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Saturday, 31 October

JEE Main 2024 · PhysicsNumerical answerNumerical valueMediumCalculation

JEE Main 4 April 2024, Shift 2, Physics Q60

Question 60 of 90 in this shift, Physics question 30 of 30, Section B.

A parallel plate capacitor of capacitance 12.5 pFpF is charged by a battery connected between its plates to potential difference of 12.0 V. The battery is now disconnected and a dielectric slab (ϵr\epsilon_r = 6) is inserted between the plates. The change in its potential energy after inserting the dielectric slab is __________ ×\times 10−1210^{-12} J.

Official answer

750

NTA final key.

Chapter
Electrostatic Potential and Capacitance
Topic
Dielectrics
Idea tested
Dielectric with Battery Disconnected

Same idea in other shifts

Asked 3× in all
  1. 25 Jul 2022, Shift 2 · Q58Two parallel plate capacitors of capacity C and 3C are connected in parallel combination and charged to a potential difference 18 V. The…MediumNumerical value
  2. 17 Mar 2021, Shift 1 · Q25A parallel plate capacitor whose capacitance C is 14 pF is charged by a battery to a potential difference V = 12 V between its plates. The…HardNumerical value

Question text from the official JEE Main paper published by NTA; answer from the final answer key. Chapter, topic and idea tags, difficulty and skill are Lumi’s. Lumi analyses 108 of the 174 JEE Main shifts held from 2017 to 2026.