A ball of radius R carries a positive charges whose volume density at a point is given as `rho=rho_(0)(1-r//R)`, Where `rho_(0)` is a constant and r is the distance of the point from the center. Assume the permittivies of the ball and the enviroment to be equal to unity.
(a) Find the magnitude of the electric field strength as a function of hte distance r both inside and outside the ball.
(b) Find the maximum intensity `E_("max")` and the corresponding distance `r_(m)`.

Answer

For `rltR, E=(kq)/(r^(2))=(rho_(0)r)/(3 epsilon_(0))[1-(3r)/(4R)]`,
For `rgtR`, E=(kq)/(r^(2))=(rho_(0)R^(3))/(12 epsilon_(0)r^(2))`

Explanation

Charge within radius r is
`q= int_(0)^(r )rho_(0) [1(-r)/(R)]4 pi r^(2)dr`
`=4 pi rho_(0)int_(0)^(r )(r^(2)-(r^3)/(R )) dr`
`=4 pi rho_(0)((r^3)/(3)-(r^4)/(4R))`…(i)
For `rltR`
`E=(kq)/(r^(2))=(rho_(0)r)/(3 epsilon_(0))[1-(3r)/(4R)]` …(ii)
For total charge, put r=R in eq.(i)
`Q=4 pi rho_(0)[(R^(3))/(3)-(R^(3))/(4)]=(4 pi rho_(0)R^(3))/(12)`
For `rgtR`,
`E=(kQ)/(r^(2))=(rho_(0)R^(3))/(12 epsilon_(0)r^(2))`...(iii)
At surface from both eqs, (ii) and (iii) put r=R,
`E_(0)=(rho_(0)R)/(12 epsilon_(0))`
Also, at `r=R//3`
`E=(rho_(0)R)/(12 epsilon_(0))`.
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