Download 18.5 ELECTRIC FIELD lines: multiple charges

OpenStax College Physics
Instructor Solutions Manual
Chapter 18
CHAPTER 18: ELECTRIC CHARGE AND
ELECTRIC FIELD
18.1 STATIC ELECTRICITY AND CHARGE: CONSERVATION OF CHARGE
1.
Common static electricity involves charges ranging from nanocoulombs to
microcoulombs. (a) How many electrons are needed to form a charge of –2.00 nC? (b)
How many electrons must be removed from a neutral object to leave a net charge of
0.500 C ?
Solution
Q  2.00  10 9 C

 1.25  1010 electrons
(a) N 
19
qe  1.60  10 C
(b) N 
 0.500  10 6 C
 3.13  1012 electrons
19
 1.60  10 C
2.
If 1.80  10 20 electrons move through a pocket calculator during a full day’s operation,
how many coulombs of charge moved through it?
Solution
Q  Nqe  1.80  10 20  1.60  10 19 C  28.8 C
3.
To start a car engine, the car battery moves 3.75  10 21 electrons through the starter
motor. How many coulombs of charge were moved?
Solution
Q  Nqe  3.75  10 21  1.60  10 19 C   600 C
4.
A certain lightning bolt moves 40.0 C of charge. How many fundamental units of
charge qe is this?
Solution
Q  N qe  N 






Q
40.0 C

 2.50  10 20
19
qe 1.60  10 C
18.2 CONDUCTORS AND INSULATORS
5.
Suppose a speck of dust in an electrostatic precipitator has 1.0000  1012 protons in it
and has a net charge of –5.00 nC (a very large charge for a small speck). How many
electrons does it have?
OpenStax College Physics
Instructor Solutions Manual
Chapter 18
Q
 5.00  10 9 C
 1.0000  1012 
 1.03  1012
qe
1.60  10 19 C
Solution
Q  N p  N e  q e  N e  N p 
6.
An amoeba has 1.00  1016 protons and a net charge of 0.300 pC. (a) How many fewer
electrons are there than protons? (b) If you paired them up, what fraction of the
protons would have no electrons?
Solution
(a) Q  N p  N e  q e  N p  N e 
(b)
N p  Ne
Np

Q
0.300  10 12 C

 1.875  10 6  1.88  10 6
19
qe
1.60  10 C
1.875  10 6
 1.88  10 10
16
1.00  10
7.
A 50.0 g ball of copper has a net charge of 2.00 C . What fraction of the copper’s
electrons has been removed? (Each copper atom has 29 protons, and copper has an
atomic mass of 63.5.)
Solution
The number of moles of copper is n 
m
50.0 g
so the number of protons, N p ,

A 63.5 g/mol
is
N p  nN A  29 protons/at om
 50.0 gm  6.02  10 23 atoms  29 protons
 

 
 1.375  10 25 protons
mol
 63.5 g/mol 
 cu atom
Since there is the same number of electrons as protons in a neutral atom, before we
remove the electrons to give the copper a net charge, we have 1.375  10 25 electrons.
Next we need to determine the number of electrons we removed to leave a net
charge of 2.00 C . We need to remove  2.00 C of charge, so the number of
electrons to be removed is given by
Q  2.00  10 6 C
N e,removed =

 1.25  1013 electrons removed.
19
qe  1.60  10 C
Lastly, the fraction of copper’s electrons removed is given by:
N e, removed
1.25  1013

 9.09  10 13
N e,initially 1.375  10 25
8.
What net charge would you place on a 100 g piece of sulfur if you put an extra electron
on 1 in 1012 of its atoms? (Sulfur has an atomic mass of 32.1.)
OpenStax College Physics
Solution
Instructor Solutions Manual
Chapter 18
The number of sulphur atoms, N s , is:
 100 g 
m
 6.02  10 23  1.875  10 24
N A  
M
 32.1g 
Nq
1.875  10 24  1.60  10 19 C
Q  s 12e 
  3.00  10 7 C
12
10
10

N s  nN A 




9.
How many coulombs of positive charge are there in 4.00 kg of plutonium, given its
atomic mass is 244 and that each plutonium atom has 94 protons?
Solution
Given Z= atomic number of an element = number of protons in elements:
m
Q  nN A q e Z    N A q e Z
M 
 4.00  10 3 gm 
 6.02  10 23 /mol 1.6  10 19 C 94   1.48  10 8 C
 
 244 gm/mol 



18.3 COULOMB’S LAW
10.
Solution
What is the repulsive force between two pith balls that are 8.00 cm apart and have
equal charges of – 30.0 nC?
F k


q1 q 2
9.00  10 9 N  m 2 /C 2  30.0  10 9 C

r2
0.0800 m 2

2
 1.266  10 3 N  1.27  10 3 N
11.
Solution
(a) How strong is the attractive force between a glass rod with a 0.700 C charge and
a silk cloth with a  0.600 C charge, which are 12.0 cm apart, using the
approximation that they act like point charges? (b) Discuss how the answer to this
problem might be affected if the charges are distributed over some area and do not
act like point charges.


6
6
(a) F  k q1 q 2  9.00  10 9 N  m 2 /C 2 0.700  10 C  0.600  10 C
r2
0.120 m2


 0.2625 N  F  0.263 N
(b) Assuming the center of mass is separated by 12.0 cm, the answer in part (a) is only
true if the charges are concentrated at the center of mass. If, however, the charges
are distributed over some area, there will be a concentration of charge along the
side closest to the oppositely charged object. This effect will increase the net force.
OpenStax College Physics
12.
Solution
Instructor Solutions Manual
Chapter 18
Two point charges exert a 5.00 N force on each other. What will the force become if
the distance between them is increased by a factor of three?
2
2
r 
qq
qq F
r
F1  k 1 2 2 , F2  k 1 22 , 2  1 2   1  or
F1 r2
r1
r2
 r2 
2
r 
1
F2  F1  1   5.00 N    0.556 N
3
 r2 
2
13.
Two point charges are brought closer together, increasing the force between them by
a factor of 25. By what factor was their separation decreased?
Solution
Using F1  k
q1 q 2
r1
2
we see that force is inversely proportional to the separation
distance squared, so that F1  k
q1 q 2
2
, F2  k
q1 q 2
2
r1
r2
Since we now the ratio of the forces, we can determine the ratio of the separation
2
F r 
r
F1
1 1

