5}=92.3\) pJ.
Work done by the capacitor on the slab \(=U_i-U_f=600-92.3=507.7\approx508\) pJ.
Answer: (1) 508 pJ
Charge stays constant, \(C'=KC\), so \(U_f=\frac{Q^2}{2KC}=\frac{U_i}{K}=\frac{600}{6.5}=92.3\) pJ.
Work done by the capacitor on the slab \(=U_i-U_f=600-92.3=507.7\approx508\) pJ.
Answer: (1) 508 pJ
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