Home
Class 11
PHYSICS
A point charge q is placed at distance ...

A point charge q is placed at distance d from the centre of a circular disc of radiius R. Find electric flux through the disc due to that charge .

Text Solution

Verified by Experts

We know that total flux original from a point charge q in all directions `(q)/(in_(0))` . This flux is originated in a solid angle 4pi . In theh given case solid angle subtended by the cone formed by the disc at the point charge is `Omega = 2pi (1-cos theta)`
So, the flux of q which is passing through the surface of the disc is .
`phi = (q)/(in_(0)) (Omega)/(4pi) = (q)/(2 in_(0)) (1-cos theta)`
From the figure , `cos theta = (d)/(sqrt(d^(2) + R^(2))) ` so `phi = (q)/(2 in_(0)) { 1- (d)/(sqrt(d^(2) + R^(2)))}`
Promotional Banner

Topper's Solved these Questions

  • GAUSS.S LAW

    AAKASH SERIES|Exercise EXERCISE (LONG ANSWER QUESTIONS)|1 Videos
  • GAUSS.S LAW

    AAKASH SERIES|Exercise EXERCISE (VERY SHORT ANSWER QUESTIONS)|10 Videos
  • FRICTION

    AAKASH SERIES|Exercise ADDITIONAL PRACTICE EXERCISE PRACTICE SHEET (ADVANCED) (MORE THAN ONE CORRECT ANSWER TYPE QUESTIONS)|1 Videos
  • GRAVITATION

    AAKASH SERIES|Exercise ADDITIONAL PRACTICE EXERCISE (LEVEL-II) PRACTICE SHEET (ADVANCED)Integer Type Questions |6 Videos

Similar Questions

Explore conceptually related problems

A charge Q is placed at a distance of 4R above the centre of a disc of radius R. The magnitude of flux through the disc is phi . Now a henispherical shell of radius R is placed over the disc such that it forms a closed surface. The flux through the curved taking direction of area vector along outward normal as positive , is