 . The separation decreased by a
distances: 1   2  so that 2 
F2  r1 
r1
F2
25 5
factor of 5.
14.
How far apart must two point charges of 75.0 nC (typical of static electricity) be to
have a force of 1.00 N between them?
Solution
qq
 kq q 
F  k 1 22  r   1 2 
r
 F 

1/ 2


 9.00  10 9 N  m 2 /C 2 75.0  10 9 C 2 


1.00 N


15.
Solution
1/ 2
 7.12  10 3 m  7.12 mm
If two equal charges each of 1 C each are separated in air by a distance of 1 km, what
is the magnitude of the force acting between them? You will see that even at a
distance as large as 1 km, the repulsive force is substantial because 1 C is a very
significant amount of charge.
q1q 2 9.0  10 9 N  m 2 /C 2 1 C2
F k 2 
 9  10 3 N
2
3
r
110 m
OpenStax College Physics
16.
Instructor Solutions Manual
Chapter 18
A test charge of  2 C is placed halfway between a charge of  6 C and another of
 4 C separated by 10 cm. (a) What is the magnitude of the force on the test
charge? (b) What is the direction of this force (away from or toward the  6 C
charge)?
Solution
9
2
2
6
6
(a) F2  4   9.00  10 N  m /C 4  10 C 2  10   28.8 N
0.05 m 2
F2  6  
9.00  10
9


N  m 2 /C 2 6  10 6 C 2  10 6
0.05 m 
2
  43.2 N
Ftot  F2  6   F2  4   43.2 N  28.8 N  14.4 N  10 N
(b) The direction of the force is away from the 6 μC charge.
17.
Bare free charges do not remain stationary when close together. To illustrate this,
calculate the acceleration of two isolated protons separated by 2.00 nm (a typical
distance between gas atoms). Explicitly show how you follow the steps in the ProblemSolving Strategy for electrostatics.
Solution
F k
q1 q 2
 ma 
r2


kq 2
9.00  10 9 N  m 2 /C 2 1.60  10 19 C
a

2
mr 2
1.67  10  27 kg 2.00  10 9 m




2
 3.45  1016 m/s 2
18.
(a) By what factor must you change the distance between two point charges to change
the force between them by a factor of 10? (b) Explain how the distance can either
increase or decrease by this factor and still cause a factor of 10 change in the force.
Solution
r
F1
r
1
F r 
 2  10  3.16
(a) F  2  1   2  , so that 2 
r
F2  r1 
r1
F2
r1
(b) If the distance increases by 3.16, then the force will decrease by a factor of 10; if
the distance decreases by 3.16, then the force will increase by a factor of 10. Either
way, the force changes by a factor of 10.
19.
Suppose you have a total charge qtot that you can split in any manner. Once split, the
separation distance is fixed. How do you split the charge to achieve the greatest force?
2
OpenStax College Physics
Instructor Solutions Manual
Chapter 18
Solution
q1q2  q1 qtot  q1   q1qtot  q1 .
This represents a quadratic equation in q1 , which would graph as a parabola with its
peak midway between the two values that cause q1 q tot  q1   0 , namely q1  0 and
0  q tot q tot
q1  q tot . Thus, the maximum value occurs at q1 

. Therefore, splitting
2
2
the charge evenly produces the greatest force.
20.
(a) Common transparent tape becomes charged when pulled from a dispenser. If one
piece is placed above another, the repulsive force can be great enough to support the
top piece’s weight. Assuming equal point charges (only an approximation), calculate
the magnitude of the charge if electrostatic force is great enough to support the
weight of a 10.0 mg piece of tape held 1.00 cm above another. (b) Discuss whether the
magnitude of this charge is consistent with what is typical of static electricity.
Solution
2
(a) F  k q1 q 2  kq  mg 
r2
r2
2



 0.0100 m 2 10.0  10 6 kg 9.80 m/s 2 
 r 2 mg 
9


q

  1.04  10 C.
9
2
2

9.00  10 N  m /C
 k 


(b) This charge is approximately 1nC, which is consistent with the magnitude of charge
of typical static electricity.
1/ 2

21.
Solution

1/ 2
(a) Find the ratio of the electrostatic to gravitational force between two electrons. (b)
What is this ratio for two protons? (c) Why is the ratio different for electrons and
protons?
(a) Since FE 
kq1q2 kq2
GMm Gm2 FE kq2

and
F

 2 , 
.
G
r2
r2
r2
r
FG Gm2
For electrons me  9.11  10 31 kg, so that







2
FE
9.00  10 9 N  m 2 /C 2 1.60  10 19 C

FG
6.67  10 11 N  m 2 /kg 9.11  10 31 kg
(b) For protons, m p  1.67  10 27 kg, so that



2
 4.16  10 42
2
FE
9.00  10 9 N  m 2 /C 2 1.60  10 19 C

 1.24  10 36
2

11
2

27
FG
6.67  10 N  m /kg 1.67  10 kg
(c) The ratio is different for electrons and protons because the masses are different.

22.

