41.
(a) We integrate the volume charge density over the volume and require the result be equal to the total
charge:
dx
dy
dz ρ = 4π
R
0
dr r
2
ρ = Q .
Substituting the expression ρ = ρ
s
r/R and performing the integration leads to
4π
ρ
s
R
R
4
4
= Q
=
⇒ Q = πρ
s
R
3
.
(b) At a certain point within the sphere, at some distance r
o
from the center, the field (see Eq. 24-8
through Eq. 24-10) is given by Gauss’ law:
E =
1
4πε
0
q
enc
r
2
o
where q
enc
is given by an integral similar to that worked in part (a):
q
enc
= 4π
r
o
0
dr r
2
ρ = 4π
ρ
s
R
r
4
o
4
.
Therefore,
E =
1
4πε
0
πρ
s
r
4
o
R r
2
o
which (using the relation between ρ
s
and Q derived in part (a)) becomes
E =
1
4πε
0
π
Q
πR
3
r
2
o
R
and simplifies to the desired result (shown in the problem statement) if we change notation r
o
→ r.