Solutions for HandIn Set #2
PHYS 212 — Spring 2007
Problem A5 (Problem 2 on Efields program)
The signs are (1) positive; (2) negative; (3) negative;
(4) negative; (5) positive.
The way to figure this out:
bring the positive test charge very close to each num
bered charge, so that the field from this charge domi
nates.
If the electric field vector points away from the
numbered charge, then it is a positive charge.
If the
vector points toward the numbered charge, then it is a
negative charge.
Problem A6 — Addition of electric fields
P
c
P
b
P
a
q
q
The directions and relative magnitudes of the fields
from the two charges (
q
1
=
q
2
=
q
= 5
×
10

6
C) are
illustrated in the figure.
a) At
y
=

3 m the fields from both charges point in the
negative
y
direction.
E
=

kq
r
2
1
+
kq
r
2
2
ˆ
j
=

kq
1
3
2
+
1
9
2
ˆ
j
= (

5
.
55
×
10
3
N/C)
ˆ
j.
b) At
y
= 3 m the fields from the charges point in op
posite directions. Because the point is equidistant from
both charges both fields have the same magnitude, and
the vector sum is zero.
c) At
y
= 9 m the fields from both charges point in the
positive
y
direction.
E
=
kq
r
2
1
+
kq
r
2
2
ˆ
j
=
kq
1
3
2
+
1
9
2
ˆ
j
= (5
.
55
×
10
3
N/C)
ˆ
j.
Problem A7 — Determining
~
E
using integrals
x
y
dq
r
d
~
E
θ
θ
The magnitude of the field due to the infinitesimal
charge
dq
located at (0
, y
) is given by
dE
=
k dq
r
2
=
kλ dy
a
2
+
y
2
.
The direction of the
vector
d
~
E
is indicated in the figure.
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 Spring '08
 Ladd
 Charge, Electric charge, Electromagnetic field, dEy

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