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QJEEM19Ph1f7b: JEE Main 2019 Physics - Class 12 JEEM19Ph1f7b Solution: A parallel plate capacitor having capacitance 12 pF is charged…

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 :

Multiple Choice Options

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

Step-by-Step Verified Solution

Initial energy of capacitor

Ui = 12 v2c

= 12 × 120×12012 = 600 J

Since battery is disconnected so charge remain same.

Final energy of capacitor

Uf = 12v2c

= 12×120×12012×6.5 = 92

W + Uf = Ui

W = 508 J

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