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Physics Bio 178.
Physics Bio 178.
of this charge — top, bottom, left, or right — is the electric field the strongest?
The weakest? Explain.
CHAPTER 16: Electric Charge and Electric Field
Questions
1.
If you charge a pocket comb by rubbing it with a silk scarf, how can you
determine if the comb is positively or negatively charged?
2.
Why does a shirt or blouse taken from a clothes dryer sometimes cling to your
body?
17.
Consider the electric field at points A, B, and C in Fig. 16–48. First draw an arrow
at each point indicating the direction of the net force that a positive test charge
would experience if placed at that point, then list the points in order of decreasing
field strength (strongest first).
18.
Why can electric field lines never cross?
19.
Show, using the three rules for field lines given in Section 16–8, that the electric
field lines starting or ending on a single point charge must be symmetrically
spaced around the charge.
3.
Explain why fog or rain droplets tend to form around ions or electrons in the air.
4.
A positively charged rod is brought close to a neutral piece of paper, which it
attracts. Draw a diagram showing the separation of charge and explain why
attraction occurs.
20.
Given two point charges Q and 2Q, a distance l apart, is there a point along the
straight line that passes through them where E = 0 when their signs are (a)
opposite, (b) the same? If yes, state roughly where this point will be.
5.
Why does a plastic ruler that has been rubbed with a cloth have the ability to pick
up small pieces of paper? Why is this difficult to do on a humid day?
21.
6.
Contrast the net charge on a conductor to the “free charges” in the conductor.
Consider a small positive test charge located on an electric field line at some
point, such as point P in Fig. 16–31a. Is the direction of the velocity and/or
acceleration of the test charge along this line? Discuss.
7.
Figures 16–7 and 16–8 show how a charged rod placed near an uncharged metal
object can attract (or repel) electrons. There are a great many electrons in the
metal, yet only some of them move as shown. Why not all of them?
22.
Sketch the electric field lines for a uniform line of charge which is infinitely long.
(Hint: Use symmetry.) Is the electric field uniform in strength?
*23.
If the electric flux through a closed surface is zero, is the electric field necessarily
zero at all points on the surface? Explain. What about the converse: If E = 0 at all
points on the surface is the flux through the surface zero?
*24.
A point charge is surrounded by a spherical gaussian surface of radius r. If the
sphere is replaced by a cube of side r, will Φ E be larger, smaller, or the same?
Explain.
8.
When an electroscope is charged, its two leaves repel each other and remain at an
angle. What balances the electric force of repulsion so that the leaves don’t
separate further?
9.
The form of Coulomb’s law is very similar to that for Newton’s law of universal
gravitation. What are the differences between these two laws? Compare also
gravitational mass and electric charge.
10.
We are not normally aware of the gravitational or electric force between two
ordinary objects. What is the reason in each case? Give an example where we are
aware of each one and why.
Problems
16–5 and 16–6
[1 mC = 10
−3
Coulomb’s Law
C , 1 µC = 10 −6 C, 1 nC = 10 −9 C.]
11.
Is the electric force a conservative force? Why or why not? (See Chapter 6.)
12.
When a charged ruler attracts small pieces of paper, sometimes a piece jumps
quickly away after touching the ruler. Explain.
1.
(I) Calculate the magnitude of the force between two 3.60- µC point charges 9.3
cm apart.
13.
Explain why the test charges we use when measuring electric fields must be
small.
2.
(I) How many electrons make up a charge of −30.0 µC?
14.
When determining an electric field, must we use a positive test charge, or would a
negative one do as well? Explain.
3.
(I) What is the magnitude of the electric force of attraction between an iron
nucleus ( q = +26e) and its innermost electron if the distance between them is
15.
Draw the electric field lines surrounding two negative electric charges a distance l
apart.
16.
Assume that the two opposite charges in Fig. 16–31a are 12.0 cm apart. Consider
the magnitude of the electric field 2.5 cm from the positive charge. On which side
1
1.5 × 10 −12 m?
4.
(I) What is the repulsive electrical force between two protons 5.0 ×10 −15 m apart
from each other in an atomic nucleus?
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Physics Bio 178.
Physics Bio 178.
5.
(I) What is the magnitude of the force a +25 µC charge exerts on a +3.0 mC charge
35 cm away?
