PHY2061
R. D. Field
Exam 1 Solutions
Page 1 of
8
February 11, 2003
PHY 2061 Exam 1 Solutions
Problem 1 (20 points):
Circle true or false for following (2 points each).
(a) (True or False)
The electrostatic force is much weaker than gravity.
(b) (True or False)
Electric field lines do not touch because they repel each other.
(c) (True or False)
Positively charged particles move from low to high potential.
(d) (True or False)
Electric field vectors are tangent to lines of constant potential.
(e) (True or False)
The vector function
z
z
y
x
x
y
z
y
x
E
ˆ
ˆ
ˆ
)
,
,
(
+
−
=
r
represents a possible
electrostatic field with magnitude given by
2
2
2
)
,
,
(
z
y
x
z
y
x
E
+
+
=
. Note that
0
≠
×
∇
E
r
r
and hence this vector function can not represent an electrostatic field.
(f) (True or False)
The scalar function V(x,y,z) = x
2
+y
2
+z
2
represents a possible electric
potential for an electrostatic field with magnitude given by
2
2
2
2
)
,
,
(
z
y
x
z
y
x
E
+
+
=
.
(g) (True or False)
The line integral of the electrostatic field around any closed loop is
always zero.
(h) (True or False)
The net electric flux through a closed surface is always positive or
zero.
(i) (True or False)
The divergence of the electric field is always zero.
(j) (True or False)
The electric field has energy but no mass.
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View Full DocumentPHY2061
R. D. Field
Exam 1 Solutions
Page 2 of
8
February 11, 2003
Problem 2 (20 points):
Four identical point charges
Q
are located at the corners of
a square with sides of length
L
that lies in the xyplane with
a
+4Q
point charge at its center and with its center at the
origin as shown in the
Figure
.
The total charge of the
system is zero.
Part A (4 points):
What is the electric potential,
V(z)
, on the zaxis a distance
z
from the origin? (
take V = 0 at infinity and express your
answer in terms of K, Q, L, and z
)
Answer:
+
−
=
2
/
1
1
4
)
(
2
2
L
z
z
KQ
z
V
Solution:
The electric potential at the point
P
is the superposition of the potentials from
each of the five charges as follows:
+
−
=
−
=
2
/
1
1
4
4
4
)
(
2
2
L
z
z
KQ
r
KQ
z
KQ
z
V
,
where I used
2
2
)
2
/
(
L
z
r
+
=
.
Part B (4 points):
Which of the following gives the electric potential,
V
, on the zaxis a distance
z = L
from
the origin
(
circle one answer and show your work and take V = 0 at infinity
)
(a)
2.2 KQ/L
(b)
4.0 KQ/L
(c)
5.7 KQ/L
(d)
0.82 KQ/L
(e)
4.7 KQ/L
(f)
0.18 KQ/L
(g)
5.4 KQ/L
(h)
none
Solution:
From part A we see that
L
KQ
L
KQ
L
z
z
KQ
z
V
L
z
734
.
0
2
/
3
1
1
4
2
/
1
1
4
)
(
2
2
≈
−
→
+
−
=
=
.
Part C (4 points):
What is the zcomponent of the electric field,
E
z
(z)
, on the zaxis a distance
z
from the
origin? (
express your answer in terms of K, Q, L, and z
)
Answer:
+
−
=
2
/
3
2
2
2
)
2
/
(
1
4
L
z
z
z
KQ
E
z
Solution:
The electric field is calculated from the electric potential as follows:
+
−
=
∂
∂
−
=
2
/
3
2
2
2
)
2
/
(
1
4
L
z
z
z
KQ
z
V
E
z
.
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 Spring '08
 FRY
 Physics, Electrostatics, Force, Gravity, Electric charge, R. D., R. D. Field

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