A ball of radius R carries a positive charge throughout its volume, such that the volume density of charge depends on distance r from the ball's centre as `rho = rho _(0) ( 1- ( r )/®)`, where `rho_(0)` is a constant. Assuming the permittivity of ball to be one , find magnitude of electric field as a function of distance r, both inside and outside the ball.
Strategy `:` The field has a spherical symmetry . For a point outside the ball `( r gt R )`

`phi =oint vec(E). bar(dA) = E xx 4 pi r^(2)`
By Gauss law , `phi= ( Q)/( epsilon_(0))`
`implies E_(out) = (Q)/( 4pi epsilon_(0)r^(2))`, where Q is total charge
For a point inside the ball ` ( r lt R )`
` phi = oint vec(E) . bar(dA) = E xx 4 pi r^(2)`
By Gauss law, `phi = ( q_(enc))/(epsilon_(0))`
`implies E_("in")= ( q_(enc))/( 4pi epsilon_(0)r^(2))`
where ` q_(enc)= ` charge in a sphere of radius `r ( lt R)`
To find the enclosed charge , consider a spherical shell of radius r and thickness dr
Its volume `dV = 4pi r^(2) dr`
Charge `dq = rho dV`
`implies dq= rho _(0) (1-(r )/(R)) 4pi r^(2) dr`
`implies q= int _(0)^(r ) rho _(0) ( 1- (r)/( R )) 4pi r^(2) dr`
`= rho _(0)4pi [ int_(0)^(r ) r^(2) dr - int_(0)^(r ) (r^(3))/(R) dr ]= rho _(0) 4pi [ ( r^(3))/( 3) - (r^(4))/( 4R)]`

Explanation

Let us now find values of inside and outside the sphere.
(i) Charge Q enclosed by the outside Gaussian sphere can be found by replacing r by R in the above expression .
`implies Q = rho _(0)4pi [ ( R^(3))/( 3) - (R^(4))/( 4R)]= ( rho _(0) 4pi R^(3))/( 12)`
`implies E_(out)= ( rho_(0)R^(3))/(12epsilon_(0)r^(2))`
(ii) Also, charge q enclosed inside a sphere of radius ` r( lt R )` is
` q = rho _(0)4pi [(r^(3))/(3) - ( r^(4))/( 4R)]`
`implies E _("in")= (-rho_(0))/(epsilon_(0)r^(2))[(r^(3))/(3) -(r^(4))/(4R)] = (rho_(0)(4rR-3r^(2)))/(12epsilon_(0)R)`