Answer, Key – Homework 8 – David McIntyre
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The due time is Central time.
Chapter 23 problems.
001
(part 1 of 1) 0 points
A solid conducting sphere is given a positive
charge
Q
.
How is the charge
Q
distributed in or on
the sphere?
1.
It is uniformly distributed throughout the
sphere.
2.
Its density decreases radially outward
from the center.
3.
It is concentrated at the center of the
sphere.
4.
Its density increases radially outward
from the center.
5.
It is uniformly distributed on the surface
of the sphere only.
correct
Explanation:
The charge distribution on conductors can
only be on the surface, and since on a spherical
surface every point is like any other surface
point, the charge distribution is uniform.
The electric field is normal to the surface of
a conductor.
The conductor is symmetrical
(since it is spherical), so the charge must be
uniform.
002
(part 1 of 1) 10 points
Three point charges are located at the vertices
of an equilateral triangle. The charge at the
top vertex of the triangle is given in the figure.
The two charges
q
at the bottom vertices of
the triangle are equal. A fourth charge 3
μ
C
is placed below the triangle on its symmetry
axis, and experiences a zero net force from
the other three charges, as shown in the figure
below.
7
.
3 m
5
.
1 m

3
.
8
μ
C
q
q
3
μ
C
Find
q .
Correct answer: 0
.
704413
μ
C.
Explanation:
Given :
Q
1
=

3
.
8
μ
C
,
Q
2
=
q ,
Q
3
=
q ,
Q
4
= 3
μ
C
,
a
= 7
.
3 m
,
and
d
=

5
.
1 m
.
a
d
Q
1
Q
2
Q
3
3
μ
C
θ
The force on
Q
4
is due to the Coulomb
forces from
Q
1
,
Q
2
, and
Q
3
. Because
Q
2
and
Q
3
have equal charge, the
x
components of
their forces cancel out (by symmetry). Thus
we only need to consider the
y
components of
the forces.
Coulomb’s law tells us
F
i
=
k
Q
i
Q
4
r
2
is the
i
th
force on
Q
4
from the
i
th charge. The forces
from
Q
2
and
Q
3
are equal to each other, and
opposite the direction of the force from
Q
1
,
since otherwise they could not cancel. Total
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Answer, Key – Homework 8 – David McIntyre
2
force on
Q
4
is
F
Q
4
=
k
Q
1
Q
4
r
2
14
+ 2
k
Q
2
Q
4
r
2
24
sin
θ
with
r
14
the distance between
Q
4
and
Q
1
,
r
24
=
r
34
the distance between
Q
4
and either
Q
2
or
Q
3
,
and
θ
indicated in the sketch above.
Remember that this force
F
Q
4
will be set equal
to zero since the problem tells us the forces
are in equilibrium.
Because
Q
1
,
Q
2
, and
Q
3
form an equilateral
triangle, of sides of length
a
, it can be seen
that
r
2
14
=
ˆ
√
3
2
a
+

d

!
2
and
r
2
24
=
a
2
4
+
d
2
.
Also,
sin
θ
=

d

r
24
=

d

q
d
2
+
a
2
4
.
Our force equation becomes
0 =
k Q
4
Q
1
‡
√
3
2
a
+

d

·
2
+
2
Q
2
‡
a
2
4
+
d
2
·

d

q
a
2
4
+
d
2
.
Rearranging, we get
Q
2
=

Q
1
2

d

h
a
2
4
+
d
2
i
3
/
2
h
√
3
2
a
+

d

i
2
=

(

3
.
8
μ
C)
2 (5
.
1 m)
×
h
(7
.
3 m)
2
4
+ (

5
.
1 m)
2
i
3
/
2
h
√
3
2
(7
.
3 m) + (5
.
1 m)
i
2
= 0
.
704413
μ
C
.
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 Spring '10
 TSOI
 Physics, Work

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