20.
(III) A +4.75 µC and a −3.55 µC charge are placed 18.5 cm apart. Where can a
third charge be placed so that it experiences no net force?
6.
(II) Two charged dust particles exert a force of 3.2 × 10 −2 N on each other. What
will be the force if they are moved so they are only one-eighth as far apart?
21.
7.
(II) Two charged spheres are 8.45 cm apart. They are moved, and the force on
each of them is found to have been tripled. How far apart are they now?
(III) Two small nonconducting spheres have a total charge of 90.0 µC. (a) When
placed 1.06 m apart, the force each exerts on the other is 12.0 N and is repulsive.
What is the charge on each? (b) What if the force were attractive?
22.
(II) A person scuffing her feet on a wool rug on a dry day accumulates a net
charge of −42 µ C. How many excess electrons does she get, and by how much
does her mass increase?
(III) A charge Q is transferred from an initially uncharged plastic ball to an
identical ball 12 cm away. The force of attraction is then 17 mN. How many
electrons were transferred from one ball to the other?
16–7 and 16–8 Electric Field, Field Lines
9.
(II) What is the total charge of all the electrons in 1.0 kg of H 2 O?
23.
10.
(II) Compare the electric force holding the electron in orbit ( r = 0.53 × 10 −10 m)
around the proton nucleus of the hydrogen atom, with the gravitational force
between the same electron and proton. What is the ratio of these two forces?
(I) What are the magnitude and direction of the electric force on an electron in a
uniform electric field of strength 2360 N C that points due east?
24.
11.
(II) Two positive point charges are a fixed distance apart. The sum of their
charges is QT . What charge must each have in order to (a) maximize the electric
force between them, and (b) minimize it?
(I) A proton is released in a uniform electric field, and it experiences an electric
force of 3.75 × 10 −14 N toward the south. What are the magnitude and direction of
the electric field?
25.
(I) A downward force of 8.4 N is exerted on a −8.8 µC charge. What are the
magnitude and direction of the electric field at this point?
8.
12.
(II) Particles of charge +75, + 48, and −85 µC are placed in a line (Fig. 16–49). The
center one is 0.35 m from each of the others. Calculate the net force on each
charge due to the other two.
26.
(I) What are the magnitude and direction of the electric field 20.0 cm directly
above an isolated 33.0 ×10 −6 C charge?
27.
13.
(II) Three positive particles of equal charge, +11.0 µC, are located at the corners of
an equilateral triangle of side 15.0 cm (Fig. 16–50). Calculate the magnitude and
direction of the net force on each particle.
(II) What is the magnitude of the acceleration experienced by an electron in an
electric field of 750 N C? How does the direction of the acceleration depend on
the direction of the field at that point?
28.
14.
(II) A charge of 6.00 mC is placed at each corner of a square 0.100 m on a side.
Determine the magnitude and direction of the force on each charge.
(II) What are the magnitude and direction of the electric field at a point midway
between a −8.0 µC and a +7.0 µC charge 8.0 cm apart? Assume no other charges
are nearby.
15.
(II) Repeat Problem 14 for the case when two of the positive charges, on opposite
corners, are replaced by negative charges of the same magnitude (Fig. 16–51).
29.
(II) Draw, approximately, the electric field lines about two point charges, +Q and
−3Q, which are a distance l apart.
16.
(II) At each corner of a square of side l there are point charges of magnitude Q,
2Q, 3Q, and 4Q (Fig. 16–52). Determine the force on (a) the charge 2Q, and (b)
the charge 3Q, due to the other three charges.
30.
(II) What is the electric field strength at a point in space where a proton
(m = 1.67 ×10 −27 kg) experiences an acceleration of 1 million “g’s”?
17.
(II) Three charged particles are placed at the corners of an equilateral triangle of
side 1.20 m (Fig. 16–53). The charges are +4.0 µC, − 8.0 µC, and −6.0 µC. Calculate
the magnitude and direction of the net force on each due to the other two.
31.
(II) An electron is released from rest in a uniform electric field and accelerates to
the north at a rate of 115 m s 2 . What are the magnitude and direction of the
electric field?
18.
(III) Two point charges have a total charge of 560 µC. When placed 1.10 m apart,
the force each exerts on the other is 22.8 N and is repulsive. What is the charge on
each?
