Verified Solution • 2025–26 Session moderate

QJEEM20Ph7e92: JEE Main 2020 Physics - Class 12 JEEM20Ph7e92 Solution: A 60 pF capacitor is fully charged by a 20 V supply. It is then …

A 60 pF capacitor is fully charged by a 20 V supply. It is then disconnected from the supply and is conneced to another uncharged 60 pF capacitor in parallel. The electrostatic energy that is lost in this process by the time the charge is redistributed between them is (in nJ) _____

Step-by-Step Verified Solution

JEE Main 2020 (Online) 7th January Evening Slot Physics - Capacitor Question 105 English Explanation
Ui = 12CV02{1 \over 2}CV_0^2

Uf = 12×2C×(V02)2{1 \over 2} \times 2C \times {\left( {{{{V_0}} \over 2}} \right)^2}

Δ\Delta E = 12CV02{1 \over 2}CV_0^2 - 12×2C×(V02)2{1 \over 2} \times 2C \times {\left( {{{{V_0}} \over 2}} \right)^2}

= CV024{{{CV_0^2} \over 4}}

= 14×60×10−12×4×102{1 \over 4} \times 60 \times {10^{ - 12}} \times 4 \times {10^2}

= 6 ×\times 10-9 = 6 nJ

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