At what distance is the electrostatic force between two protons equal to the weight of
one proton?
OpenStax College Physics
Solution
Instructor Solutions Manual
 kq2 
kq2



m
g

r

p
m g
r2
p


1/ 2
Chapter 18
 9.00  10 9 N  m 2 /C 2 1.60  10 19 C2


1.67  10 27 kg 9.80 m/s 2 






1/ 2
 0.119 m
23.
A certain five cent coin contains 5.00 g of nickel. What fraction of the nickel atoms’
electrons, removed and placed 1.00 m above it, would support the weight of this coin?
The atomic mass of nickel is 58.7, and each nickel atom contains 28 electrons and 28
protons.
Solution
The number of electrons, N e , in the nickel is
 5.00 g 
m
 6.02  10 23  1.436  10 24 .
N e  28nN A  28  N A  28
M 
 58.7 g 
Assume the number of electrons removed from the nickel is N. Then the charge of the
removed electrons is Nqe , and the charge of the remaining positively charged nickel is


kq1 q 2 k Nqe  Nqe 

. Note that there is a minus
r2
r2
sign in front of mg because the attractive force between the electrons and the
positively charged nickel will be equal and opposite in direction to the gravitational
 mgr 2
mgr 2
2
force. Continuing

N

, so that
 kq2e
kq2e
Nq p   Nqe . Thus, F  mg 



 5.00  10 3 kg 9.80 m/s 2 1.00 m2 
N 
2
 9.00  10 9 N  m 2 /C 2  1.60  10 19 C 
N 1.458  1013

 1.02  10 11
Therefore,
24
N E 1.436  10

24.
Solution


1/ 2
 1.458  1013
(a) Two point charges totaling 8.00 C exert a repulsive force of 0.150 N on one
another when separated by 0.500 m. What is the charge on each? (b) What is the
charge on each if the force is attractive?
(a) F 
kqQ  q 
Fr 2
Fr 2
2
2
so
that


qQ

q
,
q

Qq

0
k
r2
k
OpenStax College Physics
Instructor Solutions Manual
Fr 2 0.150 N 0.500 m 

 4.167  10 12 C 2
9
2
2
k
9.00  10 N  m /C
2
a  1; b  Q  8.00  10 6 C ; c 
 b b 2  4ac
q
2

8.00  10 6 c 
 4.00  10
6
 8.00  10
C  3.44  10
6
6
Chapter 18
C

2

 4 4.167  10 12 C 2

2
C
C, and Q  q  0.56  10 6 C
or q  7.44  106 C and Q  q  0.56 106 C
 7.44  10
(b) F 
6
 kqQ  q 
Fr 2
Fr 2
so
that
,
 0,
 qQ  q 2 and q 2  qQ 
2
r
k
k
a  1; b  8.00  10 6 C; c  4.167  10 12 C 2
q

 b  b 2  4ac
2
8.00  10 6 C 
6
6

2

C  4 4.167  10 12 C 2

2
C  12.98  10 6 C
6
C  8.98  10
6
C, Q  q   4.98  10 6 C.
 4.00  10
 13.0  10
 8.00  10
25.
Point charges of 5.00 C and  3.00 C are placed 0.250 m apart. (a) Where can a
third charge be placed so that the net force on it is zero? (b) What if both charges are
positive?
Solution
(a) We know that since the negative charge is smaller, the third charge should be
placed to the right of the negative charge if the net force on it is to be zero. So if
qq
we want Fnet  F1  F2  0 , we can use F  k 1 2 2 to write the forces in terms of
r
distances:
 q1 q2 
Kq1q2 Kq2q
 2  2   0 , or since r1  0.250 m  d and r2  d ,


Kq
r12
r22
 r1 r2 
5  106 C
3  106

, or d  3 0.250 m  3 d so that
2
2
d
0.250 m  d 
5
5
OpenStax College Physics
Instructor Solutions Manual
Chapter 18
3
0.250 m 

3
3
5

d 1 
 0.859 m
 5 0.250 m  and finally, d 
5
3


1
5
The charge must be placed at a distance of 0.859 m to the far side of the negative
charge.
(b) This time we know that the charge must be placed between the two positive
charges and closer to the 3 C charge for the net force to be zero. So if we want
qq
Fnet  F1  F2  0 , we can again use F  k 1 2 2 to write the forces in terms of
r
 Kq q

Kq2 q
q q 
 Kq 12  22   0
distances:  12 2 
2
r2
 r1 r2 
 r1

Or since r1 = 0.250m  r2
5  10 6 C
0.250  r2 
2

3  10 6
r2
2
, or r2 
2
3
0.250 m  r2 2 , or
5
3
0.250 m 
3
5
 0.109 m
0.250 m  r2  , and finally r2 
r2 
3
5
1
5
The charge must be placed at a distance of 0.109m from the 3 C charge.
26.
Two point charges q! and q2 are 3.00 m apart, and their total charge is 20 C . (a) If
the force of repulsion between them is 0.075 N, what are magnitudes of the two
charges? (b) If one charge attracts the other with a force of 0.525 N, what are the
magnitudes of the two charges? Note that you may need to solve a quadratic equation
to reach your answer.
Solution
(a) Q1  Q2  20 C, r  3.0 m
kQ Q
F  12 2  0.075 N
r
2
2
r F 3.0 m  0.075 N 
Q1Q2 

 7.5  10 11 C 2
9
2
2
k
9  10 N  m / C
6
Q1  Q2  20  10 C  Q1 20  10 6 C  Q1   7.5  10 11 C 2


Q12  Q1 20  10 6  7.5  10 11  0
a  1, b  20  10 6 , c  7.5  10 11
Use the quadratic formula: Q1  15 C, 5 C and Q2  20 μC  Q1  5 C or 15 C
So the two possibilities are: Q1  15 C and Q2  5 C or Q1  5 C and Q2  15 C
r 2 F 3.0 m   0.525 N 

 5.25  10 10 C 2
9
2
2
k
9  10 N  m / C
2
(b) Q1Q2 
OpenStax College Physics
Instructor Solutions Manual
Chapter 18
Same method as part (a):
Q1  Q2  20  10 6 C  Q1 20  10 6 C  Q1  5.25  10 10 C 2


Q12  Q1 20  10 6    5.25  10 10   0
a  1, b  20  10 6 , c  5.25  10 10
Use the quadratic formula. The two possibilities are:
Q1  15 C and Q2  35 C or Q1  35 C and Q2  - 15 C
18.4 ELECTRIC FIELD: CONCEPT OF A FIELD REVISITED
27.
What is the magnitude and direction of an electric field that exerts a 2.00  105 N
upward force on a  1.75 C charge?
Solution
F
2.00  10 5 N
E 
  11.4 N/C downward
q  1.75  10 6 C
28.
What is the magnitude and direction of the force exerted on a 3.50 C charge by a
250 N/C electric field that points due east?
Solution
E
29.
Calculate the magnitude of the electric field 2.00 m from a point charge of 5.00 mC
(such as found on the terminal of a Van de Graaff).
Solution
30.
Solution
31.
E