32.
(II) The electric field midway between two equal but opposite point charges is
745 N C , and the distance between the charges is 16.0 cm. What is the magnitude
of the charge on each?
19.
(III) Two charges, −Q0 and −3Q0 , are a distance l apart. These two charges are
free to move but do not because there is a third charge nearby. What must be the
charge and placement of the third charge for the first two to be in equilibrium?
33.
(II) Calculate the electric field at the center of a square 52.5 cm on a side if one
corner is occupied by a +45.0 µC charge and the other three are occupied by
−27.0 µC charges.
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Physics Bio 178.
Physics Bio 178.
34.
(II) Calculate the electric field at one corner of a square 1.00 m on a side if the
other three corners are occupied by 2.25 × 10 −6 C charges.
*45.
35.
(II) Determine the direction and magnitude of the electric field at the point P in
Fig. 16–54. The charges are separated by a distance 2a, and point P is a distance x
from the midpoint between the two charges. Express your answer in terms of Q,
x, a, and k.
(II) In Fig. 16–61, two objects, O1 and O 2 , have charges +1.0 µC and −2.0 µC,
respectively, and a third object, O 3 , is electrically neutral. (a) What is the electric
flux through the surface A1 that encloses all three objects? (b) What is the electric
flux through the surface A2 that encloses the third object only?
*46.
(II) A cube of side l is placed in a uniform field E = 6.50 ×10 3 N C with edges
parallel to the field lines. (a) What is the net flux through the cube? (b) What is
the flux through each of its six faces?
*47.
(II) The electric field between two square metal plates is 130 N C . The plates are
1.0 m on a side and are separated by 3.0 cm. What is the charge on each plate
(assume equal and opposite)? Neglect edge effects.
*48.
(II) The field just outside a 3.50-cm-radius metal ball is 2.75 × 10 2 N C and points
toward the ball. What charge resides on the ball?
*49.
(II) A solid metal sphere of radius 3.00 m carries a total charge of −3.50 µC. What
is the magnitude of the electric field at a distance from the sphere’s center of (a)
0.15 m, (b) 2.90 m, (c) 3.10 m, and (d) 6.00 m? (e) How would the answers differ
if the sphere were a thin shell?
*50.
(III) A point charge Q rests at the center of an uncharged thin spherical
conducting shell. (See Fig. 16–33.) What is the electric field E as a function of r
(a) for r less than the inner radius of the shell, (b) inside the shell, and (c) beyond
the shell? (d) Does the shell affect the field due to Q alone? Does the charge Q
affect the shell?
36.
37.
(II) Two point charges, Q1 = 25 µC and Q2 = +50 µC, are separated by a distance of
12 cm. The electric field at the point P (see Fig. 16–55) is zero. How far from Q1
is P?
(II) (a) Determine the electric field E at the origin 0 in Fig. 16–56 due to the two
charges at A and B. (b) Repeat, but let the charge at B be reversed in sign.
38.
(II) Use Coulomb’s law to determine the magnitude and direction of the electric
field at points A and B in Fig. 16–57 due to the two positive charges (Q = 7.0 µC)
shown. Are your results consistent with Fig. 16–31b?
39.
(II) You are given two unknown point charges, Q1 and Q2 . At a point on the line
joining them, one-third of the way from Q1 to Q2 , the electric field is zero (Fig.
16–58). What is the ratio Q1 Q2 ?
40.
(III) Determine the direction and magnitude of the electric field at the point P
shown in Fig. 16–59. The two charges are separated by a distance of 2a. Point P is
on the perpendicular bisector of the line joining the charges, a distance x from the
midpoint between them. Express your answers in terms of Q, x, a, and k.
41.
(III) An electron (mass m = 9.11× 10 −31 kg ) is accelerated in the uniform field
E ( E = 1.45 ×10 4 N C) between two parallel charged plates. The separation of the
plates is 1.10 cm. The electron is accelerated from rest near the negative plate and
passes through a tiny hole in the positive plate, Fig. 16–60. (a) With what speed
does it leave the hole? (b) Show that the gravitational force can be ignored.
42.
*16–11
*51.
