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.