A ball of radius `R` carries a positive charge whose volume density depends only on a separation `r` from the ball's centre as `rho = rho_(0) (1 - r//R)`, where `rho_(0)` is a constant. Asumming the permittivites of the ball and the enviroment to be equal to unity find :
(a) the magnitude of the electric field strength as a function of the distance `r` both inside and outside the ball :
(b) the maximum intensity `E_(max)` and the corresponding distance `r _(m)`.

Explanation

(a) We assume the ball to be divided into infinite number of concentric thin shells of thickness dr. Let us assume such a shell at a radial distance `'r'`.
Volume of this shell = `dV=4pir^(2)dr`
Charge in the shell = `rhodV=4pir^(2)rhodr=4pir^(2)rho_(0)(1-(r)/(R))dr`
Net charge enclosed in the sphere of radius 'r'
`q=intrhodv=underset(0)overset(r)int4pirho_(0)r^(2)(1-(r)/(R))dr=4pirho_(o)((r^(3))/(3)-(r^(4))/(4R))`
Electric field at radial distance of 'r'
`E=(q)/(4piepsilon_(0)r^(2))=(1)/(4piepsilon_(0)r^(2))xx4pirho_(0)((r^(3))/(3)-(r^(4))/(4R))=(rho_(0)r)/(3epsilon_(0))(1-(3r)/(4R))`
Now for those points outside the sphere, we have to take into account the total charges contained in the sphere of radius R.
Total charge `Q=underset(0)overset(R)intrho_(0)(1-(r)/(R))4pir^(2)dr=(pirho_(0)R^(3))/(3)`
Electric field, using Gauss's law, at a point which is at a distance `r(gtR)` from the centre,
`intE.ds=(Q)/(epsilon_(0))impliesE=(Q)/(4piepsilon_(0)r^(2))=(rho_(0)R^(3))/(12epsilon_(0)r^(2))`

(b) Again let us consider the expression for electric field within the sphere of radius R:
`E=(rho_(0)r)/(3epsilon_(0))(1-(3r)/(4R))`.
For maximum electric field:
`(dE)/(dr)=0implies1-(6r)/(4R)=0impliesr=(2R)/(3)`
Putting, `r=(2R)/(3)`, value of maximum electric field, `E_(max)=(rho_(0)R)/(9in_(0))`,