(III) An electron moving to the right at 1.0% the speed of light enters a uniform
electric field parallel to its direction of motion. If the electron is to be brought to
rest in the space of 4.0 cm, (a) what direction is required for the electric field, and
(b) what is the strength of the field?
DNA
(III) The two strands of the helix-shaped DNA molecule are held together by
electrostatic forces as shown in Fig. 16–44. Assume that the net average charge
(due to electron sharing) indicated on H and N atoms is 0.2e and on the indicated
C and O atoms is 0.4e. Assume also that atoms on each molecule are separated by
1.0 × 10 −10 m. Estimate the net force between (a) a thymine and an adenine; and (b)
a cytosine and a guanine. For each bond (red dots) consider only the three atoms
in a line (two atoms on one molecule, one atom on the other). (c) Estimate the
total force for a DNA molecule containing 105 pairs of such molecules.
*16–10 Gauss’s Law
*43.
*44.
General Problems
(I) The total electric flux from a cubical box 28.0 cm on a side is
1.45 ×10 3 N ⋅ m 2 C . What charge is enclosed by the box?
(II) A flat circle of radius 18 cm is placed in a uniform electric field of magnitude
5.8 × 10 2 N C What is the electric flux through the circle when its face is (a)
perpendicular to the field lines, (b) at 45° to the field lines, and (c) parallel to the
field lines?
5
52.
How close must two electrons be if the electric force between them is equal to the
weight of either at the Earth’s surface?
53.
A 3.0-g copper penny has a positive charge of 38 µC. What fraction of its
electrons has it lost?
54.
A proton (m =1.67×10 kg) is suspended at rest in a uniform electric field E.
Take into account gravity at the Earth’s surface, and determine E.
−27
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Physics Bio 178.
Physics Bio 178.
55.
Measurements indicate that there is an electric field surrounding the Earth. Its
magnitude is about 150 N C at the Earth’s surface and points inward toward the
Earth’s center. What is the magnitude of the electric charge on the Earth? Is it
positive or negative? [Hint: the electric field outside a uniformly charged sphere is
the same as if all the charge were concentrated at its center.]
64.
A large electroscope is made with “leaves” that are 78-cm-long wires with tiny
24-g spheres at the ends. When charged, nearly all the charge resides on the
spheres. If the wires each make a 30° angle with the vertical (Fig. 16–64), what
total charge Q must have been applied to the electroscope? Ignore the mass of the
wires.
56.
(a) Given the local electric field of 150 N C what is the acceleration experienced
by an electron near the surface of the Earth? (b) What about a proton? (c)
Calculate the ratio of each acceleration to g = 9.8 m s 2 .
65.
Dry air will break down and generate a spark if the electric field exceeds about
3 × 10 6 N C . How much charge could be packed onto a green pea (diameter 0.75
cm) before the pea spontaneously discharges? [Hint: Eqs. 16–4 work outside a
sphere if r is measured from its center.]
57.
A water droplet of radius 0.018 mm remains stationary in the air. If the
downward-directed electric field of the Earth is 150 N C , how many excess
electron charges must the water droplet have?
66.
58.
Estimate the net force between the CO group and the HN group shown in Fig. 16–
62. The C and O have charges ± 0.40e, and the H and N have charges ± 0.20e,
where e = 1.6 ×10 −19 C. [Hint: Do not include the “internal” forces between C and
O, or between H and N.]
Two point charges, Q1 = −6.7 µC and Q2 = 1.8 µC are located between two
oppositely charged parallel plates, as shown in Fig. 16–65. The two charges are
separated by a distance of x = 0.34 m. Assume that the electric field produced by
the charged plates is uniform and equal to E = 73,000 N C . Calculate the net
electrostatic force on Q1 and give its direction.
67.
A point charge ( m = 1.0 g) at the end of an insulating string of length 55 cm is
observed to be in equilibrium in a uniform horizontal electric field of 12,000 N C ,
when the pendulum’s position is as shown in Fig. 16–66, with the charge 12 cm
above the lowest (vertical) position. If the field points to the right in Fig. 16–66,
determine the magnitude and sign of the point charge.
68.
A point charge of mass 0.210 kg, and net charge +0.340 µC, hangs at rest at the
end of an insulating string above a large sheet of charge. The horizontal sheet of
uniform charge creates a uniform vertical electric field in the vicinity of the point
charge. The tension in the string is measured to be 5.67 N. Calculate the
magnitude and direction of the electric field due to the sheet of charge (Fig. 16–
67).
