Electrostatics 5 Question 13

14. A parallel plate capacitor having capacitance 12pF 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) 560pJ

(b) 508pJ

(c) 692pJ

(d) 600pJ

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Answer:

Correct Answer: 14. (b)

Solution:

  1. Energy stored in a charged capacitor is given by

U=12CV2=12Q2C

Here, C=12×1012 F and V=10 V.

U=12×12×1012×100U=6×1010 J

After insertion of slab, capacitance will be

C=KC and final energy, 

U=12Q2C=12Q2KCU=1KU=16.5×6×1010 J

So, energy dissipated in the process will be equal to work done on the slab, i.e.

ΔU=UU=116.5×6×1010 JΔU=5.56.5×6×1010 J5.08×1010 J or 508pJ



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