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Chapter 21: Electrostatics
Chapter 21: Electrostatics
In Class Exercises
21.1. d 21.2. a 21.3. e 21.4. e 21.5. c 21.6. b 21.7. a 21.8. a 21.9. c 21.10. e
Multiple Choice
21.1. b 21.2. b 21.3. b 21.4. d 21.5. b 21.6. b 21.7. a 21.8. a 21.9. c 21.10. b
Questions
21.11. The given quantities are the charge of the two particles, Q 1 Q and Q 2 (^) Q .They are separated by a
distance d. The Coulomb force between the changed particles is
2 1 2 2 2
Q Q Q
F k k d d
. If the change on
each particle is doubled so that Q 1 (^) ^ 2 Q Q 2 and the separation distance is d 2 d the then the
Coulomb Force is given by:
2 2
2 2
Q Q
F k k d d
so the force is the same as it was in the initial situation.
21.12. The gravitational force between the Sun and the Earth is S E g (^2)
M M
F G
r
where G is the gravitational
constant and is equal to 11 2 2 6.67 10 N m / kg , M (^) Sis the mass of the Sun ( 30 1.989 10 kg) and M (^) Eis
the mass of the Earth ( 24 5.974 10 kg). The Coulomb force is given by the equation 1 2 C (^2)
Q Q
F k r
where k is Coulomb’s constant ( k = 9 2 2 8.99 10 N m / C ). In this question Q 1 (^) Q 2 Q and is the
charge given to the Earth and Sun to cancel out the gravitational force. 2 S E C g 2 2 S E
kQ GM M G F F Q M M r r k
Therefore,
11 2 2 30 24 17 9 2 2
6.67 10 N m / kg (1.989 10^ kg)(5.974 10^ kg) 2.97 10 C. 8.99 10 N m / C
Q
I can get the number of elementary charges, n , by dividing Q by
19 1.602 10 C
(the charge of one
electron): 17 36 19
2.97 10 C
1.602 10 C
n
To estimate the number of elementary change of either sign for the Earth I can assume the mass of the Earth is due to the mass of the protons, neutrons and electrons of which it is primarily composed. If I assume that the Earth’s mass is due to the proton and neutron masses primarily (became an electrons mass is much smaller than a protons) and I assume that there are an equal number of protons and neutrons than I can get the number of protons by dividing the Earth’s mass
by two times the mass of a proton. The mass of a proton is
27 m (^) P 1.6726 10 kg,
so you can estimate
the number of elementary charges on the Earth, n E by:
24 E 51 E (^27) P
5.97 10 kg 3.57 10. 1.67 10 kg
m n m
So the
percentage of the Earth’s changes that would be required to cancel out the gravitational force
is
14 n / n E 100% 5.18 10 %,
a very small percentage.
21.13. One reason that it took such a long time to understand the electrostatic force may have been
because it was not observed as frequently as the gravitational force. All massive objects are acted on
Bauer/Westfall: University Physics, 1E
by the gravitational force; however, only objects with a net charge will experience an electrostatic force.
21.14. The accumulation of static charge gives the individual hairs a charge. Since like charges repel and
because the electrostatic force is inversely proportional to the charges separation distance squared, the hairs arrange themselves in a manner in which they are as far away from each other as possible. In this case that configuration is when the hairs are standing on end.
21.15. The given quantities are the charge which is Q 1 (^) Q 2 Q and the separation distance of 2. d The
third charge is Q 3 (^) 0.2 Q and it is positioned at d. Charge Q 3 is then displaced a
distance x perpendicular to the line connecting the positive charges. The displacement x d .The
question asks for the force, F , on charge Q 3 (^) .For x d the question also asks for the approximate
motion of the negative charge.