69.
What is the total charge of all the electrons in a 15-kg bar of aluminum? What is
the net charge of the bar? (Aluminum has 13 electrons per atom and an atomic
mass of 27 u.)
70.
Two small, identical conducting spheres A and B are a distance R apart; each
carries the same charge Q. (a) What is the force sphere B exerts on sphere A? (b)
An identical sphere with zero charge, sphere C, makes contact with sphere B and
is then moved very far away. What is the net force now acting on sphere A? (c)
Sphere C next makes contact with sphere A and is then moved far away. What is
the force on sphere A in this third case?
71.
Given the two charges shown in Fig. 16–68, at what position(s) x is the electric
field zero? Is the field zero at any other points, not on the x axis?
72.
Two point charges, +Q and −Q of mass m, are placed on the ends of a massless
rod of length L, which is fixed to a table by a pin through its center. If the
apparatus is then subjected to a uniform electric field E parallel to the table and
perpendicular to the rod, find the net torque on the system of rod plus charges.
59.
In a simple model of the hydrogen atom, the electron revolves in a circular orbit
around the proton with a speed of 1.1× 10 6 m s . Determine the radius of the
electron’s orbit. [Hint: See Chapter 5 on circular motion.]
60.
Suppose that electrical attraction, rather than gravity, were responsible for holding
the Moon in orbit around the Earth. If equal and opposite charges Q were placed
on the Earth and the Moon, what should be the value of Q to maintain the present
orbit? Use these data: mass of Earth = 5.98 ×10 24 kg, mass of Moon = 7.35 × 10 22 kg,
radius of orbit = 3.84 ×10 8 m. Treat the Earth and Moon as point particles.
61.
An electron with speed v 0 = 21.5 × 10 6 m s is traveling parallel to an electric field of
magnitude E = 11.4 × 10 3 N C . (a) How far will the electron travel before it stops?
(b) How much time will elapse before it returns to its starting point?
62.
A positive point charge Q1 = 2.5 × 10 −5 C is fixed at the origin of coordinates, and a
negative charge Q 2 = −5.0 × 10 −6 C is fixed to the x axis at x = +2.0 m. Find the
location of the place(s) along the x axis where the electric field due to these two
charges is zero.
63.
A small lead sphere is encased in insulating plastic and suspended vertically from
an ideal spring ( k = 126 N m) above a lab table, Fig. 16–63. The total mass of the
coated sphere is 0.800 kg, and its center lies 15.0 cm above the tabletop when in
equilibrium. The sphere is pulled down 5.00 cm below equilibrium, an electric
charge Q = −3.00 × 10 −6 C is deposited on it and then it is released. Using what you
know about harmonic oscillation, write an expression for the electric field
strength as a function of time that would be measured at the point on the tabletop
(P) directly below the sphere.
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Physics Bio 178.
73.
Physics Bio 178.
Four equal positive point charges, each of charge 8.0 µC, are at the corners of a
square of side 9.2 cm. What charge should be placed at the center of the square so
that all charges are at equilibrium? Is this a stable or unstable equilibrium (Section
9–4) in the plane?
CHAPTER 16: Electric Charge and Electric Field
Answers to Questions
1.
A plastic ruler is suspended by a thread and then rubbed with a cloth. As discussed in section
16-1, the ruler is negatively charged. Bring the charged comb close to the ruler. If the ruler
is repelled by the comb, then the comb is negatively charged. If the ruler is attracted by the
comb, then the comb is positively charged.
2.
The clothing gets charged by frictional contact in the tumbling motion of the dryer. The air
inside the dryer is dry, and so the clothes can sustain a relatively large static charge. That
charged object will then polarize your clothing, and be attracted to you electrostatically.
3.
Water is a polar molecule – it has a positive region and a negative region. Thus it is easily
attracted to some other charged object, like an ion or electron in the air.
4.
The positively charged rod slightly polarizes the molecules in the paper. The negative
charges in the paper are slightly attracted to the part of the paper closest to the rod, while the
positive charges in the paper are slightly repelled from the part of the paper closest to the
rod. Since the opposite charges are now closer together and the like charges are now farther
apart, there is a net
− +
attraction between the rod + + + + + + + + + +
− +
and the paper.
