Phy 122 – Assignment 1
A.
1. Opposites attract, so the electron is pulled toward the origin:
The field’s direction is that of the force a positive charge would feel:
(It’s a property of the point in space, not of what is located there. It’s
similar with gravity: Earth’s gravitational field in a certain place is the
same whether an elephant or a bread crumb is located there.)
2. Coulomb’s law gives the force between point-like charges, or charged spheres. With a
different charge distribution, like these plates, E would be given by a different formula.
3. Convert to metric: (1.0 x109
Electric field: E
r
2
k
q
E
k
lb
C
)
1N
N
4.444x109
.225lb
C
q
r2
9
(8.988x10 )
(26)(1.602x10
19
9
4.444x10
)
= 8.424x10
-18
-9
Take square root: r = 2.90 x 10 m ans.

B. 1. Field lines point away from + charges, and toward -. So, since E points toward it,
this is a negative charge.
E
k
q
r
2
q
Er 2
Answer: -2.78x10
( 40)(.252 )
9
k
8.988x10
-10
2.78x10
10
C
C
2. Field from A: 40.0 N/C, south (given)
Field from B: E
k
q
(8.988x109 )
r2
Total: 40 + 10 = 50.0 N/C ans.
2.78x10
10
(.50m) 2
10.0 N / C , south
C. a.
Left.
(Opposites attract, so the electron is pulled toward the ball.)
F
b. E
1.98 10
11
N

1.24 108 N / C , Right. ( E ’s direction is defined with a
q 1.60 10 19 C
positive charge in mind. A positive charge at the electron’s location would be repelled by
the +2.7 μC.)

c. Same answer as (b). E is a property of the point in space, not of what is located there.
D.
F1 is the force that q feels from q1. F2 is the force that q feels from q2. The directions
come from the fact that opposite charges attract: q is pulled toward q1, and q is pulled
toward q2. The magnitude of each force comes from Coulomb’s law:
F1
k
F2
k
q1 q
r2
q2 q
r2
8.988 x109
8.988 x109
(5.00 x10 6 )(5.00 x10 6 )
(1.00m) 2
(3.00 x10 6 )(5.00 x10 6 )
(1.50m) 2
0.2247 N
0.0599 N
Add them, remembering to call right positive and left negative:
(-.2247) + (.0599) = -.1649
Answer: .165 N to the left
E.