Problem
4.
A 2-g ping-pong ball rubbed against a wool jacket acquires a net positive charge of
1
μ
C
. Estimate the
fraction of the ball’s electrons that have been removed.
Solution
If half the ball’s mass is protons, their number (equal to the original number of electrons) is
1 g
=
m
p
. The
number of electrons removed is
1
μ
C
=
e
,
so the fraction removed is
(
1
μ
C
=
e
)
(
1
g
=
m
p
)
=
10
!
6
C
"
1
.
67
"
10
!
24
g
1
.
6
"
10
!
19
C
"
1 g
=
1
.
04
"
10
!
11
(a hundred billionth).
Problem
6.
Find the ratio of the electrical force between a proton and an electron to the gravitational force between
the two. Why doesn’t it matter that you aren’t told the distance between them?
Solution
At all distances (for which the particles can be regarded as classical point charges), the Coulomb force is
stronger than the gravitational force by a factor of:
F
elec
F
grav
=
ke
2
r
2
!
"
#
$
%
&
r
2
Gm
p
m
e
!
"
#
$
%
&
=
(9
!
10
9
N
"
m
2
/C
2
)(1.6
!
10
#
19
C)
2
(6.67
!
10
#
11
N
"
m
2
/kg
2
)(1.67
!
10
#
27
kg)(9.11
!
10
#
31
kg)
º
2.3
!
10
39
.
The spacial dependence of both forces is the same, and cancels out.
Problem
9.
Two charges, one twice as large as the other, are located 15 cm apart and experience a repulsive force
of 95 N. What is the magnitude of the larger charge?
Solution
The product of the charges is
q
1
q
2
=
r
2
F
Coulomb
=
k
=
(0.15 m)
2
(95 N)
=
(9
!
10
9
N
"
m
2
/C
2
)
=
2.38
!
10
#
10
C
2
.
If one charge is twice the
other,
q
1
=
2
q
2
, then
1
2
q
1
2
=
2.38
!
10
"
10
C
and
q
1
= ±
21.8
μ
C.
Problem
11. A proton is on the
x-
axis at
x
=
1.6 nm.
An electron is on the
y
-axis at
y
=
0.85 nm.
Find the net
force the two exert on a helium nucleus (charge
+
2
e
) at the origin.
Solution
A unit vector from the proton’s position to the origin is
!
î
,
so the Coulomb force of the proton on the
helium nucleus is
F
P,He
=
k
(
e
)(2
e
)(
!
î
)
=
(1.6 nm)
2
=
!
0.180
î
nN.
(Use Equation 23-1, with
q
1
for the
proton,
q
2
for the helium nucleus, and the approximate values of
k
and
e
given.) A unit vector from the
electron’s position to the origin is
!
ˆ
j
, so its force on the helium nucleus is
F
e,He
=
k
(
!
e
)(
2
e
)(
!
ˆ
j
)
=
(
0
.
85 nm
)
2
=
0
.
638
ˆ
j
nN
.
The net Coulomb force on the helium nucleus is the sum

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