Browse by Stream
QnA
Home
QnA
Go To Your Study Dashboard

Get Answers to all your Questions

header-bg qa
  • Home
  • NCERT
  • Show that the force on each plate of a parallel plate capacitor has a magnitude equal to (½) QE, where Q is the charge on the capacitor, and E is the magnitude of electric field between the plates. Explain the origin of the factor ½.

2.28 Show that the force on each plate of a parallel plate capacitor has a magnitude equal to \left (\frac{1}{2} \right ) QE, where Q is the charge on the capacitor, and E is the magnitude of electric field between the plates. Explain the origin of the factor \left (\frac{1}{2} \right )

Answers (1)

best_answer

Let

The surface charge density of the capacitor = \sigma

Area of the plate = A

NOW,

As we know,

Q=\sigma A\:and \: E=\frac{\sigma}{\epsilon _0}

When the separation is increased by \Delta x,

work done by external force= F\Delta x

Now,

Increase in potential energy :

\Delta u=u*A\Delta x

By work-energy theorem,

F\Delta x=u*A\Delta x

F=u*A=\frac{1}{2}\epsilon _0E^2A

putting the value of \epsilon _0

F=\frac{1}{2}\frac{\sigma}{E}E^2A=\frac{1}{2}\sigma AE=\frac{1}{2}QE

Origin of 1/2 lies in the fact that field is zero inside the conductor and field just outside is E, Hence it is the average value of E/2 that contributes to the force.

Posted by

Pankaj Sanodiya

View full answer

Crack CUET with india's "Best Teachers"

  • HD Video Lectures
  • Unlimited Mock Tests
  • Faculty Support
cuet_ads

Similar Questions

Trending Articles/News

anam-khan

CBSE Class 12 board exams 2025 to have 50% competency-based questions; no change in 10th
anam-khan

CBSE Class 12 board exams 2024 over; expected result date, trends of last 10 years
anam-khan

CBSE Class 12 accountancy board exam analysis 2024 out; paper rated easy, balanced

Latest Question