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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 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 pJEnergy stored in a charged capacitor is given by U=12CV2=Q22C Here, C=12×10−12 F; V=10Volts ⇒Ui=12×12×10−12×(100) ∴Ui=6×10−10 J After insertion of slab, final energy of capacitor Uf=UK=6×10−106.5 The energy dissipated during the process is ΔU=Ui−Uf ⇒(ΔU)Lost=(1−16.5)×6×10−10 ∴(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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