5.
The plastic ruler has gained some electrons from the cloth and thus has a net negative
charge. This charge polarizes the charge on the piece of paper, drawing positives slightly
closer and repelling negatives slightly further away. This polarization results in a net
attractive force on the piece of paper. A small amount of charge is able to create enough
electric force to be stronger than gravity, and so the paper can be lifted.
On a humid day this is more difficult because the water molecules in the air are polar. Those
polar water molecules are able to attract some fraction of the free charges away from the
plastic ruler. Thus the ruler has a smaller charge, the paper is less polarized, and there is not
enough electric force to pick up the paper.
6.
The net charge on the conductor is the unbalanced charge, or excess charge after neutrality
has been established. The net charge is the sum of all of the positive and negative charges in
the conductor. If a neutral conductor has extra electrons added to it, then the net charge is
negative. If a neutral conductor has electrons removed from it, then the net charge is
positive. If a neutral conductor has the same amount of positive and negative charge, then
the net charge is zero.
Free charges in a conductor refer to those electrons (usually 1 or 2 per atom) that are so
loosely attracted to the nucleus that they are “free” to be moved around in the conductor by
an external electric force. Neutral conductors have these free electrons.
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Physics Bio 178.
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7.
15.
8.
9.
For each atom in a conductor, only a small number of its electrons are free to move. For
example, every atom of copper has 29 electrons, but only 1 or 2 from each atom are free to
move easily. Also, not even all of the free electrons move. As electrons move toward a
region, causing an excess of negative charge, that region then exerts a large repulsive force
on other electrons, preventing them from all gathering in one place.
The force of gravity pulling down on the leaves, tending to return them to the vertical
position.
The magnitude of the constant in Newton’s law is very small, while the magnitude of the
constant in Coulomb’s law is quite large. Newton’s law says the gravitational force is
proportional to the product of the two masses, while Coulomb’s law says the electrical force
is proportional to the product of the two charges. Newton’s law only produces attractive
forces, since there is only one kind of gravitational mass. Coulomb’s law produces both
attractive and repulsive forces, since there are two kinds of electrical charge.
10. For the gravitational force, we don’t notice it because the force is very weak, due to the very
small value of G, the gravitational constant, and the small value of ordinary masses. For the
electric force, we don’t notice it because ordinary objects are electrically neutral to a very
high degree. We notice our weight (the force of gravity) due to the huge mass of the Earth,
making a significant gravity force. We notice the electric force when objects have a static
charge (like static cling from the clothes dryer), creating a detectable electric force.
11. The electric force is conservative. You can “store” energy in it, and get the energy back.
For example, moving a positive charge close to another stationary positive charge takes
work (similar to lifting an object in the Earth’s gravitational field), but if the positive charge
is then released, it will gain kinetic energy and move away from the “stored energy” location
(like dropping an object in the Earth’s gravitational field). Another argument is that the
mathematical form of Coulomb’s law is identical to that of Newton’s law of universal
gravitation. We know that gravity is conservative, and so we would assume that the electric
force is also conservative. There are other indications as well. If you move a charge around
in an electric field, eventually returning to the starting position, the net work done will be 0
J. The work done in moving a charge around in an electric field is path independent – all
that matters is the starting and ending locations. All of these are indications of a
conservative force.
12. The charged plastic ruler has a negative charge residing on its surface. That charge polarizes
the charge in the neutral paper, producing a net attractive force. When the piece of paper
then touches the ruler, the paper can get charged by contact with the ruler, gaining a net
negative charge. Then, since like charges repel, the paper is repelled by the comb.
13. The test charge creates its own electric field, and so the measured electric field is the sum of
the original electric field plus the field of the test charge. By making the test charge small,
the field that it causes is small, and so the actual measured electric field is not much different
than the original field to measure.
14. A negative test charge could be used. For purposes of defining directions, the electric field
might then be defined as the OPPOSITE of the force on the test charge, divided by the test
charge. Equation (16-3) might be changed to E = − F q , q < 0 .
