Step 1
Given,
∅=ar2+b Where a and b are cont.
r = distance from center.
Step 2
we can solve this by using gauss's law = E⋅4πr2=qin∈0
Step 3
Solve,
given, ∅=ar2+b
Electrical filed E = d∅dr=−2ar
now putting the value of E on gauss's law
E⋅4πr2=qin∈0
(−2ar)⋅4πr2=qin∈0
−8π∈0ar3=qin
dqindr=−24π∈0ar2
now,
V=43πr3
dVdr=4πr3
Changingdensityη=dqindrxdrdV
=−24π∈0ar24πr2=−6∈0a
Hence, the charge inside the ball is
−6∈0a