Answer :

Given, C1 = 600 pF = 600 x 10-12 F

V1 = 200 V


Energy stored in the capacitor,


U1 = (1/2) C1 (V1)2 = (1/2) × 600 × 10-12 × (200)2


= 12 × 10-6 J


When this charged capacitor is connected to another uncharged capacitor C2 ( = 600 pF) ,they will share charges, till potential differences across their plates become equal.


Total charge on the two capacitors,


q = C1V1 + C2V2 = 600 × 10-12 × 200 + 0


= 12 × 10-8 C


Total capacitance of the two capacitors,


C = C1 + C2 = 600 pF + 600 pF


= 1200 pF


= 1200 x 10-12 F


Therefore, common potential of the two capacitors,



Energy stored in the combination of the two capacitors,


U2 = (1/2)CV2 = (1/2) x 1200 x 10-12 x (100)2


= 6 x 10-6 J


Therefore, energy lost by the capacitor C1( in the form of heat and electromagnetic radiation),


U1 – U2 = 12 x 10-6 – 6 x 10-6 = 6 x 10-6 J


Rate this question :

How useful is this solution?
We strive to provide quality solutions. Please rate us to serve you better.
Try our Mini CourseMaster Important Topics in 7 DaysLearn from IITians, NITians, Doctors & Academic Experts
Dedicated counsellor for each student
24X7 Doubt Resolution
Daily Report Card
Detailed Performance Evaluation
caricature
view all courses
RELATED QUESTIONS :

What is the area NCERT - Physics Part-I

A spherical capacNCERT - Physics Part-I

A small sphere ofNCERT - Physics Part-I

A spherical capacNCERT - Physics Part-I

A cylindrical capNCERT - Physics Part-I

A. Find equivalenPhysics - Board Papers

An electrical tecNCERT - Physics Part-I

A parallel plate NCERT - Physics Part-I

Three capacitors NCERT - Physics Part-I