F
 F  qE  3.50  10 6 C 250 N/C  8.75  10 4 N
q



kQ 9.00  10 9 N  m 2 /C 2 5.00  10 3 C

 1.13  10 7 N/C
2
r2
2.00 m
(a) What magnitude point charge creates a 10,000 N/C electric field at a distance of
0.250 m? (b) How large is the field at 10.0 m?
(a) E 
kQ
r 2 E 0.250 10000 N/C

Q


 6.94  10 8 C
2
9
2
2
k
r
9.00  10 N  m /C
(b) E 
kQ 9.00  10 9 N  m 2 /C 2 6.94  10 8 C

 6.25 N/C
2
r2
10.0 m
2





Calculate the initial (from rest) acceleration of a proton in a 5.00  106 N/C electric
field (such as created by a research Van de Graaff). Explicitly show how you follow the
steps in the Problem-Solving Strategy for electrostatics.
OpenStax College Physics
Instructor Solutions Manual


Chapter 18

Solution
qE 1.60  10 19 C 5.00  10 6 N/C
F  ma  qE  a 

 4.79  1014 m/sec 2
 27
m
1.67  10 kg
32.
(a) Find the direction and magnitude of an electric field that exerts a 4.80  10 17 N
westward force on an electron. (b) What magnitude and direction force does this field
exert on a proton?
Solution
F  4.80  10 17 C

 300 N/C (eas t)
q  1.60  10 19 C
(b) F  qE  1.60 10 19 C 300 N/C  4.80 10 17 N (east)
(a) E 


18.5 ELECTRIC FIELD LINES: MULTIPLE CHARGES
33.
(a) Sketch the electric field lines near a point charge  q . (b) Do the same for a point
charge  3.00q .
Solution (a)
+q
(b)
– 3q
34.
Sketch the electric field lines a long distance from the charge distributions shown in
Figure 18.26(a) and (b).
OpenStax College Physics
Instructor Solutions Manual
Chapter 18
Solution (a)
– 2q
(b)
field of
two opposite
charges
No net charge is seen from far away.
35.
Solution
36.
Figure 18.47 shows the electric field lines near two charges q1 and q2 . What is the
ratio of their magnitudes? (b) Sketch the electric field lines a long distance from the
charges shown in the figure.
q1
23

  1.9 : 1
q2  12
(b) A long distance away, the two charges appear as one charge, with total charge
q  q1  q2  (23 12)e  11e , where e is some basic unit of charge. The field will
look like that of one of the point charges in Figure 18.23, but with 11 lines directed
away.
(a)
Sketch the electric field lines in the vicinity of two opposite charges, where the
negative charge is three times greater in magnitude than the positive. (See Figure
18.47 for a similar situation.)
Solution
18.7 CONDUCTORS AND ELECTRIC FIELDS IN STATIC EQUILIBRIUM
OpenStax College Physics
37.
Instructor Solutions Manual
Chapter 18
Sketch the electric field lines in the vicinity of the conductor in Figure 18.48, given the
field was originally uniform and parallel to the object’s long axis. Is the resulting field
small near the long side of the object?
Solution
–
– –
–
–
––
–
+
–
+
++
+
+
+ +
+
Yes, the field is smaller near the long side of the object.
38.
Sketch the electric field lines in the vicinity of the conductor in Figure 18.49 given the
field was originally uniform and parallel to the object’s long axis. Is the resulting field
small near the long side of the object?
Solution
––
–
––
–
–
–
–
+
+
++
+
+
+ +
+
Yes, the field is smaller near the long side of the object.
39.
Solution
40.
Sketch the electric field between the two conducting plates shown in Figure 18.50,
given the top plate is positive and an equal amount of negative charge is on the
bottom plate. Be certain to indicate the distribution of charge on the plates.
+
+
+
+
+
++
– –
–
–
–
–
–
Sketch the electric field lines in the vicinity of the charged insulator in Figure 18.51
noting its nonuniform charge distribution.
Solution
+
+ +
41.
+
+
+
+ + + ++ + +
What is the force on the charge located at x  8.00 cm in Figure 18.52(a) given that
q  1.00 μC ?
OpenStax College Physics
Solution
FK
Instructor Solutions Manual
q3q8
qq
 K 82 11
2
r 3, 8
r 8,11



Chapter 18

  2 1.00  10 6 C 2  2 1.00  10 6 C 2 
 9.00  10 N.m /C 


2
0.0300 m 2 
 0.0500 m 
 12.8 N to the right.

9
2
2

42.
Find the total electric field at x  1.00 cm in Figure 18.52(b) given that q  5.00 nC .
(b) Find the total electric field at x  11.00 cm in Figure 18.52(b). (c) If the charges are
allowed to move and eventually be brought to rest by friction, what will the final
charge configuration be? (That is, will there be a single charge, double charge, etc.,
and what will its value(s) be?)
Solution
(a) Since E  K
q
, and there is negative point charge at x  1.00 cm , the electric field
r2
at x  1.00 cm will be negative infinity: E x 1.00cm   
q
q
q
q
(b) F  K 2 1  K 2 5  K 2 8  K 2 14 and since q  5.00  109 C
r 1,11
r 5,11
r 8,11
r 11,14
  10.0  10 9 C 5.00  10 9 C 


2
0.0600 m 2 
 9.00  10 9 N  m 2   0.100 m 

E  
 2.12  10 5 N/C
2
9
9


C

  15.0  10 C   5.00  10 C


2
0.0300 m 2 
 0.0300 m 
(c) The two charges will come together to be a charge of  q and the two right
charges will come together to form one charge of  q , so eventually the charges
will come together to form one charge of  q .
43.
(a) Find the electric field at x  5.00 cm in Figure 18.52(a), given that q  1.00μC . (b)
At what position between 3.00 and 8.00 cm is the total electric field the same as that
for  2q alone? (c) Can the electric field be zero anywhere between 0.00 and 8.00 cm?
(d) At very large positive or negative values of x, the electric field approaches zero in
both (a) and (b). In which does it most rapidly approach zero and why? (e) At what
position to the right of 11.0 cm is the total electric field zero, other than at infinity?
(Hint: A graphing calculator can yield considerable insight in this problem.)
Solution
 q3
q8
q11 
(a) E  K  2  2  2 
 r3,5 r5,8 r5,11 
 9.00  10 9 N  m 2
 