F F 13 F 23 ,
where F 13
is the force Q 3 feels due to Q 1 and F 23
is the force Q 3 feels due to charge
Q 2 (^) .Because Q 1 and Q 2 have the same sign and are of equal charge there is no net force in the ˆ x ‐
direction. The forces in the y ˆ ‐direction are given by:
1 3 (^13 ) 1
sin
Q Q
F k r
and 2 3 (^23 ) 2
sin ,
Q Q
F k r
where 2 2 r 1 d x and 2 2 r 2 d x and the negative signs denote that there will be an attraction
between the positive and negative charges. To simplify we can substitute sin 1 x / r 1 and
sin 2 x / r 2 into force equations. So we can write the force equation as:
1 3 2 3 2 2 2 2 2 2
3 (^13 23 2 2 12 ) 2 2 3/
kQ Q (^) x kQ Q (^) x kxQ F F F Q Q d x (^) d x d x d x (^) d x
Substituting Q 1 (^) Q 2 Q and Q 3 (^) 0.2 Q gives:
(^2 )
2 2 3/2^2 2 3/ 2^2 2 3/ 2
kx 0.2 Q^2 k^ 0.2 Q^ x 0.4 kQ x F Q Q d x d x d x
Since x d ,it is reasonable to use the approximation 2 2 3/ 2 2 3/2 3 ( d x ) ( d ) d .Hence,
2
3
0.4 kQ x F d
This solution is similar in form to Hooke’s law which describes the restoring force due to the compression or expansion of a spring, F spring kx where k is the spring constant. The motion of
the negative charge can therefore be approximated using simple harmonic motion.
21.16. As the garment is dried it acquires a charge from tumbling in the dryer and rubbing against other
clothing. When I put the charged garment on it causes a redistribution of the charge on my skin and this causes the attractive electric force between the garment and my skin.
Bauer/Westfall: University Physics, 1E
The electric lines flow from the positive charge to the negative charge as is shown in the sketch below.
There is nowhere on the line between the charged particles that I could place a test charge without it moving. This is due to the eclectic charges on the line having opposite charge, so a test charge (of either sign) that is placed between these two charges would be attracted by one and repelled by the other.
In order for the test charge to feel no net force it would have to be at a location where the force it
felt due to the charge Q 2 (^) 4 C.For convenience I can say that the charge Q 1 (^) 2 Cis located
at x 1 (^) 0 , and charge Q 2 (^) 4 C is located at x (^) 2 L and charge Q 3 is located at a position, x 3 which is
between 0 and L. I can equate the expressions for the electric force on Q 3 due to Q 1 and the electric
force on Q 3 due to Q 2 to solve for x 3 as these forces would have to balance for the charge Q 3 to feel
no net force.
13 23
1 3 2 3 2 2 3 3 2 2 1 3 2 3 2 2 2 1 3 3 2 3 2 2 1 2 3 1 3 1
F F
kQ Q kQ Q
x L x
Q L x Q x
Q x x L L Q x
Q Q x Q x L Q L
Note that in the second step of the calculation above, it is shown that the sign and magnitude of Q 3
will not impact the answer. I can solve using the quadratic equation: (^2 2 2 2 2 2 ) 1 1 1 2 1 3 1 2
2 4 4( )( ) 2(2 C) 4(2 C) 4(4C )
2( ) 4C
Q L Q L Q Q Q L L L L
x L L Q Q
The correct answer is x (^) 3 0.414 L because this point is between Q 1 and Q 2.
21.24. When a positively charged rod is brought near to an isolated neutral conductor without touching it
the rod will experience an attractive force. The electric charge on the rod induces a redistribution of charge in the conductor. The net effect of this distribution is that electrons move to the side of the conductor nearest to the rod. The positively charged rod is attracted to this region.
21.25. Using a metal key to touch a metal surface before exiting the car, which will discharge any charge I
carry. When I begin to fuel a car, I can touch the gas pump and the car before pumping the gas, discharging myself. If I get back into the car, I can re‐charge myself, and when I again get out of the
Chapter 21: Electrostatics
car and touch the fuel nozzle without grounding myself first, I can get a spark, which might ignite the gasoline.
Problems
21.26. The charge of each electron is 19 1.602 10 C. The total number n of electrons required to give a total
charge of 1.00 C is obtained by dividing the total charge by the charge per electron:
18 19
1.00 C
6.18 10 electrons. 1.602 10 C/electron
Q
n e
21.27. The number of atoms or molecules in one mole of a substance is given by Avogadro’s number,
23 1 N (^) A 6.022 10 mol.
