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
25
2
2
2
2
2
2
4
3.52
10
N.
2
3
A x
B
A x
C
A x
D
A x
e
e
e
e
e
e
F
F
F
F
k
k
k
d
d
d

→
→
→
=
+
+
=


= +
×
(b) We now consider the net force on
B
.
We note that
A
B
F
→
r
This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
This is the end of the preview.
Sign up
to
access the rest of the document.
 Spring '08
 Goksu
 Physics, Acceleration, Force, Electric charge, net force, F elec, qB r2 =k

Click to edit the document details