C2

 4.00  10 7 N/C
 1.00  10 6 C  2.00  10 6 C 1.00  10 6 C 
 


2
0.0300 m 2 0.0600 m 2 
  0.0200 m 
OpenStax College Physics
Instructor Solutions Manual
Chapter 18
(b) For the total electric field to have the same value as it would be for the  2q charge
alone, we need to find the place where the two positive charges alone create a
zero electric field. This point is halfway between the two points or at x  7.00 cm.
(c) No, the electric field cannot be zero anywhere between 0.00cm and 8.00cm.
(d) At very large positive or negative values of x, the electric field approaches zero
because the distance from the charges approaches infinity. The electric field
approaches zero more rapidly in Figure 18.52 because the net charge is zero.
(e) For a position greater than 11.0 cm, we must find a value of x that satisfies:
1
2
1
where x has units of centimeters. We can then
y


2
2
x  3 x  8 x  112
graph this function, using a graphing calculator or graphing program, to determine
the values of x that yield y  0.
From the graph we find that the electric field is zero at x  30.6 cm.
44.
(a) Find the total Coulomb force on a charge of 2.00 nC located at x  4.00 cm in
Figure 18.52(b), given that q  1.00μC . (b) Find the x-position at which the electric
field is zero in Figure 18.52(b).
Solution
(a) The net force is given by
q q
q
Kq q Kq q Kq q Kq q
q 
F  21  25  28  14
 Kq 12  25  82  142  , to the right.
2
r1
r5
r8
r14
r14 
 r1 r5 r8
Substituting in the values given:
 9.00  10 9 N.m 2 
2.00  10 9 C 
F  
2
C




 2.00  10 6 C
1.00  10 6 C
3.00  10 6 C



2
2
2 
0.0500 m  0.0400 m  0.0800 m  0.0400 m  
 0.0400 m  0.0100 m 


 1.00  10 6 C


2
 0.140 m  0.0400 m 

= - 0.252 N(right) or 0.252 N to the left
(b) The only possible location where the total electric field could be zero is between
5.00 and 8.00 cm. For the total electric field to be zero between 5.00 and 8.00 cm,
Kq Kq Kq Kq
we know that E  2 1  2 5  2 8  214  0 .
r1
r5
r8
r14
Dividing by common factors and ignoring units (but remembering x has a unit of
cm), we can get a simplified expression:

2
1
3
1 



y 
2
2
2
x  1 x  5 x  8 x  142 

OpenStax College Physics
Instructor Solutions Manual
Chapter 18
We can graph this function, using a graphing calculator or graphing program to
determine the values of x that yield y  0
Therefore the total electric field is zero at 6.07 cm.
45.
(a) Using the symmetry of the arrangement, determine the direction of the force on q
in the figure below, given that q a  qb  7.50 μC and qc  qd  7.50 μC . (b)
Calculate the magnitude of the force on the charge q , given that the square is 10.0 cm
on a side and q  2.00 μC .
Solution
(a) Due to symmetry the net force on q will be straight down, since q a and q b are
positive and q c and q d are negative and all have the same magnitude. q a and q b
will force the charge downward and q c and q d will pull the charge downward also.
(b) Since the square is 10.0 cm on a side, the distance to the charge q will be
10.0 cm
2  7.071 cm so
2
6
6
qqa 8.99 109 N  m 2 2.00 10 C 7.50 10 C
Fa  k 2 
 26.97 N
2
r
C2
 0.07071 m 



45 degrees below x-axis.
The total force will be:
Fy  4  Fa sin 45  4   26.97 N  sin 45  76.3 N (downward)
46.
(a) Using the symmetry of the arrangement, determine the direction of the electric
field at the center of the square in Figure 18.53, given that q a  qb  1.00 μC and
qc  qd  1.00 μC . (b) Calculate the magnitude of the electric field at the location of
q , given that the square is 5.00 cm on a side.
Solution
(a) The electric field at the center of the square will be straight up, since q a and q b are
positive and q c and q d are negative and all have the same magnitude.
(b) Since the square is 5.00cm on a side, the distance to the center of the square is
5.00 cm
Q
2  3.536 cm , so using E  k 2 , the electric field due to upper left
2
r
hand charge is:

q
9.00  10 9 N  m 2 /C 2  1.00  10 6 C
E a  K 2a 
 7.20  10 6 N/C,
2
r
0.03536 m 

45 degrees below x-axis, or Ea  7.20 106 N/C, 135 above x-axis.
The other three fields will have the same magnitude, only in different directions.
From part (a) we know that the only component that we need to calculate is the y-


OpenStax College Physics
Instructor Solutions Manual
Chapter 18
component, since the x-component will cancel. The total electric field at the center
will then be:
E y  4  Ea sin 45  4  7.20 106 N/C sin 135  2.04 107 N/C(upward )


47.
Find the electric field at the location of q a in Figure 18.53 given that
qb  qc  qd  2.00 nC , q  1.00 nC , and the square is 20.0 cm on a side.
Solution


q
9.00  10 9 N  m 2 /C 2 2.00  10 9 C
E a  K 2a 
 450 N/C in the x-dir; Ea = 450N/C
2
r
0.200 m 
in the +y direction.

