×

A ball of radius $$R$$ is uniformly charged with the volume density $$\rho$$. Find the flux of the electric field strength vector across the ball’s section formed by the plane located at a distance $$r < R$$ from the centre of the ball.

Note by Harsh Shrivastava
6 months, 1 week ago

Sort by:

That's easy......The q means that take a plane at a distance r from the centre and pass it through the sphere.The plane intercepts the sphere and forms a circular cross section.We need to find the flux of the electric field through this area.As usual move up a distance s and DX and rotate it to get a circular strip of area $$dA= 2πxdx$$.Now $$d(\phi)=EdAcos\alpha$$.Where $$cos\alpha$$=$$r/√(r^2+x^2)$$.$$E=kQ/R^3*√r^2+x^2$$ as its an internal point.So the req eq is $$d(\phi)=kQ/R^3*√(r^2+x^2)*2πxdx*r/√(r^2+x^2)$$.Now integrate from $$x=0$$to $$x=√(R^2-r^2)$$.And $$Q=\rho *4/3πR^3$$.To get the ans as $$(1/3)\rho rπ/\epsilon(R^2-r^2)$$

- 6 months, 1 week ago

Oh thanks a lot, and upvoted!

- 6 months, 1 week ago

- 6 months, 1 week ago

And upvote.it..ha ha....just joking...

- 6 months, 1 week ago

×