0

Guest

Courses New

A parallel plate capacitor of plate area A and plate separation d is charged to a potential difference V and then the battery is disconnected. A slab of dielectric constant K is then inserted between the plates of the capacitor so as to fill the whole space between the plates. Find the work done on the system in the process of inserting the slab.

A parallel plate capacitor of plate area A and plate separation d is charged to a potential difference V and then the battery is disconnected. A slab of dielectric constant K is then inserted between the plates of the capacitor so as to fill the whole space between the plates. Find the work done on the system in the process of inserting the slab.

Grade:11

1 Answers

Arun
25758 Points
4 years ago
 
Initial energy – final energy = work done
And charge remains conserved = CV = A*epsilon*V/D
Therefore, W = ColdV2/2 – Q2/2Cnew= A*epsilon*V2/2D – A*epsilon*V2/2KD

Think You Can Provide A Better Answer ?

Provide a better Answer & Earn Cool Goodies See our forum point policy

Other Related Questions on Electrostatics

View all Questions »

ASK QUESTION

Get your questions answered by the expert for free

Answer and earn gift vouchers

Win Gift vouchers upto Rs 500/-

View courses by askIITians

Design classes One-on-One in your own way with Top IITians/Medical Professionals

Click Here Know More

Complete Self Study Package designed by Industry Leading Experts

Click Here Know More

Live 1-1 coding classes to unleash the Creator in your Child

Click Here Know More

a Complete All-in-One Study package Fully Loaded inside a Tablet!

Click Here Know More