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RECITATION 2
11.
(a) With
a
understood to mean the magnitude of acceleration, Newton’s second and third laws lead to
(
29
(
29
7
2
7
2
6.3 10 kg
7.0 m s
4.9 10 kg.
9.0 m s
B
B
A
A
B
m a
m a
m


×
=
⇒
=
=
×
(b) The magnitude of the force on the first particle (A) is
(
29
(
29
(
29
2
2
2
2
2
B
A
7
9
2
2
2
11
so
(
)
6.3 10 kg
7.0 m s
8.99 10 N m
C
(0.0032 m)
7.1 10
C.
A
B
B
A
A
A
q
q
q
F
m a
k
k
r
r
q
F
q
k
→
→


=
=
=
=
=
×
=
×
⋅
=
×
Therefore, 
q
 = 7.1 x 10
–11
C
30.
We take rightwards to be the positive
x
direction and label the particles:
A
(for the one with the +2
e
charge),
B
(
e
),
C
(+
e
) and
D
(+4
e
).
(a)
We consider the net force on
A
.
B
A
F
→
r
points rightward (and therefore has a
positive x
component)
since
A
is “attracted” to
B
.
C
A
F
→
r
and
D
A
F
→
r
both point leftward (with a
negative x
component) since
A
is
“repelled” by
C
and
D
.
Thus, using
d
= 0.0200 m, the
x
component of the net force,
(
29
(
29
(
29
(
29
(
29
(
29
(
29
(
29
net
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This note was uploaded on 06/01/2008 for the course PHYS 232 taught by Professor Goksu during the Spring '08 term at Millersville.
 Spring '08
 Goksu
 Physics, Acceleration, Force

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