The faraday unit is F N e A ,where e is the elementary charge of an electron
or proton and is equal to
19 1.602 10 C. To calculate the number of coulombs in 1.000 faraday you
can multiply N (^) Aby the elementary charge: 23 19 1.000 F N e A (6.022 10 atoms/mol)(1.602 10 C) 96470 C.
2 5 1 dyne 1 g cm / s 1 10 N and it is a unit of force. An electrostatic unit or esu is defined as follows:
Two point charges, each of 1 esu and separated by one centimeter exert a force of exactly one dyne on each other. Coulomb’s law gives the magnitude of the force on one charge due to another, which
is
2 F k q q 1 2 (^) / r (where
9 2 2 k 8.99 10 N m / C , q 1 and q 2 are electric charges and r is the
separation distance between charges.) (a) By substituting the values given in the question into Coulomb’s law, the relationship between the esu and the Coulomb can be determined: 2 2 5 5 10 2 9 2 2
(1 esu) (0.01 m) (1 10 N) 1 10 N 1 esu 3.34 10 C (0.01 m) 8.99 10 N m / C
k
(b) The result of part (a) shows that 10 1 esu 3.34 10 C. The elementary charge on an electron or
proton is 19 e 1.602 10 C. To get the relationship between the esu and elementary charge, can
divide 1 esu by the charge per electron (or proton). 10 9 19
3.34 10 C
1 esu 2.08 10 1.602 10 C/
e e
21.29. The given quantities are the current,
3 I 5.00 10 A
and the exposure time, t 10.0 s. One coulomb is equal to1 A s. To calculate the number of electrons that flow through your skin at this current and
during this time, multiply I by t and then divide by the elementary charge per electron which is 19 1.602 10 C.
3
17 19
5.00 10 A 10.0 s 0.0500 A s 0.0500 C;
0.0500 C 3.12 10 electrons. 1.602 10 C /
I t
e
21.30. THINK: Consider a mass, m 1.00 kgof water. To calculate how many electrons are in this mass, a
relationship must be found between mass, the number of water atoms presents and their charge.
Let denote the number of electrons.
SKETCH:
Chapter 21: Electrostatics
DOUBLE CHECK: The calculated answer has the correct units of charge. The value seems reasonable considering the values that were provided in the question.
21.32. The charges obtained by the student performing the experiment are listed here: 19 3.26 10 C, 19 6.39 10 C, 19 5.09 10 C, 19 4.66 10 C, 19 1.53 10 C. Dividing the above values by the smallest
measured value will give the number of electrons, ne found in each measurement.
The number of
electrons ne must be rounded to their closest integer value because charge is quantized. Dividing the
observed charge by the integer number of electrons gives the charge per electron. Taking the average of the observed charge/integer value data the average charge on an electron is calculated to
be 19 (1.60 0.03) 10 C. The error in a repeated measurement of the same quantity is:
error
standard deviation
number of measurements
N
21.33. THINK: An intrinsic silicon sample is doped with phosphorous. The level of doping is 1
phosphorous atom per one million silicon atoms. The density of silicon is
3
S 2.33 g/cm and its
atomic mass is m (^) S 28.09 g/mol.The phosphorous atoms act as electron donors. The density of
copper is
3
C 8.96 g/cm and its atomic mass is m C 63.54 g/mol.
SKETCH:
RESEARCH: Avogadro’s number is
23 1 N (^) A 6.022 10 mol.
It gives the number of atoms or
molecules per mole of a substance. Density, m / V ,where m massand V volume.
SIMPLIFY:
Observed charge ne Integer va lu e
Observed charge (integer value)
19 3.26 10 C
2.13^2
19 1.63 10 C
19 6.39 10 C 4.17^4 19 1.60 10 C
19 5.09 10 C
3.32^3
19 1.69 10 C
19 4.66 10 C 3.04^3 19 1.55 10 C
19 1.53 10 C
1 1
19 1.53 10 C
Bauer/Westfall: University Physics, 1E
(a) There will be 1 conduction electron per
6 1.00 10 silicon atoms.The number of silicon atoms
per
3
cm is n S S / m S N A. The number of conduction electrons per
3 cm is
6 n e (^) n S / (1.00 10 ).
(b) The number of copper atoms is n C C / m C N A. The number of conduction electrons in the
copper is n C. The ratio of conduction electrons in silicon to conduction electrons in copper is n e (^) / n C.
