Physics Worksheet With Answers - Terry Honan - Blinn College Page 3
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8Chapter B - Problems
3
Solution to B.4
Half of the field lines of a point charge will pass below a horizontal plane containint the charge and half will pass above it. If a
charge sits above a horizontal plane then any half the field lines will eventually pass through the plane. Thus
Q
1
F
=
F
=
.
plane
total
2 ε
2
0
We can neglect the contribution of any horizontal field lines because they will not contribute to the flux. On a sphere the equator
doesn't contribute to its area.
For an infinite plane this is independent of the distance of the charge from the plane.
Problem B.5
A point charge Q sits at the center of a cube with sides of length . What is the electric flux through one face of the cube?
Solution to B.5
Q
The total flux through all six faces of the cube is, by Gauss's law, given by: F
=
. Since the charge is at the center of the cube
total
ε
0
we know, by symmetry, that each face will have the same flux. Thus F
= 6 F
and
total
face
Q
F
=
. Note that this answer is independent of the length of the sides.
face
6 ε
0
Problem B.6
(a) An insulating sphere with a 12 cm radius has a uniform charge of 216 mC. What is the charge inside spherical Gaussian surfaces
of radius 4 cm, 6 cm and 15 cm?
(b) A conducting sphere with a 12 cm radius has a net charge of 216 mC. What is the charge inside spherical Gaussian surfaces of
radius 4 cm, 6 cm and 15 cm?
Solution to B.6
For a sphere of radius R = 12 cm and uniform charge Q = 216 mC, we can find the charge inside a spherical surface of radius r by
considering the two cases:
(a) For r < R the fraction of the charge is the fraction of the volume.
4
3
p
r
V
r
3
inside
3
r < R ï Q
= Q
= Q
= Q J
N
inside
4
V
R
3
p R
total
3
This applies to the 4 cm and 6 cm cases:
3
4
r = 4 cm ï Q
= 216 J
N
= 8 mC
inside
12
3
6
r = 6 cm ï Q
= 216 J
N
= 27 mC
inside
12
For r > R all af the charge is inside the Gaussian surface.
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