q
9.00  10 9 N  m 2 /C 2 2.00  10 9 C
Ed  K d2 
 225 N/C, 135 above the x-axis.
2
r
2  0.200 m






q 9.00  10 N  m /C
Eq  K 2 
r
9
2
2

1.00 10

9
C
2  0.100 m

2
 450 N/C, 45 below the x-axis.
Summing gives:
E x  E b cos 180  E c cos 90  E d cos 135  E q cos 45
 450 N/C  0  159 N/C  318 N/C  291 N/C and
E y  E b sin 180  E c sin 90  E d sin 135  E q sin  45
 0  450 N/C  159 N/C  318 N/C  291 N/C,
So the electric field at the location of q is
E
 291 N/C2  291 N/C2
 412 N/C, 45 degrees above the x-axis.
48.
Find the total Coulomb force on the charge q in Figure 18.53, given that q  1.00 μC ,
qa  2.00 μC , qb  3.00 μC , qc  4.00 μC , and qd  1.00 μC . The square is 50.0 cm
on a side.
Solution
 0.500 m 
r  2
  0.3536 m.
2 

6

qa q
C 1.00  10 6 C
9
2
2 1.00  10
Fd  k 2  9.00  10 N  m /C
 0.0720 N, 45
r
0.3536 m 2
above the x-axis.

Fb  3  Fd  0.216 N, 45 above the +x-axis, Fc  4  Fd  0.288 N, 45 below the -x
axis, and Fa  2  Fd  0.144 N, 45 below the +x-axis.


Summing up the components gives:


OpenStax College Physics
Instructor Solutions Manual
Chapter 18
Fx  Fa cos 45  Fb cos 45  Fc cos 135  Fd cos135
 0.102 N  0.153 N  0.204 N  0.0509 N  0.00 N.
Fy  Fa sin  45  Fb sin 45  Fc sin  135  Fd sin 135
 0.102 N  0.153 N  0.204 N  0.0509 N  0.102 N
So the total coulomb force on the charge q is 0.102 N, in the  y  direction .
49.
(a) Find the electric field at the location of qa in Figure 18.54, given that
qb  10.00 μC and qc  5.00 μC . (b) What is the force on qa , given that
qa  1.50 nC ?
Solution
6

qb
C
9
2
2 10.0  10
 1.44  10 6 N/C, 60
(a) E b  k 2  9.00  10 N  m /C
2
r
0.250 m


above the -x-axis and
6

qc
C
9
2 5.00  10
Ec  k 2  9.00  10 N  m
 7.2  10 5 N/C, 60 below the –x-axis.
2
r
0.250 m


Summing up the components gives:
E x  E b cos 120  E c cos 120
 7.20  10 5 N/C  3.60  10 5 N/C  1.08  10 6 N/C, and
E y  E b sin 120  E c sin  120  1.25  10 5 N/C,
So the total electric field at the location of qa is 1.25  10 6 N/C and

6.24  105 N/C
 30.0 above the –x-axis, or E  1.25  10 6 N/C, 30.0
6
1.08  10 N/C
above the –x-axis.
  tan 1


(b) Fa  qa E  1.50  10 9 C 1.25  10 6 N/C  1.88  10 3 N, 30.0 above the –x-axis.



50.
(a) Find the electric field at the center of the triangular configuration of charges in
Figure 18.54, given that qa  2.50 nC , qb  8.00 nC , and qc  1.50 nC . (b) Is there
any combination of charges, other than qa  qb  qc , that will produce a zero
strength electric field at the center of the triangular configuration?
Solution
(a) To determine the electric field at the center, we first must determine the distance
from each of the charges to the center of the triangle. Since the triangle is
equilateral, the center of the triangle will be halfway across the base and 1/3 of
OpenStax College Physics
Instructor Solutions Manual
Chapter 18
the way up the height. To determine the height use the Pythagorean theorem, or
the height is given by h  25.0 cm  12.5 cm  21.7 cm . So the distance from
each charge to the center of the triangle is 2/3 of 21.7 cm, or
2
Q
r  21.7 cm   14.4 cm. Since E  k 2 ,
3
r

 2.50  10 9 C 
q
  1085 N/C at a 90 angle
Ea  k 2a  9.00  10 9 N  m 2 /C 2 
2 
r


0
.
144
m


below the horizontal,

  8.00  10 9 C 
q
  3472 N/C at a 30 angle
E b  k 2b  9.00  10 9 N  m 2 /C 2 
2

r
 0.144 m  
below the horizontal, and
2




C
  681.0 N/C at a 30 angle



0
.
144
m


above the horizontal. Adding the vectors by components gives:

q
Ec  k 2c  9.00  10 9 N  m 2 /C 2
r

 1.50 10
2
9
2
E x  E a cos 90  E b cos 30  E c cos 30
E x  0 N/C  3472 N/C 0.860   681.0 N/C 0.8660   3597 N/C
E y  E a sin  90  E b sin  30  E c sin 30
E y  1085 N/C - 3472 N/C 0.5000   681 N/C 0.5000   2481 N/C
So that the electric field is given by:
E
Ex  E y 
2
2
3597 N/C2   2481 N/C2
 4370 N/C and

 2481 N/C
 34.6 , or E  4.37  10 3 N/C, 34.6
Ex
3597 N/C
below the horizontal.
  tan 1
Ey
 tan 1
(b) No, there are no combinations other than qa  q b  qc that will produce a zero
strength field at the center of the triangular configuration because of the vector
nature of the electric field.
OpenStax College Physics
Instructor Solutions Manual
Chapter 18
18.8 APPLICATIONS OF ELECTROSTATICS
51.
Solution
(a) What is the electric field 5.00 m from the center of the terminal of a Van de Graaff
with a 3.00 mC charge, noting that the field is equivalent to that of a point charge at
the center of the terminal? (b) At this distance, what force does the field exert on a
2.00 μC charge on the Van de Graaff’s belt?
(a) E 



kQ 9.00  10 9 N  m 2 /C 2 3.00  10 3 C

 1.08  10 6 N/C
2
2
r
5.00 m 
(b) F  qE  2.00  10 6 C1.08  10 6 N/C  2.16 N
52.
Solution
(a) What is the direction and magnitude of an electric field that supports the weight of
a free electron near the surface of Earth? (b) Discuss what the small value for this field
implies regarding the relative strength of the gravitational and electrostatic forces.
(a) F  mg  qE  E 



