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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 placed between the plates. The work done by the capacitor on the slab is:

A
560 pJ
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B
508 pJ
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C
692 pJ
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D
600 pJ
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Solution

The correct option is B 508 pJ
Energy stored in a charged capacitor is given by

U=12CV2=Q22C

Here, C=12×1012 F; V=10Volts

Ui=12×12×1012×(100)

Ui=6×1010 J

After insertion of slab, final energy of capacitor

Uf=UK=6×10106.5

The energy dissipated during the process is

ΔU=UiUf

(ΔU)Lost=(116.5)×6×1010

(ΔU)Lost=508pJ

As we know that
work done on the slab = Energy dissipated

Work done on the slab =508 pJ.

Hence, option (b) is correct.
Why this question ?
Tip: When a dielectric is inserted inside a capacitor, the charge & final energy of the capacitor decreases. The work done on slab by capacitor dissipates in form of heat produced.

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