CALCULATE:
(a)
3 23 1 22 3 C
2.33 g/cm 6.022 10 mol 4.995 10 /cm 28.09 g/mol
n
22 16 3 e (^6)
4.995 10 conduction electrons / cm 1.00 10
n
(b)
3 23 1 22 3 C
8.96 g/cm 6.022 10 mol 8.4918 10 /cm 63.54 g/mol
n
16 e^7 22 C
n
n
ROUND: There were three significant figures provided in the question so the answers should be:
(a) 16 3 n e (^) 5.00 10 conduction electrons / cm
(b) There are 7 5.88 10 conduction electrons in the doped silicon sample for every conduction
electron in the copper sample.
DOUBLE CHECK: It is reasonable that there are approximately 7 5 10 less conduction electrons in
the doped silicon sample compared to the copper sample.
21.34. The force between the two charged spheres is F 1 k
qa qb
d 1 2 initially. After the spheres are moved the
force is (^2 ) 2
q qa b F k d
Taking the ratio of the force after to the force before gives:
F 2 / F 1 k
qa qb
d 2
2
/ k
qa qb
d 1
2
d 1 2 / d 2 2 4. The new distance is then d 2 d 1 2 / 4 d 1 / 2 4 cm.
21.35. The charge on each particle is q. When the separation distance is d 1.00 m,the electrostatic force
is F 1.00 N.The charge q is found from 2 2 2 F kq q 1 2 (^) / d kq / d .Then,
2 2 5 9 2 2
(1.00 N)(1.00 m) 1.05 10 C. 8.99 10 N m / C
Fd q k
The sign does not matter, so long as each particle has a charge of the same sign, so that they repel.
21.36. In order for two electrons to experience an electrical force between them equal to the weight of one
of the electrons, the distance d separating them must be such that. 2 2 F g (^) F Coulomb (^) m ge ke / d .Then,
9 2 2 19 2 2
31 2
8.99 10 N m / C 1.602 10 C 5.08 m e (9.109 10^ kg)(9.81 m/s )
ke d m g
21.37. In solid sodium chloride, chloride ions have a charge
19 q Cl (^) e 1.602 10 C,
while sodium ions
have a charge
19 q Na (^) e 1.602 10 C.
These ions are separated by about d 0.28 nm.The Coulomb
force between the ions is
9 2 2 19 2 Cl Na 9 9 2 9 2
8.99 10 N m / C (1.602 10 C) 2.94285 10 N 2.9 10 N. (0.28 10 m)
kq q F d
Bauer/Westfall: University Physics, 1E
(^2 1 2 212 )
2.0 μC 4.0 μC 8.99 10 N m /C 1.8 N 0.200 m
q q q q F k r k x x x r r
The 4.0 μCcharge pulls the 2.0 μC charge to the right.
21.41. THINK: The two identical spheres are initially uncharged. They are connected by an insulating
spring of equilibrium length L 0 (^) 1.00 mand spring constant k 25.0 N/m. Charges q and q are
then placed on metal spheres 1 and 2, respectively. Because the spring is insulating, the charges cannot neutralize across the spring. The spring contracts to new length L 0.635 m,due to the
attractive force between the charges spheres. Determine the charge q. If someone coats the spring
with metal to make it conducting, find the new length of the spring. SKETCH:
RESEARCH: The magnitude of the spring force is F S (^) k S x. The magnitude of the electrostatic force
is
2 F kq q 1 2 (^) / r. For this isolated system, the two forces must be in balance, that is F S F. From this
balance, the charge q can be determined. The spring constant is denoted by k (^) Sto avoid confusion
with the Coulomb constant, k.
SIMPLIFY:
2 2 1 2 S^0 S S 2 S (^02)
kq q kq k^ L^ L^ L F F k x k L L q r (^) L k
CALCULATE:
2 5 9 2 2
25.0 N/m 0.635 m (1.00 m 0.635 m) 2.02307 10 C 8.99 10 N m / C
q
If someone were to coat the spring such that it conducted electricity, the charge on the two spheres would distribute themselves evenly about the system. If the charges are equal in magnitude and opposite in sign, as they are in this case, the net charge in the system would be zero. Then the electrostatic force between the two spheres would be zero, and the spring would return to its equilibrium length of 1.00 m.
