, 
3
4
3
r
q
Z e
ρ
π
=
+
Substituting for the charge density r  obtained earlier, we have
3
3
r
q
Z e
Z e
R
=
−
Gauss’s law then gives,
2
3
0
1
( )
;
4
Z e
r
E r
r
R
r
R
ε
 
 
=
−
<
 
 
 
 
π
The electric field is directed radially outward (ii) r > R:  In this case, the total charge enclosed by the Gaussian
spherical surface is zero since the atom is neutral Thus, from Gauss’s
law,
E (r) × 4 p  r 2 = 0  or  E (r) = 0;    r > R
At r = R, both cases give the same result: E = 0 SUMMARY
1