Arun
Last Activity: 5 Years ago
(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
2
ρdr=4πr
2
ρ
0
(1−
R
r
)dr
Net charge enclosed in the sphere of radius r
q=∫pdv=
0
∫
r
4πρ
0
r
2
(1−
R
r
)dr=4πρ
0
(
3
r
3
−
4R
r
4
)
Electric field at radial distance r
E=
4πε
0
r
2
q
=
4πε
0
r
2
1
×4πρ
0
(
3
r
3
−
4R
r
4
)=
3ε
0
ρ
0
r
(1−
4R
3r
)
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−
R
r
)4πr
2
dr=
3
πρ
0
R
3
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
2
Q
=
12ε
0
r
2
ρ
0
R
3
(b) Again let us consider the expression for electric field within the sphere of radius R:
E=
3ε
0
ρ
0
r
(1−
4R
3r
)
For maximum electric field:
dr
dE
=0⟹1−
4R
6r
=0⟹r=
3
2R
Putting, r=
3
2R
, value of maximum electric field, E
max
=
9∈
0
ρ
0
R