Physics 21 Spring 2008
January 7, 2008
H
W

1
S
o
l
u
t
i
o
n
s
11. (HRW 213)
A particle of charge
is 12.0 cm distant from a second
particle of charge
. Calculate the magnitude of the electrostatic force
between the particles.
Solution:
The magnitude of the mutual force of attraction at
r
= 0.120 m is
()
66
12
92
2
22
3.00 10 C 1.50 10 C
8.99 10 N m C
2.81N.
(0.120 m)
qq
Fk
r
−−
××
==
×
⋅
=
12. (HRW 2110)
Three particles are fixed on an x axis. Particle 1 of charge
is at
, and particle 2 of charge
is at
. If their net electrostatic force on
particle 3 of charge
is to be zero, what must be the ratio
when particle 3 is at
(a)
and (b)
?
Solution:
(a) The individual force magnitudes (acting on
Q
) are, by Eq. 211,
00
11
44
/2
qQ
aa
πε
=
−
which leads to 
q
1
 = 9.0 
q
2
. Since
Q
is located between
q
1
and
q
2
, we conclude
q
1
and
q
2
are likesign. Consequently,
q
1
/
q
2
= 9.0.
(b) Now we have
3/2
=
−
which yields 
q
1
 = 25 
q
2
. Now,
Q
is not located between
q
1
and
q
2
, one of them must
push and the other must pull. Thus, they are unlikesign, so
q
1
/
q
2
= –25.
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 Spring '08
 Kim
 Atom, Charge, Force, Electric charge, HRW

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