A ball of radius `R` carries a positive charge whose volume density depends according 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) Let us consider a sphere of radius `r lt R` then charge, inclosed by the considered sphere,
`q_("inclosed") = int_(0)^(r ) 4pi r^(2) d r rho = int_(0)^( r) 4pi r^(2) rho _(0) (1 - (r )/(R )) dr` (1)
Now, applying Gauss's theorem,
`E, 4pi r^(2) = (q_("inclosed"))/(epsilon_(0))`, (where `E`, is the projection of elctric field along the radail line.)
`(rho_(0))/(epsilon_(0)) int_(0)^(r ) 4pi r^(2) (1- (r )/(R )) dr`
or, `E_(r) = (rho_(0))/(3 epsilon_(0)) [r^(2) - (3r^(2))/(4R)]`
And for a point outside the sphere `r gt R`.
`q_("inclosed") = int_(0)^(R ) 4pi r^(2) d r rho_(0) (1- (r )/( R))` ( as there is no charge outside the ball)
Again from Gauss's theorem,
`E_(r) 4pi r^(2) = int_(0)^(R ) (4pi r^(2) dr rho_(0)(1 - (r )/( R)))/(epsilon_(0))`
or, `E_(r) = (rho_(0))/(r^(2) epsilon_(0)) [(R^(3))/(3) - (R^(4))/(4R)] = (rho_(0) R^(3))/(12 r^(2) epsilon_(0))`
(b) As-magnitude of electric field decreases with increasing `r` from `r gt R`, field will be maximum for `r lt R`. Now, for `E_(r)`, to be maximum,
`(d)/(dr) (r - (3r^(2))/(4R)) = 0` or `1 - (3r)/(2R) = 0` or `r = r_(m) = (2R)/(3)`
Hence `E_(max) = (rho_(0) R)/(9 epsilon_(0))`