ROUND: To three significant figures, 5 q 2.02 10 C.
DOUBLE CHECK: Dimensional analysis confirms that the answer is in coulombs, the appropriate unit for charge.
21.42. THINK: A point‐like charge of q 1 (^) 3 q is located at x 1 (^) 0,and a point‐like charge of q 2 q is
located on the x ‐axis at x (^) 2 D ,where D 0.500 m.Find the location on the x ‐axis x (^) 3 where will a
third charge q 3 (^) q 0 experiences no net force from the other two charges.
SKETCH:
Chapter 21: Electrostatics
RESEARCH: The magnitude of the electrostatic force is
2 F kq q 1 2 (^) / r. The net force on the third
charge q 3 is zero when the sum of the forces from the other two charges is zero:
F net,3 (^) F 13 (^) F 23 (^) 0 F 13 (^) F 23 .The two forces F 13 and F 23 must be equal in magnitude, but
opposite in direction. Consider the following three possible locations for the charge q 3. Note that
this analysis is independent of the charge of q 3. In the case x (^) 3 x 1 0,the two forces F 13 and F 23 will
be opposite in direction but they cannot be equal in magnitude: the charge q 1 at x 1 is greater in
magnitude than the charge q 2 at x (^) 2 and x (^) 3 would be closer to x 1. (Remember that the electrostatic
force increases as the distance between the charges decreases.) This makes the magnitude of F 13 greater than that of F 23. In the case 0 m x 3 0.500 m, the two forces are in the same direction
and therefore cannot balance. In the case x 3 (^) x 2 D , the two forces are opposite in direction, and
in direct opposition to the first situation, the force F 13 and F 23 can now be balanced. The solution will
have a positive x position, or more accurately, the third charge q 3 must be placed near the smaller
fixed charge, q 2 , without being between the two fixed charges q 1 and q 2
SIMPLIFY: Since x 3 (^) x 2 , consider only the magnitudes of the forces. Since only the magnitudes of the forces
are compared, only the magnitudes of the charges need be considered.
1 3 2 3 2 2 2 2 13 23 2 2 1 3 2 2 3 3 3 (^3 3 )
kq q kq q F F q x x q x q x D qx x (^) x x
(^2 2 2 ) 3 x 3 (^) D x (^) 3 0 2 x (^) 3 6 x D 3 3 D 0
Solving for x 3 :
2 2
3
D D D
x
CALCULATE:
2 2
3
6(0.500 m) 36(0.500 m) 24(0.500 m) 1.1830 m, 0.3170 m 4
x
ROUND: Since x (^) 3 x 2 , x 3 (^) 1.18 m.
DOUBLE CHECK: The solution fits the expected location that was determined above (where x 3 (^) x 2 ).
21.43. THINK: Identical point charges
6 Q 32 10 C
are placed at each of the four corners of a rectangle of
dimensions L 2.0 mby W 3.0 m.Find the magnitude of the electrostatic force on any one of the
charges. Note that by symmetry the magnitude of the net force on each charge is equal. Choose to compute the net electrostatic force on Q 4.
SKETCH:
Chapter 21: Electrostatics
RESEARCH: The magnitude of the force between two charges is
2 3 F 12 (^) kq q r 1 2 12 / r 12 (^) kq q r 1 2 12 (^) / r 12.
The total force on charge q 3 is the sum of all the forces acting on it. The magnitude of F 3 is found
from (^) 2 2 1/ F 3 (^) F 1 (^) F 2 ,and the direction (^) is found from (^) 1
tan Fy / Fx.
