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Physics 112
Homework 1 (solutions)
(2004 Fall)
Solutions to Homework Questions 1
Chapt15, Problem-1:
A 4.5 x 10–9 C charge is located 3.2 m from a –2.8 x 10–9 C charge. Find
the electrostatic force exerted by one charge on the other.
Solution:
Since the charges have opposite signs, the force is !attractive .
The magnitude so the force is given by Coulomb’s Law, so making the substitutions we get
F=
k e q1q2
r
2
(
)(
)
)9
)9
#
N " m 2 & 4.5 ! 10 C 2 .8 ! 10 C
"8
= % 8.99 ! 109
= 1.1! 10 N
(
2
2
C
$
'
(3.2 m )
Chapt15, Problem-3:
An alpha particle (charge = +2.0e) is sent at high speed toward a gold
nucleus (charge = +79e). What is the electrical force acting on the alpha particle when it is 2.0 x 10–14 m
from the gold nucleus?
Solution:
Since the charges have opposite signs, the force is !repulsive
The magnitude so the force is given by Coulomb’s Law, so making the substitutions we get
F=
k e( 2 e ) (79 e )
r2
(
)19
2
#
& (158 ) 1.60 !10 C
9 N "m
= % 8.99 ! 10
(
2
C2 '
$
2 .0 !10 )14 m
(
)
)
2
= 91 N
( repulsion )
Chapt15, Problem-5:
The nucleus of 8Be, which consists of 4 protons and 4 neutrons, is very
unstable and spontaneously breaks into two alpha particles (helium nuclei, each consisting of 2 protons and
2 neutrons). (a) What is the force between the two alpha particles when they are 5.00 x 10–15 m apart, and (b)
what will be the magnitude of the acceleration of the alpha particles due to this force? Note that the mass of
an alpha particle is 4.0026 u.
Solution:
Since the charges have opposite signs, the force is !repulsive
The magnitude so the force is given by Coulomb’s Law, so making the substitutions we get
k e ( 2e )2 #
N "m2&
((
(a) F =
= %% 8.99 !10 9
r2
C2 '
$
* #
)19 C & 2 , 4 $ 1.60 !10
' //
,+
.
= 36.8 N
2
# 5.00 ! 10)15 m &
$
'
(b) The mass of an alpha particle is m = 4.0026 u , where u = 1.66 ! 10"27 kg is the unified mass unit.
Applying Newton’s 2nd law, the acceleration of either alpha particle is then
a=
F
36.8 N
27
2
=
= 5.54 !10 m s
"27
m 4.0026 1.66 !10 kg
(
)
Of course from Newton’s 3rd law, both alpha particles experience the same force, and hence undergo the
same acceleration.
1
Physics 112
Homework 1 (solutions)
Chapt15, Problem-7:
(2004 Fall)
Suppose that 1.00 g of hydrogen is separated into electrons and protons.
Suppose also that the protons are placed at Earth’s North Pole and the electrons are placed at the South
Pole. What is the resulting compressional force on Earth?
Solution:
1.00 g of hydrogen contains Avogadro’s number of atoms, each containing one proton and one electron. Thus,
each charge has magnitude q = N A e . The distance separating these charges is r = 2 RE , where RE is the
Earth’s radius. Thus applying Coulomb’s Law,
F=
(
k e NA e
2
)
2
(2 R E )
[(
)(
23
)19
#
N " m 2 & 6.02 !10 1.60 ! 10 C
= % 8.99 ! 109
(
2
C2 '
$
4 6.38 ! 106 m
(
Chapt15, Problem-8:
2
)]
5
= 5.12 !10 N
)
An electron is released a short distance above Earth’s surface. A second
electron directly below it exerts an electrostatic force on the first electron just great enough to cancel the
gravitational force on it. How far below the first electron is the second?
Solution:
The magnitude of the repulsive force between electrons must equal the weight of an electron, Thus by
using Coulomb;’s Law and Newton’s 2nd Law (applied to gravity), we have
ke e2 r 2 = m e g
so rearranging this expression, and then making the substitutions, we get
ke e2
r=
=
me g
9
2
2
2
(8.99 !10 N" m C )(1.60 !10 C )
(9.11 !10 kg )( 9.80 m s )
#19
2
#31
= 5.08 m
Chapt15, Problem-11:
Three charges are arranged
as shown in Figure P15.11. Find the magnitude and direction
of the electrostatic force on the charge at the origin.
Solution:
In the sketch to the right, FR is the resultant of the
forces F6 and F3 that are exerted on the charge at
the origin by the 6.00 nC and the –3.00!nC charges
respectively. Applying Coulomb’s Law to each, we get
F6
(
)(
)
(
)(
)
)9
)9
#
N" m2 & 6.00 !10 C 5.00 !10 C
F6 = % 8.99 !109
C2 ('
$
(0.300 m )2
5.00 nC
!
