State Gauss 'law in electrostatics.
Consider an overall neutral sphere of radius R. This sphere has a point charge +Q at its centre and this positive charge is surrounded by a uniform density `rho` of negative charges up to a radius R.
Use Gauss, law to obtain expression for the electric field , of this sphere , at a point distant r, from its centre ,where
` r lt R, `
` r gt R `
Show that these two expressions give identical results, for the electric field, at r= R.

Explanation

Consider a sphere of radius R and having point charge +Q at its centre point. Negative charges density of sphere be `rho. ` then total negative charge on sphere ` =(4)/(3) pi R^(3) rho =-Q `
` rArr " " rho =-(3Q)/( 4 pi R^(3)) `
For a point `P_1 ` situated at a distance r (where r `lt R ) ` frm its centre, considering a sphere of radius .r. as the Gaussian surface , we have
` rArr " " phi _in =int oversetto (E) . oversetto (ds) = E . 4 pi r^(2) (1)/(in_0) ("charged enclosed") `
` " " (1)/( in_0) [Q +rho .(4)/(3) pi r^(3) ]`
`rArr " "E. 4 pi r^(2) =(1)/( in_0) [Q -(3Q)/( 4 pi R^(3)) .(4)/(3) pi r^(3) ] =(1)/(in_0) [Q -(Qr^(3))/( R^(3)) ]=(Q)/( in_0) [1-(r^(3))/( R^(3)) ]`
`rArr " " E= ( Q)/( 4 pi in _0) [(1)/( r^(2)) -(r)/( R^(3)) ]`
For r= R, we have ` E_("surface") =0.`
(b) For a point `P_2` situated at a distance r (where `r gt R) ` we again consider a spere of radius r as the Gaussian surface. Now total charges enclosed in the surface.
`=Q +rho .(4)/(3) pi R^(3) =Q- ( 3Q)/( 4 pi R^(3) ) .(4)/(3) pi R^(3) =Q -Q = 0 `
This shown that `phi_in =E. 4pi r^(2) =(1)/(in_0) ` (charge enclosed) `=(1)/(in_0) (0) =0 `
`rArr " " E= 0 `
and field at surface of sphere `E_("surface") =0` .