SIMPLIFY:
net, 3 13 23
1 3 13 2 3 23 3 3 13 23
1 3 3 1 3 1 2 3 3 2 3 2 2 2 3/2^2 2 3/ 2 3 1 3 1 3 2 3 2
1 3 2 3 3 3 2 2 3/ 2^2 3 2 3
( ) ˆ^ ( ) ˆ^ ( ) ˆ^ ( )ˆ
F F F
kq q r kq q r
r r
kq q x x x y y y kq q x x x y y y
x x y y x x y y
kq q kq q y y x x y y y x y
CALCULATE:
^
^
9 2 2 8 8
net, 3 (^3)
9 2 2 8 8
2 2 3/ 2
5 5 5
8.99 10 N m / C (1.4 10 C)(2.1 10 C) 0.24 m ˆ 0.24 m
8.99 10 N m / C ( 1.8 10 C)(2.1 10 C) 0.18 ˆ^ m 0.24ˆ m
0.18 m 0.24 m
(4.5886 10 N)ˆ (2.265 10 N) ˆ (3.0206 10 N)
F y
x y
y x
5 5
2.265 10 N ˆ^ 1.568 10 N ˆ
y
x y
2 2 5 2 5 2 5 net, 3 5 1 1 5
(2.265 10 N) (1.568 10 N) 2.755 10 N
1.568 10 N
tan tan 34.69 above the horizontal 2.265 10 N
x y
y
x
F F F
F
F
^
Bauer/Westfall: University Physics, 1E
ROUND: With 2 significant figures in each given value, the final answers should be rounded to
5 5 5 F net, 3 2.265 10 N x ˆ^ 1.568 10 N y ˆ 2.8 10 N
and 35 .
DOUBLE CHECK: Due to the attraction between q 2 and q 3 and that q 1 is directly underneath q 3 ,
the x component of F net, 3
has to be positive.
21.45. THINK: A positive charge Q is on the y ‐axis at a distance a from the origin and another positive
charge q is on the x ‐axis at a distance b from the origin. (a) Find the value(s) of b for which the x ‐
component of the force on q is a minimum. (b) Find the value(s) of b for which the x ‐component of
the force on q is a maximum.
SKETCH:
RESEARCH: The electrostatic force is
2 F kqQr / r. The x ‐component of this force
is
2
Fx ( kqQ / r )cos .The values of b for which Fx is a minimum can be determined by inspection;
the values of b for which Fx is a maximum can be found by calculating the extrema of Fx , that is,
taking the derivative of Fx with respect to b , setting it to zero, and solving for b.
SIMPLIFY:
2 3 3/ 2 2 x cos
kqQ kqQb kqQb F r r (^) a b
a) Minima: By inspection, the least possible value of Fx is zero, and this is attained only when b 0.
b) Maxima: 0
dFx
db
2 2 2 2 2 5/ 2 2 3/ 2^2 2 5/ 2
2 2 2
kqQ kqQ a^ b^ kqQb kqQ a b b a b a b
a a b b b
^
CALCULATE: Reject the negative solution, since distances have to be positive:. 2
a b
Bauer/Westfall: University Physics, 1E
9 net, 2 2 2 3/2^2 2 3/ 2
8 7
(1.00) 17.0 ˆ^ 5.00 ˆ (3.00)(19.0 ˆ 16.0 )ˆ
1.2181 10 ˆ^ 7.2469 10 ˆ.
x y (^) x y F
x y
^ ^ ^
^ ^ ^
Then, the units of F net, 2
are:
2 2 net, 2 (^) 3/2 3/ 2 2 2 2
(mC)(mm mm) (mC)(mm mm) N m / C (mC) N
mm mm mm mm
F
^
^ ^ ^
^ ^ ^
Altogether , (^)
8 7 F net, 2 1.2181 10 N x ˆ^ 7.2469 10 N y ˆ.
The magnitude of the force is
2 2 8 2 7 2 8 F net, 2 (^) Fx Fy 1.2181 10 N 7.2469 10 N 1.4174 10 N
ROUND: (^)
8 7 8 F net, 2 1.22 10 N x ˆ^ 7.25 10 N ˆ y and F net, 2 1.42 10 N.
DOUBLE CHECK: The charges are large and the separation distance are small, so F net, 2 should be
very strong.
21.48. THINK: the masses of the beads are
5 m 10.0 mg 1.00 10 kg
and they have identical charge. They
are a distance d 0.0200 mapart. The coefficient of static friction between the beads and the
surface is 0.200.Find the minimum charge q needed for the beads to start moving.
SKETCH:
RESEARCH: Assume the surface is parallel to the surface of the Earth. The frictional force is f N ,
where N mg .The electrostatic force is
2 2 F kq / d .The beads will start to move as soon as F is
greater than f ,enabling one bead to move away from the other. Then the minimum charge q can be
found by equating f and F.