11
16. The electric field is strongest to the right of the positive charge, because the individual fields
from the positive charge and negative charge both are in the same direction (to the right) at
that point, so they add to make a stronger field. The electric field is weakest to the left of the
positive charge, because the individual fields from the positive charge and negative charge
are in opposite directions at that point, and so they partially cancel each other. Another
indication is the spacing of the field lines. The field lines are closer to each other to the right
of the positive charge, and further apart to the left of the positive charge.
17. At point A, the net force on a positive test charge would be down and to the left, parallel to
the nearby electric field lines. At point B, the net force on a positive test charge would be up
and to the right, parallel to the nearby electric field lines. At point C, the net force on a
positive test charge would be 0. In order of decreasing field strength, the points would be
ordered A, B, C.
18. Electric field lines show the direction of the force on a test charge placed at a given location.
The electric force has a unique direction at each point. If two field lines cross, it would
indicate that the electric force is pointing in two directions at once, which is not possible.
19. From rule 1:
A test charge would be either attracted directly towards or repelled
directly away from
a point charge, depending on the sign of the point charge. So the field lines
must be directed either radially towards or radially away from the point
charge.
From rule 2: The magnitude of the field due to the point charge only depends on the
distance from the point charge. Thus the density of the field lines must be the
same at any location around the point charge, for a given distance from the
point charge.
From rule 3:
If the point charge is positive, the field lines will originate from the
location of the
point charge. If the point charge is negative, the field lines will end at the
location of the point charge.
Based on rules 1 and 2, the lines are radial and their density is constant for a given distance.
This is equivalent to saying that the lines must be symmetrically spaced around the point
charge.
20. If the two charges are of opposite sign, then E = 0 at a point closer to the weaker charge,
and on the opposite side of the weaker charge from the stronger charge. The fields due to
the two charges are of opposite direction at such a point. If the distance between the two
charges is l, then the point at which E = 0 is 2.41 l away from the weaker charge, and 3.41 l
away from the stronger charge.
12
Physics Bio 178.
Physics Bio 178.
If the two charges are the same sign, then E = 0 at a point between the two charges, closer to
the weaker charge. The point is 41% of the distance from the weaker charge to the stronger
charge.
21. We assume that there are no other forces (like gravity) acting on the test charge. The
direction of the electric field line gives the direction of the force on the test charge. The
acceleration is always parallel to the force by Newton’s 2nd law, and so the acceleration lies
along the field line. If the particle is at rest initially and then released, the initial velocity
will also point along the field line, and the particle will start to move along the field line.
However, once the particle has a velocity, it will not follow the field line unless the line is
straight. The field line gives the direction of the acceleration, or the direction of the change
in velocity.
F =k
2.
3.
1.
Q1Q2
r
2
= (8.988 ×109 N ⋅ m2 C2 )
2
= 13.47 N ≈ 13 N
2
−2
(1.602 ×10 C)( 26 × 1.602 ×10
(1.5 ×10 m)
−19
−19
−12
C)
2
= 2.7 ×10−3 N
Use Coulomb’s law to calculate the magnitude of the force.
Q1Q2
r2
(1.602 ×10 C)
C )
( 5.0 ×10 m )
−19
(
= 8.988 ×10 N ⋅ m
9
2
2
−15
2
2
= 9.2 N
Use Coulomb’s law to calculate the magnitude of the force.
( 25 ×10−6 C)( 3.0 ×10−3 C) = 5.5 ×103 N
QQ
F = k 1 2 2 = (8.988 ×109 Nim2 C 2 )
2
r
( 3.5 ×10−1 m )
6.
Since the magnitude of the force is inversely proportional to the square of the separation
1
distance, F ∝ 2 , if the distance is multiplied by a factor of 1/8, the force will be multiplied
r
by a factor of 64.
(
)
F = 64 F0 = 64 3.2 ×10 −2 N = 2.0 N
7.
Since the magnitude of the force is inversely proportional to the square of the separation
1
distance, F ∝ 2 , if the force is tripled, the distance has been reduced by a factor of 3 .
r
r
8.45 cm
r= 0 =
= 4.88 cm
3
3
8.
Use the charge per electron and the mass per electron.
 1 electron 
−6
14
14
−42 ×10 C 
 = 2.622 ×10 ≈ 2.6 ×10 electrons
−19
 −1.602 × 10 C 
)
( 2.622 ×10 e )  9.11×110
e
14
−

13
−6
)
(
Use Coulomb’s law to calculate the magnitude of the force.