 mg  9.11 10 31 kg 9.80 m/s 2

 5.58  10 11 N/C
19
q
 1.60  10 C
Electric field is towards the surface of earth.
(b) The coulomb force is extraordinarily stronger than gravity.
53.
A simple and common technique for accelerating electrons is shown in Figure 18.55,
where there is a uniform electric field between two plates. Electrons are released,
usually from a hot filament, near the negative plate, and there is a small hole in the
positive plate that allows the electrons to continue moving. (a) Calculate the
acceleration of the electron if the field strength is 2.50  104 N/C . (b) Explain why the
electron will not be pulled back to the positive plate once it moves through the hole.
Solution
(a) F  ma  qE 
qE
 1.60  10 19 C  2.50  10 4 N/C
a

m
9.11  10 31 kg



 4.391  1015 m/sec 2  4.39  1015 m/sec 2
(b) There is no field outside the plates.
54.
Earth has a net charge that produces an electric field of approximately 150 N/C
downward at its surface. (a) What is the magnitude and sign of the excess charge,
noting the electric field of a conducting sphere is equivalent to a point charge at its
center? (b) What acceleration will the field produce on a free electron near Earth’s
surface? (c) What mass object with a single extra electron will have its weight
supported by this field?
OpenStax College Physics
Solution
Instructor Solutions Manual

Chapter 18

KQ
r2E
6.37  10 6  150 N/C
Q

 6.76  10 5 C
2
9
2
2
K
r
9.00  10 N  m /C
F

ma

qE

(b)
qE
 1.60  10 19 C  150 N/C 
a

 2.63  1013 m/s 2 (upwards)
31
m
9.11  10 kg
(a) E 

2

(c) F  mg  qE  m 


 qE   1.60  10 19 C 150 N/C

 2.45  10 18 kg
2
g
9.80 m/s
55.
Point charges of 25.0 C and 45.0 C are placed 0.500 m apart. (a) At what point
along the line between them is the electric field zero? (b) What is the electric field
halfway between them?
Solution
9 
9
K  q1 
Kq1
9
1
9
5 
5
, so that 2 

x  d  x,
(a) q 2  q1 , E1  E 2  2  
2
2
5
5
x
x
d  x 
d  x 
And finally solving for x gives:
d
0.500 m
x

 0.214 m from 25.0 C charge.
9
9
1
1
5
5
(b)
 Kq1 Kq2
4 K q 2  q1  4 9.00  10 9 N  m 2 /C 2 20.0  10 6 C
E  E1  E 2 



2
2
d2
0.500 m 2
d
d
2
2
E  2.88  106 N/C towards the larger charge.
   



56.
What can you say about two charges q1 and q 2 , if the electric field one-fourth of the
way from q1 to q2 is zero?
Solution
If the electric field is zero 1/4 from the way of q1 and q 2 , then we know from
q
3x 
Q
Kq
Kq2
so that 2  2  9
E  k 2 that E1  E2  21 
2
q1
x
x
r
3x 
2
The charge q2 is 9 times larger than q1 .
57.
Integrated Concepts Calculate the angular velocity  of an electron orbiting a proton
in the hydrogen atom, given the radius of the orbit is 0.530  10 10 m . You may assume
that the proton is stationary and the centripetal force is supplied by Coulomb
attraction.
OpenStax College Physics
Solution
Fc 
Instructor Solutions Manual
 kq 2 
mv 2 m(r ) 2


 mr 2 , so that    3 
r
r
 mr 
kq1 q 2
r2

Chapter 18


 9.00  10 9 N  m 2 / C 2 1.60  10 19 C 2 


3
31
10

 9.11  10 kg 0.530  10 m



12
(where q1  q 2  q )
12
 4.12  1016 rad s
58.
Integrated Concepts An electron has an initial velocity of 5.00  10 6 m/s in a uniform
2.00  105 N/C strength electric field. The field accelerates the electron in the direction
opposite to its initial velocity. (a) What is the direction of the electric field? (b) How far
does the electron travel before coming to rest? (c) How long does it take the electron
to come to rest? (d) What is the electron’s velocity when it returns to its starting point?
Solution
(a) The field is in the direction of the electron’s initial velocity.
2
(b) v 2  v02  2ax  x   v0 v  0. Also, F  ma  qE  a  qE
2a
m





2
v2m
5.00  10 6 m s 9.11  10 31 kg
x 0 
 3.56  10  4 m
2qE
2  1.60  10 19 C 2.00  10 5 N C
v0
v0 m
v  0
(c) v  v0  at  t    
a
qE
t  
5.00  10 m s9.11 10
 1.60  10 C2.00  10
6
19
(d) v  2ax 
12

 2qEx 
 

 m 


31
5

kg
 1.42  10 10 s
NC
12



 2  1.60  10 19 C 2.00  10 5 N C  3.56  10  4 m 


9.11  10 31 kg


6
  5.00  10 m/s (opposite its initial velocity)
12
59.
Integrated Concepts The practical limit to an electric field in air is about 3.00 10 6 N/C
. Above this strength, sparking takes place because air begins to ionize and charges
flow, reducing the field. (a) Calculate the distance a free proton must travel in this field
to reach 3.00% of the speed of light, starting from rest. (b) Is this practical in air, or
must it occur in a vacuum?
Solution
(a) F  qE  ma  a 

qE
v2
v 0  0
, and v 2  v02  2ax  x 
m
2a

 

0.0300 3.00  108 m s 1.67  10 27 kg  0.141 m
v2m
x

2qE
2 1.60  10 19 C 3.00  10 6 N C
(b) No, it is impractical in air. A proton would collide with an air molecule before it
could travel this far.