SIMPLIFY:
2 2 2 /
kq F f mg q d mg k d
CALCULATE:
(^2 5 ) 10 9 2 2
0.0200 m (0.200)(1.00 10 kg)(9.81 m / s ) 9.3433 10 C 8.99 10 N m / C
q
ROUND: All of the given values have three significant figures, so
10 q 9.34 10 C.
DOUBLE CHECK: The units of the solution are those of charge. This is a reasonable charge required to overcome the frictional force.
21.49. THINK: The ball’s mass is m 1 (^) 0.0300 kg;its charge is q 1 (^) 0.200 μ C .The ball is suspended a
distance of d 0.0500 m above an insulating floor. The second small ball has mass
m 2 (^) 0.0500 kgand a charge q 2 (^) 0.400 μC.Determine if the second ball leaves the floor. Find the
tension T in the string when the second ball is directly beneath the first ball. SKETCH:
Chapter 21: Electrostatics
RESEARCH: The electrostatic force between two charges is 2 F kq q 1 2 (^) / r. The force of gravity is
F g (^) mg .The ball will leave the floor if the electrostatic force between the two balls is greater that
the force of gravity, that is if F F g, and if the charges are opposite. The tension in the rope can be
found by considering all of the vertical forces acting on the first ball.
SIMPLIFY: The electrostatic force is:
2
F kq q 1 2 / d .The gravitational force is: F g m 2 g . The
forces acting on m 1 in the y ‐direction sum to: 0 T F coulomb (^) m g 1 .So the tension is T F coulomb (^) m g 1.
CALCULATE: (^)
9 2 2 6 6 2 F 8.99 10 N m / C ( 0.200 10 C)(0.400 10 C)/(0.0500 m) 0.28768 N,
2 F g (^) (0.0500 kg)( 9.81 m/s ) 0.4905 N,
2 T 0.28768 N (0.0300 kg)( 9.81 m/s ) 0.58198 N.
Since F g (^) F ,the second ball does not leave the ground.
ROUND: With all given values containing three significant figures, round the tension to T 0.582 N. DOUBLE CHECK: The balls are not quite close enough to overcome the force of gravity, but the
magnitude of F coulomb is comparable to F g , despite the small charges (on the order of 7 10 C ).
21.50. THINK: A q 1 (^) 3.00 mC charge and a q 2 (^) 4.00 mC charge are fixed in position and separated by
d 5.00 m.Take the position of q 1 to be at x (^) 1 0,and position of q 2 to be at x (^) 2 5.00 m.(a) Find
the location, x (^) 3 ,of a q 3 (^) 7.00 mC charge so that the net force on it is zero. (b) Find the
location, x (^) 3 ,of a q 3 (^) 7.00 mC charge so that the net force on it is zero.
SKETCH:
RESEARCH: The electrostatic force between two charges is
2 F kq q 1 2 (^) / r. The net force on a third
charge is zero: F net,3 (^) F 13 (^) F 23 (^) 0 F 13 (^) F 23 .The two forces must be equal in magnitude, but
opposite in direction. Consider the following three possible locations for the charge q 3. Note that
this analysis is independent of the charge of q 3 : At x 3 (^) 5.00 m,the two forces F 13 and F 23 will be
opposite in direction but they cannot be equal in magnitude: the charge q 2 at x (^) 2 5.00 mis greater
in magnitude than the charge q 1 at x 1 (^) 0 and x (^) 3 would be closer to x (^) 2. (Remember that the
electrostatic force increases as the distance between the charges decreases.) This makes the magnitude of F 23 greater than that of F 13. Next, consider values of x 3 satisfying: 0 m x 3 5.00 m.
