Verified Solution • 2025–26 Session moderate

QJEEM18Phad7c: JEE Main 2018 Physics - Class 12 JEEM18Phad7c Solution: A parallel plate capacitor of capacitance 90 pF is connected to …

A parallel plate capacitor of capacitance 90 pF is connected to a battery of emf 20 V. If a dielectric material of dielectric constant K = 5/3 is inserted between the plates, the magnitude of the induced charge will be :

Multiple Choice Options

  • A 0.9 n C
  • B 1.2 n C
  • C 0.3 n C
  • D 2.4 n C

Step-by-Step Verified Solution

Charge on Capacitor initially,

Qi = CV

After inserting dielectric of dielectric constant = K,

new capacitance, Qf = (KC) ∨\vee

∴   \therefore\,\,\, Induced charges on dielectric

= Qf −- Qi

= KCV −- CV

= (K −- ) CV

= (53−1)\left( {{5 \over 3} - 1} \right) ×\times 90 ×\times 10 −- 12 ×\times 20

= 1.2 ×\times 10 −- 9 C

= 1.2 nC

Related Learning Resources & Exam Preparation

JEE Main PYQ Question Bank
Browse year-wise authentic previous years' questions with solutions
Authentic CBT Mock Tests
Simulate official exam environment with live timer and percentile scoring