SERWAY 24.18
A solid sphere of radius 40.0 cm has a total positive charge of 26 μC uniformly distributed throughout its volume. Calculate the magnitude of the electric field (a) 0 cm,
(b) 10.0 cm, (c) 40.0 cm, and (d) 60.0 cm from the center of the sphere.
(a) Our task in part (a) is to find the E-field at the very center of the charge distribution.
Because the charge is uniformly distributed throughout the sphere, the center of the sphere has equal amounts of charge on all sides. I think it should be intuitive to you that, given equal charge on all sides, the E-field at the center of the sphere is 0. This is similar to the idea that, if you could stand at the center of the Earth, you would feel no gravitational ‘pull,’ because you’d be surrounded by equal amounts of mass on all sides, and their gravitational fields would cancel out. Regardless, let’s do the math.
Our Gaussian surface of choosing is so small that it doesn’t contain any volume; it’s a spherical surface with a radius = 0. Because the Gaussian sphere is infinitesimally small, it contains no charge. Thus, writing Gauss’s Law, we obtain:

E
 

E

 d A
 q
 enc
0


E
 d

A

0

0

EdA cos
 
E
 dA

0
0
E

0 at the center of the charge distribution.
(b) Now, we’re looking at a point 10 cm (0.100 m) from the center of the orange sphere. Our Gaussian surface of choosing needs to pass through this point and have the same symmetry as the charge distribution. So, we choose a Gaussian spherical shell, concentric with the charge distribution and having a radius of 0.100 m, as shown. Let’s write Gauss’s Law:

E
 

E

 d A
 q
 enc
0
Now, attack the nasty part first. Rewrite the dot product in terms of the vectors’ magnitudes.

EdA cos
  q
 enc
0
Regardless where you stand on the Gaussian surface, the electric field points radially outward, and the normal vectors describing the surface point radially outward – thus they are always parallel. For every infinitesimal area inside the integral, θ = 0 and cos(θ) = 1. Therefore:


EdA
 q enc

0
Also, regardless where you stand on the Gaussian surface, the charge distribution appears to be the same to you. Therefore, the magnitude of the E-field is constant at all points on the Gaussian surface. We can pull it out of the surface integral.
E
 dA
 q
 enc
0
Now, our task is to determine the amount of charge enclosed in our Gaussian surface.
Well, we know there is a total of 26 μC uniformly distributed throughout the orange sphere, but our Gaussian sphere only takes up a fraction of the volume of the overall orange sphere. To get the amount of charge enclosed, we take the total charge and multiply it by the ratio of the Gaussian volume to the orange sphere’s volume. q enc

V
Gaussian
Q
V
Orange Sphere

Q
4
3
  3 r
Gaussian
4
3
  3 r
Orange Sphere

Q r
3
Gaussian
3 r
Orange Sphere
So, Gauss’s Law becomes:
E
 dA

3 r
Gaussian
Q
3 r
Orange Sphere

0
But, the surface integral on the left-hand-side just represents the total surface area of the Gaussian surface,

  2
4 r
Gaussian

. Plugging this in…
E

4
  2 r
Gaussian


3 r
Gaussian
Q
3 r
Orange Sphere

0
Cancelling common factors…
Eqn. (1) E

Q
 r
Gaussian
4

0
3 r
Orange Sphere


 8 .
99

10
9
Nm
2
C
2




26 .
0

10


6 μC
0 .
400 m




3
0 .
100 m

E

3 .
65

10
5
N
C
(c) You can look at part (c) in a few equivalent ways. I’m choosing here to look at it exactly the same way I looked at part (b). The picture here is exactly the same, except

now, the radius of the Gaussian surface equals the radius of the orange insulator, and the Gaussian surface encompasses the total 26 μC of charge.
The same analysis applies as for part (b). Let’s rewrite equation (1), but now I’m going to use 0.400 m for the radius of the Gaussian surface:
E

Q
4

0
 r
Gaussian
3 r
Orange Sphere


 8 .
99

10
9
Nm
2
C
2




26 .
0

10


6 μC

0 .
400 m


3

0 .
400 m

E

1 .
46

10 6
N
C
(d) Now, the radius of the Gaussian surface is bigger than the radius of the orange insulator. All of the insulator’s charge is enclosed inside the Gaussian surface we choose:

E
 

E


 d A

E


 d A
 q enc

0
Q

0

EdA cos
 
Q

0

EdA

Q

0
E
 dA

E

4
  r 2


Q

0
Q

0
E
E


 8 .
99

10
9

Nm
2
C
2 k e


Q
2
 r
26 .
0

10

0 .
600 m

6

2
μC

E

6 .
49

10
5
N
C