The two forces are in the same direction and therefore cannot balance. At x (^) 3 0 m,the two forces
are opposite in direction, and in direct opposition to the first situation, the force F 13 and F 23 can
now be balanced. The solution will have a negative position, or more accurately, the third charge q 3
Chapter 21: Electrostatics
SIMPLIFY: Balancing the forces in the vertical ( z ) direction yields
coulomb (^2)
sin. 4 4
e e
kqq F W m g r
Solving for q :
2 3/ 2 2 3 2 2 3/ 2 1 ( )^2 . 4 sin 4 4 4
e e e e
e e e
d m g z m gr m gr m g L z q
kq kq z kq z kez
CALCULATE:
2 3/ 31 2 2
9 2 2 19
29 10
(10.0 μm) 9.109 10 kg (9.81 m / s ) (15 nm) 2
4 8.99 10 N m / C (1.602 10 C)(15 nm)
3.6562 10 C, or - 2.282 10
q
e
ROUND: With 2 significant figures in z ,^
29 10 q 3.7 10 C 2.3 10 e.
DOUBLE CHECK: The gravitational force on an electron is extremely small, on the order of
30 10 N.
The force charges q need only an extremely small charge to balance the gravitational force on the
electron.
21.52. THINK: A uniformly charged thin rod of length L has a total charge Q. Find an expression for the
electrostatic force strength acting on an electron, whose magnitude of charge is e, is positioned on the axis of the rod at distance d from the center.
SKETCH:
RESEARCH: The electrostatic force between two charges is 2 F kqQ / r. The net electrostatic force
acting on a charge q is the sum of all the electrostatic forces acting on q. In the event of a
continuous and linear distribution of charge of length L and total charge Q , the force due to an
infinitesimal amount of charge dp from the distribution acting on the charge q is: 2 dF kq dq / x ,
where dp ( Q / L dx ) dx .( is the linear charge density.) In this case, the total force on the
electron is then / 2
/2 2
d L
d L
ke F dx x
where the integration runs over the length of the rod, starting from the point closest to the electron
d^ ^ L / 2and ending with the point farthest from the electron^ d^ L / 2 .
SIMPLIFY:
/ (^2 2) / 2 2 2 2 2
/2 /
/2 /
d
d L d
d L d
L
L
L d L
ke ke L keQ F dx ke dx ke x ke x x d^ L^ d^ L d L d L
(^) (^) ^ ^
CALCULATE: Not applicable ROUND: Not applicable
DOUBLE CHECK: The answer is in the correct units of force:
2
2
2
N m C C C N. m
F
21.53. THINK: A negative charge q is located and fixed at (0, 0). A positive charge q is initially at
( , 0). x The positive charge will accelerate towards the negative charge. Use the binomial expansion
Bauer/Westfall: University Physics, 1E
to show that when the positive charge moves a distance x closer to the negative charge, the
force on it increases by 2 3
F 2 kq / x.
SKETCH:
RESEARCH: The Coulomb force is 2 F 21 (^) kq q r 1 2 (^) 21 / r 21 ,
where r 21
is the unit vector that points from
charge 2 to charge 1. To first order, the binomial expansion is (1 ) 1 n x nx for x 1.
SIMPLIFY: The initial force on q (when it was at ( , 0) x was
2
2
kq F x x
After moving closer to q
by 1 the new force on q is
2 2 2 2
2 2 2 2
kq kq kq F x x x x x x x x
^ (^)
Using the binomial
expansion,
2 2
2 2
kq kq F x x x x x x
^ ^ ^
(to first order in ). Then,
2
3
2 kq F F F x x
^
and
2
3
kq F x
as desired.
CALCULATE: Not applicable. ROUND: Not applicable.
DOUBLE CHECK: The charge in force has the correct units for force:
2
2 2
N m C C m C m N. m
^ F^ ^
21.54. THINK: Two charges, both q^ , are located and fixed at coordinates^ (^ d^ ,0)and^ ( ,0) d^ in the^ x^ y^ plane.
A positive charge of the same magnitude q and of mass m is placed at coordinate (0,0). The positive
charge is then moved a distance d along the +y direction and then released. It will oscillate
between co‐ordinates (0, )and (0, ). Find the net force F net acting on the positive charge when it is
moved to (0, )and use the binomial expansion to find an expression for the frequency of the
resulting oscillation. SKETCH:
RESEARCH: The Coulomb force is 2 F 21 (^) kq q r 1 2 21 / r 21 ,
where F 21
is the force on the charge 1 by
charge 2, and r 21
points from charge 2 to charge 1. To first order, the binomial expansion is, in
