A ball of radius `R` is uniformly charged with the volume density `rho`. Find the flux if the electric field strength vector across the balls section formed by the plane located at a distance `r_(0) lt R` from the centre of the ball.

Explanation

Let us consider a ring element of radius `x` and thickness`dx`, as shown in the fig.Now , flux over the considered element,
`d varphi = vec(E). vec(dS) = E_(r ) dS cos theta`
But `E_(r ) - (rho r)/(3 epsilon_(0))` from Gauss's theorem,
and `dS = 2pi x dx, cos theta = (r_(0))/(r )`
Thus `d Phi = (rho r)/(3 epsilon_(0)) 2pi xx dx (r_(0))/(r ) = (rho r_(0))/( 3 epsilon_(0)) 2pi x dx`
Hence sought flux
`sqrt(R^(2) - r_(0)^(2))`
`Phi = (2pi rho r_(0))/(3 epsilon_(0)) int_(0) x dx = (2pi rho r_(0))/(3 epsilon_(0)) ((R^(2) - r_(0)^(2)))/(2) = (pi rho r_(0))/(3 epsilon_(0)) (R^(2) - r_(0)^(2))`