2


OpenStax College Physics
60.
Instructor Solutions Manual
Chapter 18
Integrated Concepts A 5.00 g charged insulating ball hangs on a 30.0 cm long string in
a uniform horizontal electric field as shown in Figure 18.56. Given the charge on the
ball is 1.00 C , find the strength of the field.
Solution
Let T  tension force in the string.
Tsin   FE  qE
Tcos  mg
tan  
E
61.
Solution
qE
mg



mgtan 
5.00 10 3 kg 9.80 m s 2 tan 8.00

 6.89 103 N C
6
q
1.00 10 C
Integrated Concepts Figure 18.57 shows an electron passing between two charged
metal plates that create an 100 N/C vertical electric field perpendicular to the
electron’s original horizontal velocity. (These can be used to change the electron’s
direction, such as in an oscilloscope.) The initial speed of the electron is 3.00  106 m/s ,
and the horizontal distance it travels in the uniform field is 4.00 cm. (a) What is its
vertical deflection? (b) What is the vertical component of its final velocity? (c) At what
angle does it exit? Neglect any edge effects.
y
x
+
+
+
+
vf

d
v0
–
–
–
L
–
OpenStax College Physics
Instructor Solutions Manual
qE
m
(a) F  ma y  qE  a y 
L  v0 x t  t 
L
1
1
, and since d  v0 y t  a y t 2  d  a y t 2 v0 y  0 
v0 x
2
2
1
1  qE  L
d  ayt 2  

2
2  m  v0 x



2
 1.60  10 C 100 N C0.0400m 

29.11  10 kg 3.00  10 m s 
 v  a t  v  a t v  0
19
31
(b) v y
0y
y
v y  at 
Chapter 18
y
2
6
y

0y
2
 1.56  10 3 m

aL qEL  1.60  10 19 C  100 N C0.0400m 


3.00  106 m s9.11 1031 kg 
v0 x v0 x m
 2.342  105 m s  2.34  105 m s
(c) tan  
 vy 
 2.342 105 m s 
  4.46
   tan 1    tan 1 
6
v0 x
v
3
.
00

10
m
s


0
x
 
vy
62.
Integrated Concepts The classic Millikan oil drop experiment was the first to obtain an
accurate measurement of the charge on an electron. In it, oil drops were suspended
against the gravitational force by a vertical electric field. (See Figure 18.58.) Given the
oil drop to be 1.00 m in radius and have a density of 920 kg/m 3 : (a) Find the weight
of the drop. (b) If the drop has a single excess electron, find the electric field strength
needed to balance its weight.
Solution
Let  = density of oil drop, and V = volume of oil drop.
3
4
4
(a) W  mg  Vg   r 3 g   1.00  10 6 m 920 kg m 3 9.80 m s 2
3
3
14
 3.777  10 N  3.78  10 14 N

(b) W  FE  qE  E 



W 3.777  10 14 N

  2.36  10 5 N C
q  1.60  10 19 C
63.
Integrated Concepts (a) In Figure 18.59, four equal charges q lie on the corners of a
square. A fifth charge Q is on a mass m directly above the center of the square, at a
height equal to the length d of one side of the square. Determine the magnitude of q
in terms of Q , m , and d , if the Coulomb force is to equal the weight of m . (b) Is this
equilibrium stable or unstable? Discuss.
Solution
(a) Because of symmetry, the only force that does not cancel is the vertical
OpenStax College Physics
Instructor Solutions Manual
Chapter 18
component of the force, so we need only calculate the vertical component effect
of one of the corner charges and multiply by four to get the total force. The
distance from any corner charge q to the charge Q is
2
2
3
d  d 
r      d2 
d
2
2 2
qQ 2 qQ
,
 k
r2 3 d 2
acting along the line from q to Q. So, the component of the force acting in the
d
2

vertical direction is: Fv , q  F sin  , where sin  
, so that the total force
3
3
d
2
2 2 qQ
qQ
in the vertical direction is given by Fv  4  F sin   4
k 2  2.177k 2 .
33 d
d
Now, if this force is balanced by the force of gravity, then Fv  Fg and
Therefore, the magnitude of the force due to one charge is: F  k
2
2.177k
64.
Solution
mgd
qQ
.
 mg so that q  0.459
2
kQ
d
Unreasonable Results (a) Calculate the electric field strength near a 10.0 cm diameter
conducting sphere that has 1.00 C of excess charge on it. (b) What is unreasonable
about this result? (c) Which assumptions are responsible?


kQ 9.00 10 9 N  m 2 C 2 1.00 C 

 9.00 1011 N C
2
2
r
0.100 m 
(b) The field is too large to be in air.
(c) 1.00 C is too great for an excess charge on the sphere's surface.
(a) E 
65.
Unreasonable Results (a) Two 0.500 g raindrops in a thunderhead are 1.00 cm apart
when they each acquire 1.00 mC charges. Find their acceleration. (b) What is
unreasonable about this result? (c) Which premise or assumption is responsible?
Solution
(a) F  ma 
kq1 q 2

r2



2
9.00 10 9 N  m 2 C 2 1.00 10 3 C
kq 2
2

1.80 1011 m s
2
2
3
mr
0.500 10 kg 0.0100 m 
(b) The resulting acceleration is unreasonably large; the raindrops would not stay
together.
(c) The assumed charge of 1.00 mC is much too great; typical static electricity is on
the order of 1 C or less.
a


OpenStax College Physics
66.
Solution
Instructor Solutions Manual
Chapter 18
Unreasonable Results A wrecking yard inventor wants to pick up cars by charging a
0.400 m diameter ball and inducing an equal and opposite charge on the car. If a car
has a 1000 kg mass and the ball is to be able to lift it from a distance of 1.00 m: (a)
What minimum charge must be used? (b) What is the electric field near the surface of
the ball? (c) Why are these results unreasonable? (d) Which premise or assumption is
responsible?
(a) F  mg 
kq 2
 mg 
, so that q  r 

2
r
 k 
12
1.04310 3 C 1.10 10 6 C




 1000 kg  9.80 m s 2 
 1.00 m 
9
2
2 
 9.00 10 N  m C 

kQ 9.00 10 9 N  m 2 C 2 1.10 10 6 C

 2.45  10 5 N/C
r2
(0.200 m) 2
(c) It would be impractical to place that much excess charge on a wrecking ball. Even
so, the necessary E-field strength is 245 MV/m , much too great to be created in
air.
(d) It is unreasonable that one can lift heavy objects with static electricity.
(b) E 
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