howard (cah3459) – oldhomework 01 – Turner – (56705)
1
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001
10.0 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 fgure.
The two charges
q
at the bottom vertices oF
the triangle are equal. A Fourth charge 2
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 fgure
below.
7
.
4m
5m

5
.
7
μ
C
qq
2
μ
C
±ind
q.
Correct answer: 1
.
05392
μ
C.
Explanation:
Let :
Q
1
=

5
.
7
μ
C
,
Q
2
=
q,
Q
3
=
Q
4
=2
μ
C
,
a
=7
.
,
and
d
=

.
a
d
Q
1
Q
2
Q
3
2
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
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
=
±
√
3
2
a
+

d

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

d

r
24
=

d

³
d
2
+
a
2
4
.
Our Force equation becomes
0=
kQ
4
Q
1
±
√
3
2
a
+

d

²
2
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2
+
2
Q
2
±
a
2
4
+
d
2
²

d

³
a
2
4
+
d
2
.
Rearranging, we get
Q
2
=

Q
1
2

d

±
a
2
4
+
d
2
²
3
/
2
´
√
3
2
a
+

d

µ
2
=

(

5
.
7
μ
C)
2 (5 m)
×
¶
(7
.
4 m)
2
4
+(

5 m)
2
·
3
/
2
¸
√
3
2
(7
.
4 m) + 5 m
¹
2
=
1
.
05392
μ
C
.
Note:
Neither the sign nor the magnitude
of the charge
Q
4
(given in the problem as
2
μ
C) enter into this equation.
Actually, “the resultant force on
Q
4
is zero”
means that the resultant electric Feld is zero.
Because the electric Fled is independent of the
test charge
Q
4
, the answer is also independent
of the sign or magnitude of the charge
Q
4
.
002
(part 1 of 2) 10.0 points
Three pointcharges (

q
,+
q
, and

q
) are
placed at the vertices of an equilateral triangle
(see Fgure below).
a
60
◦

q

q
+
q
ˆ
ı
ˆ
The magnitude of the electric force on the
charge at the bottom lefthand vertex of the
triangle due to the other two charges is given
by
1.
±
±
F
±
=
1
2
√
3
kq
2
a
2
2.
±
±
F
±
=
2
a
2
correct
3.
±
±
F
±
=
√
2
3
2
a
2
4.
±
±
F
±
=
√
3
2
2
a
2
5.
±
±
F
±
=
2
√
3
2
a
2
6.
±
±
F
±
=
3
√
2
2
a
2
7.
±
±
F
±
=
1
2
2
a
2
8.
±
±
F
±
=
1
√
2
2
a
2
9.
±
±
F
±
=
√
2
2
a
2
10.
±
±
F
±
=
√
3
2
a
2
Explanation:

q

q
+
q
ˆ
ı
ˆ
In this case, each of these forces has a mag
nitude
F
21
=
F
31
=
2
a
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 Fall '08
 Turner
 Charge, Electric charge, Howard, KE, Q1 Q2, Q4 Q2 Q4

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