( 3.60 ×10 C)
( 9.3× 10 m )
5.
24. The electric flux depends only on the charge enclosed by the gaussian surface, not on the
shape of the surface. Φ E will be the same for the cube as for the sphere.
Solutions to Problems
)
Use Coulomb’s law to calculate the magnitude of the force.
F =k
+
23. Just because the electric flux through a closed surface is zero, the field need not be zero on
the surface. For example, consider a closed surface near an isolated point charge, and the
surface does not enclose the charge. There will be electric field lines passing through the
surface, but the total electric flux through the surface will be zero since the surface does not
enclose any charge. The same number of field lines will enter the volume enclosed by the
surface as leave the volume enclosed by the surface.
On the contrary, if E = 0 at all points on the surface, then there are no electric field lines
passing through the surface, and so the flux through the surface is zero.
(
= 8.988× 109 N ⋅ m2 C2
Use the charge per electron to find the number of electrons.
 1 electron 
14
−30.0 ×10−6 C 
 = 1.87 ×10 electrons
−19
 −1.602 × 10 C 
F =k
++++++++++++++++++++
r2
(
4.
22. Since the line of charge is infinitely long, it has no preferred left or right direction. Thus by
symmetry, the lines must point radially out away from the center of the line. A side view
and an end view are shown for a line of positive charge. As seen from the end view, the
field is not uniform. As you move further away from the line of charge, the field lines get
further apart, indicating that the field gets weaker as you move away from the line of charge.
Q1Q2
−
−31
kg 
−16
 = 2.4 ×10 kg

14
Physics Bio 178.
Physics Bio 178.
9.
Convert the kg of H2O to moles, then to atoms, then to electrons. Oxygen has 8 electrons
per atom,
and hydrogen has 1 electron per atom.
13. The forces on each charge lie along a line connecting the charges. Let the
variable d represent the length of a side of the triangle, and let the variable Q
represent the charge at each corner. Since the triangle is equilateral, each angle is
60o.
2
2
2
Q
Q
Q
F12 = k 2 → F12 x = k 2 cos 60o , F12 y = k 2 sin 60o
d
d
d
 1 mole H 2 O   6.02 × 1023 molec.   10 e   −1.602 × 10−19 C 
1.0 kg H 2 O = (1.0 kg H 2 O ) 




−2
1 mole
e
 1.8 × 10 kg 
  1 molec.  

F13 = k
= −5.4 × 107 C
10.
Take the ratio of the electric force divided by the gravitational force.
QQ
2
k 12 2
8.988 × 109 N ⋅ m2 C2 1.602 × 10−19 C
FE
kQ1Q2
r
=
=
=
= 2.3 ×10 39
m1m2 Gm1m2
FG
6.67 × 10−11 N ⋅ m2 kg2 9.11× 10−31 kg 1.67 × 10−27 kg
G 2
r
(
(
)(
)(
)(
)
)
The electric force is about 2.3 × 1039 times stronger than the gravitational force for the
given scenario.
11.
(a)
between the
charges is FE = k
then FE =
k
r2
(
q ( QT − q )
r
2
=
k
r
2
( qQ
T
−q
2
) = rk Q
2
2
T
 q  q 
 − 
 QT  QT 

→ F13 x = −k
F1x = F12 x + F13 x = 0
F1 = F1 x + F1 y = 3k
2
2
Q
cos 60o , F13 y = k
F1 y = F12 y + F13 y = 2k
2
d2
d2
Q2
(
Q2
d
2
= 3 8.988 × 10 N ⋅ m C
9
2
d2
)
d
Q2
Q3
d
sin 60o
sin 60 o = 3k
2
Q1
d
Q2
d2
(11.0 ×10 C )
−6
( 0.150 m )2
2
= 83.7 N
The direction of F1 is in the y-direction . Also notice that it lies along the bisector of the
opposite side of the triangle. Thus the force on the lower left charge is of magnitude
charge is of magnitude 83.7 N , and will point 30o below the + x axis .
14. Determine the force on the upper right charge, and then use the
symmetry of the configuration to determine the force on the other three
charges. The force at the upper right corner of the square is the vector
sum of the forces due to the other three charges. