PHY2049 Spring 2010
Profs. P. Avery, A. Rinzler, S. Hershfield
Exam 1 Solution
1. Four parallel infinite sheets of charge spaced by 4 cm between adjacent sheets
are shown edgewise in the figure. Their surface charge densities are as indicated
with
σ
= 2
.
0
μ
C/m
2
. What is the magnitude of the electric field (in N/C) at
point P located midway between two sheets as shown? (Hint: Electric fields
are vectors that add as such.)
A
B
C
D
P
+
gid1
gid1
2
gid2
+
3
gid1
gid1
4
gid2
Answer:
4
.
5
×
10
5
Solution:
Sheet A produces an electric field of magnitude
σ/
(2
ǫ
o
) going away from it, which means the electric field
is pointing towards the bottom of the page at point P. Sheet B produces an electric field of magnitude 2
σ/
(2
ǫ
o
) going
towards it, which implies that the electric field is pointing towards the bottom of the page at point P. Similarly, at point
P the electric fields due to sheets C and D are 3
σ/
(2
ǫ
o
) toward the top of the page and 4
σ/
(2
ǫ
o
) toward the bottom of
the page. Adding these four fields together produces a net electric field of 4
σ/
(2
ǫ
o
) toward the bottom of the page.
2. Suppose we have an insulating spherical ball of uniform charge density
ρ
and radius
R
. At what radius or radii from the
center of the sphere is the electric field strength reduced by a factor of 16 from the electric field strength at the surface?
Answer:
R/
16 and 4
R
Solution:
Applying Gauss’s law, the flux through a sphere of radius
r < R
is
Φ = (
vector
E
·
ˆ
n
)4
πr
2
=
Q
enc
ǫ
o
=
1
ǫ
o
4
π
3
r
3
ρ
⇒
vector
E
·
ˆ
n
=
1
3
ǫ
o
rρ.
The flux through a sphere of radius
R > r
is
Φ = (
vector
E
·
ˆ
n
)4
πr
2
=
Q
enc
ǫ
o
=
1
ǫ
o
4
π
3
R
3
ρ
⇒
vector
E
·
ˆ
n
=
1
3
ǫ
o
R
3
r
2
ρ.
The electric field at the surface,
r
=
R
, is
vector
E
·
ˆ
n
=
ρR/
(3
ǫ
o
). To determine where the electric field is 1
/
16 of the value at
the surface, set
vector
E
·
ˆ
n
for the
r < R
and
r > R
cases equal to (1
/
16)
ρR/
(3
ǫ
o
) and solve for
r
.
3. A nonuniform electric field given by
vector
E
= (5
.
5
ˆ
i

2
.
1
ˆ
j
+ (4
.
6
z
2

3)
ˆ
k
) N/C
pierces a cube with sides 3 m, as shown in the figure. The cube has its rear
corner at the origin. What is the total charge inside the cube?
y
z
x
Answer:
+3
.
3 nC
Solution:
By Gauss’s law the net flux is equal to
Q
enc
/ǫ
o
. The flux through each face of the cube is (
vector
E
·
ˆ
n
)(3
m
)
2
, where
the normal vector points outwards. For the front face the normal component of the electric field is (
vector
E
·
ˆ
n
) = 5
.
5 N/C,
while for the back face it is (
vector
E
·
ˆ
n
) =

5
.
5 N/C. These two fluxes will cancel in computing the net flux. Similarly, the
flux for the right and left faces will cancel. Only the fluxes through the top and bottom faces do not cancel. The flux
through the top face (
z
= 3) is (4
.
6
·
3
2

3) 3
2
, and the flux through the bottom face is

(4
.
6
·
0
2

3) 3
2
. Combine these
to find the charge enclosed is
ǫ
o
4
.
6
·
3
4
.
This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
4. Members of a space mission discover a mysterious alien artifact consisting of a solid conducting sphere of radius 4 cm
floating inside a conducting shell with inner and outer radii 10 cm and 12 cm, respectively. Scientists measure the electric
This is the end of the preview.
Sign up
to
access the rest of the document.
 Spring '08
 Any
 Physics, Electron, Charge, Electric charge, NC

Click to edit the document details