A ball of radius R carries a positive charge whose volume density at a point is given by Where ρ 0 is a constant and r is distance from the centre. The value of maximum electric field due to this configuration is

							
(a) We assume the ball to be divided into infinite number of concentric thin shell of thickness dr. Let us assume such a shell at a radial distance r.
Volume of this shell =pdV=4πr 
2ρdr=4πr2
ρ0 (1− Rr)dr
Net charge enclosed in the sphere of radius r
q=∫pdv= 0∫r? 
4πρ02 (1− Rr?)dr=4πρ
0? ( 3r 3?  − 4Rr 4? )
Electric field at radial distance r
E= 4πε0 r
2q?  = 4πε 
0 r 21
×4πρ 0?( 3r 3?  − 4Rr 4 )= 3ε 0? 
 ρ 0? r (1− 4R3r)
Now for those points outside the sphere, we have to take into account the total charge contained in the sphere of radius R.
Total charge Q= 0∫R ρ 0?  (1− Rr?  )4πr 2 dr= 3πρ 0? R3
Electric field, using Gauss's laws, at a point which is at distance r(>R) from the centre;
∫E.ds= ε 0 Q? 
?E= 4πε 0
r 2Q? = 2ε 
0 r 2ρ 0? R3
(b) Again let us consider the expression for electric field within the sphere of radius R:
E= 3ε 0ρ 0 r?  (1− 4R3r)
For maximum electric field:
drdE?  =0?1− 4R6r
=0?r= 32R
?Putting, r= 3
2R
 , value of maximum electric field, E 
max? = 9∈ 0?ρ 0? R