A system consits of a uniformly charged sphere of radius R and a surrounding medium filled by a charge with the volume density `rho=alpha/r`, where `alpha` is a positive constant and r is the distance from the centre of the sphere. Find the charge of the sphere for which the electric field intensity E outside the sphere is independent of R.

Answer

`alpha//2 epsilon_(0)`

Explanation

Consider a spherical surface of radius r(ltR) having center at the centre of the ball. If E is the magnitude of electric field
strength at the surface, then the electric flux through this surface
is
`int_(s) vecE*dvecS = int_(s) EdS cos0^@ = Eint_(s) dS = E4pir^2`
By Gauss theorem,
`int_(s) vecE*d vecS = 1/epsilon_0 (Q_(encl osed))`
If q is charge on ball, then
`Q_(en clo sed) = q + int_(R)^(r) rho dv = q+ int_(R)^(r) alpha/x * 4pix^2dx`
` = q + 2pialpha (r^2-R^2)`
From Eq. (i), we get
`E4 pir^2 = 1/epsilon_0 [q+2pialpha(r^2-R^2)]`
or `E = 1/(4piepsilon_0r^2) [ q+2pialpha(r^2-R^2)]`
`alpha/(2epsilon_0) + 1/epsilon_0 (q/(4pir^2)-alpha/2 R^2/r^2)`
This will be independent of r if the second term on RHS is zero, i.e.,
`q/(4pir^2) - (alphaR^2)/(2r^2) = 0 or q = 2pialphaR^2` ....... (iii)
Then the electric field strength E will be `alpha//2epsilon_0`.