FR
)9
)9
2
#
& 3.00 ! 10 C 5.00 ! 10 C
9 N" m
F3 = % 8.99 !10
= 1.35 ! 10)5 N
(
2
C2 '
$
(0.100 m )
FR =
or
(F6 )2 + ( F3 )2 = 1.38 !10 "5 N
# F3 &
( = 77.5° ,
$ F6 '
at ! = tan "1 %
"5
FR = 1.38 !10 N at 77.5° below " x axis
2
0.300 m
0.100 m
F3
-3.00 nC
= 3.00 ! 10)6 N
From the Superposition Principle, the resultant is
6.00 nC
Physics 112
Chapt15, Problem-13:
Homework 1 (solutions)
(2004 Fall)
Three point charges are located
at the corners of an equilateral triangle as in Figure P15.13.
Calculate the net electric force on the 7.00 µC charge.
Solution:
The forces on the 7.00 µC charge are shown in the sketch to
the right. Applying Coulomb’s Law to calculate each force,
we get
(
)(
2
7.00 ! 10)6 C 2.00 ! 10)6 C
#
9 N "m &
F1 = % 8.99 ! 10
2
C2 ('
$
(0.500 m )
= 0.503 N
2
7.00 !10 )6 C 4.00 ! 10)6 C
#
9 N" m &
F2 = % 8.99 !10
2
C2 ('
$
(0.500 m )
(
)(
)
y
)
+ 7.00 µC
F2
= 1.01 N
60.0°
x
–
+
2.00 µC 0.500 m -4.00 µC
From the superposition principle, we known
!Fx = ( F1 + F2 ) cos 60.0° = 0.755 N ,
and
F1
!Fy = ( F1 " F2 ) sin 60.0° = "0.436 N
So the resultant force on the 7.00 mC charge is
$ #Fy '
= 0.872 N at ! = tan "1 &
) = "30.0° ,
% #Fx (
or FR = 0.872 N at 30.0° below the + x axis
FR =
(!Fx )2 + ( !Fy )
2
Chapt15, Problem-16:
A charge of 6.00x10–9 C and a charge of –3.00x10–9 C are separated by
a distance of 60.0 cm. Find the position at which a third charge of 12.0 x 10–9 C can be placed so that the
net electrostatic force on it is zero.
Solution:
The required position is shown in the
sketch to the right. Note that this places q
closer to the smaller charge, which will
allow the two forces to cancel. Applying Coulmob’s
Law, and requiring that F6 = F3 gives
k e (6.00 nC ) q
(x + 0.600 m )
2
=
6.00 nC
F3
0.600 m
k e ( 3.00 nC) q
2
, or 2 x 2 = ( x + 0.600 m )
x2
Solving for x gives the equilibrium position as
x=
0.600 m
2!1
q
–3.00 nC
= 1.45 m beyond the - 3.00 nC charge
3
x
F6
Physics 112
Homework 1 (solutions)
Chapt15, Problem-19:
(2004 Fall)
An airplane is flying through a thundercloud at a height of 2 000 m.
(This is a very dangerous thing to do because of updrafts, turbulence, and the possibility of electric
discharge.) If there are charge concentrations of +40.0 C at height 3 000 m within the cloud and –40.0 C at
height 1 000 m, what is the electric field E at the aircraft?
Solution:
We shall treat the concentrations as point charges. Then, the resultant field consists of two contributions,
one due to each concentration.
The contribution due to the positive charge at 3000 m altitude is
2
#
& ( 40.0 C)
9 N" m
5
E+ = ke 2 = % 8.99 !10
(
2
2 = 3.60 !10 N C
r $
C ' ( 1000 m )
q
(downward)
The contribution due to the negative charge at 1000 m altitude is
E ! = ke
2
$
' ( 40.0 C)
9 N# m
=
8.99
"10
= 3.60 "105 N C
&
)
2
2
r %
C ( ( 1000 m ) 2
q
(downward)
From the Superposition Principle, the resultant field is then
5
E = E+ + E ! = 7.20 !10 N C ( downward )
Chapt15, Problem-27:
In Figure P15.27,
determine the point (other than infinity) at which
the total electric field is zero.
Solution:
If the resultant field is zero, the
+y
contributions from the two charges must
r1 = d
1.0 m
be in opposite directions and also have
+x
equal magnitudes. Choose the line
connecting the charges as the x-axis, with
q1 = –2.5 µC
q2 = 6.0 µC
E2
E1
the origin at the –2.5!µC charge. Then, the
two contributions will have opposite
r2 = 1.0 m + d
directions only in the regions x < 0 and
x > 1.0 m . For the magnitudes to be equal, the point must be nearer the smaller charge. Thus, the point of
zero resultant field is on the x-axis at x < 0 .
