This preview shows pages 1–3. Sign up to view the full content.
lew (dl9564) – hw 2 – Opyrchal – (121006)
1
This printout should have 14 questions.
Multiplechoice questions may continue on
the next column or page – fnd all choices
beFore answering.
001
10.0 points
Three point charges, two positive and one
negative, each having a magnitude oF 34
μ
C
are placed at the vertices oF an equilateral
triangle (22 cm on a side).
a
q
+q
+q
The
Coulomb
constant
is
8
.
98755
×
10
9
N
·
m
2
/
C
2
.
What is the magnitude
F
oF the electro
static Force on one oF the positive charges?
Correct answer: 214
.
661 N.
Explanation:
Let :
q
= 34
μ
C
,
a
= 22 cm
,
and
k
e
= 8
.
98755
×
10
9
N
·
m
2
/
C
2
.
The magnitude oF electrostatic Force is
F
=
k
e

q
1
 
q
2

r
2
The answer is independent oF the exact orien
tation oF the triangle, but let’s choose a par
ticular orientation (
i.e.
, choose a coordinate
system) to use For our calculations. Consider
the arrangement oF charges where the nega
tive point charge is positioned at the origin
and one oF the positive charges is located at
x
=
a
. Then consider the electrostatic Force
on the upper positive charge, which we call
A
;
(
i.e.
,
A
be in the frst quadrant). ±irst, note
that by symmetry and by the Fact that neg
ative charge exerts an attractive Force while
the positive charge exerts a repulsive Force,
F
y
= 0
.
This leaves us with only the
x
component oF
the Force on
A
. Using the principle oF super
position, and recalling that For an equilateral
triangle,
θ
= 60
◦
:
F
= 2
p
k
e
q
2
a
2
P
cos 60
◦
=
k q
2
a
2
since cos 60
◦
=
1
2
, so
F
=
k
e
q
2
a
2
=
(
8
.
98755
×
10
9
N
·
m
2
/
C
2
)
×
(
3
.
4
×
10
−
5
C
)
2
(0
.
22 m)
2
=
214
.
661 N
.
002
(part 1 oF 2) 10.0 points
Two charges are located on a horizontal axis.
The
Coulomb
constant
is
8
.
98755
×
10
9
N
·
m
2
/
C
2
.
1
.
9
μ
C
1
.
9
μ
C
p
1
.
5 m
2 m
2 m
Determine the electric feld at
p
on a verti
cal axis as shown in the fgure above. Up is
the positive direction.
Correct answer: 3278
.
66 V
/
m.
Explanation:
Let :
x
= 2 m
,
y
= 1
.
5 m
,
q
= 1
.
9
μ
C
,
and
k
e
= 8
.
98755
×
10
9
N
·
m
2
/
C
2
.
E
θ
q
q
y
x

x
This preview has intentionally blurred sections. Sign up to view the full version.
View Full Documentlew (dl9564) – hw 2 – Opyrchal – (121006)
2
The distance from each point charge to the
point of interest on the
y
axis is
r
=
r
x
2
+
y
2
=
R
(2 m)
2
+ (1
.
5 m)
2
= 2
.
5 m
.
Therefore, the electric Feld due to
one
of the
point charges is
E
=
k
e
q
r
2
=
(8
.
98755
×
10
9
N
·
m
2
/
C
2
)(1
.
9
×
10
−
6
C)
(2
.
5 m)
2
= 2732
.
22 V
/
m
.
Now, we need to vector add the contribu
tions from each charge. By symmetry, the
x
components cancel, and we only have twice
the
y
components left. To Fnd the
y
compo
nents, we note that the angle
θ
is
θ
= arcsin
p
1
.
5 m
2
.
5 m
P
= 36
.
8699
◦
,
so the total electric Feld is
E
tot
= 2 (2732
.
22 V
/
m) sin36
.
8699
◦
=
3278
.
66 V
/
m
,
where
E
tot
points upward.
This is the end of the preview. Sign up
to
access the rest of the document.
 Fall '08
 KEN
 Charge

Click to edit the document details