Requiring equal magnitudes gives
Thus, (1.0 m + d )
ke q1
r12
=
ke q2
r22
or
2 .5 µC
6.0 µC
=
2 .
d2
(1.0 m + d )
2 .5
=d
6.0
Solving for d yields
d = 1.8 m ,
or
1.8 m to the left of the ! 2 .5 µC charge
4
Physics 112
Chapt15, Problem-28:
Homework 1 (solutions)
(2004 Fall)
Figure P15.28 shows
the electric field lines for two point charges separated
by a small distance. (a) Determine the ratio q1 /q2 . (b)
What are the signs of q 1 and q2 ?
Solution:
The magnitude of q2 is three times the magnitude of q1
because 3 times as many lines emerge from q2
as enter q1 .
q2 = 3 q1
Then,
(a)
(b)
q1 q2 =!1 3
q2 > 0 because lines emerge from it,
and q1 < 0 because lines terminate on it
Chapt15, Problem-35:
If the electric field strength in air exceeds 3.0 x 106 N/C, the air becomes a
conductor. Using this fact, determine the maximum amount of charge that can be carried by a metal sphere
2.0 m in radius.
Solution:
For a uniformly charged sphere, the field is strongest at the surface.
k eqmax
,
R2
2
6
R 2 Emax ( 2.0 m ) 3.0 !10 N C
"3
qmax =
=
= 1.3 !10 C
ke
8.99 !109 N" m2 C2
Thus, Emax =
or
(
)
Chapt15, Problem-41:
A 40-cm-diameter loop is rotated in a uniform electric field until the
position of maximum electric flux is found. The flux in this position is measured to be 5.2 x 105 N · m2/C.
Calculate the electric field strength in this region.
Solution:
From the definition of Electric Flux
Thus, E =
! E = EAcos"
and ! E = ! E , max when ! = 0°
5
2
! E , max ! E , max 4 5.2 #10 N$ m C
6
=
=
= 4.1! 10 N C
2
A
"d 2 4
" (0.40 m )
(
)
5
Physics 112
Homework 1 (solutions)
Chapt15, Problem-57:
(2004 Fall)
Two 2.0-g spheres are suspended by
10.0-cm-long light strings (Fig. P15.57). A uniform electric field is
applied in the x direction. If the spheres have charges of –5.0 x 10–8 C
and +5.0 x 10–8 C, determine the electric field intensity that enables
the spheres to be in equilibrium at ! = 10°.
Solution:
The sketch to the right gives a free-body diagram of the
positively charged sphere. Here, F1 = k e q
2
r 2 is the attractive
force exerted by the negatively charged sphere and F2 = qE
is exerted by the electric field.
10°
T
mg
cos 10°
2
ke q
!Fx = 0 " F2 = F1 + T sin10° or qE = 2 + mg tan10°
r
!Fy = 0 " T cos10° = mg or T =
x
At equilibrium, the distance between the two spheres is r = 2 ( Lsin 10° ) .
ke q
2
4 ( Lsin 10° )
(8.99 ! 10
=
9
+
F2
F1
Thus,
E=
y
mg tan10°
q
mg
)(
N " m 2 C2 5.0 !10 #8 C
2
4 [ ( 0.100 m ) sin10° ]
) + (2.0 !10
#3
)(
)
kg 9.80 m s 2 tan10°
(5.0 !10
#8
C
)
,
5
and so the electric field strength required is E = 4.4 !10 N C .
Chapt15, Conceptual-2:
Hospital personnel must wear special conducting shoes when they work
around oxygen in an operating room. Why? Contrast with what might happen when personnel wear rubbersoled shoes.
Solution:
To avoid making a spark. Rubber-soled shoes acquire a charge by friction with the floor and could discharge
with a spark, possibly causing an explosive burning situation, where the burning is enhanced by the oxygen.
Chapt15, Conceptual-12:
Is it possible for an electric field to exist in empty space? Explain
Solution:
An electric field once established by a positive or negative charge extends in all directions from the
charge. Thus, it can exist in empty space if that is what surrounds the charge.
Chapt15, Conceptual-14:
Would life be different if the electron were positively charged and the
proton were negatively charged? Does the choice of signs have any bearing on physical and chemical
interactions? Explain
Solution:
No. Life would be no different if electrons were positively charged and protons were negatively charged.
Opposite charges would still attract, and like charges would still repel. The designation of charges as
positive and negative is merely a definition.
6
Physics 112
Chapt15, Conceptual-16:
Homework 1 (solutions)
(2004 Fall)
Why should a ground wire be connected to the metal support rod for
a television antenna?
Solution:
The antenna is similar to a lightning rod and can induce a bolt to strike it. A wire from the antenna to the
ground provides a pathway for the charges to move away from the house in case a lightning strike does
occur.
7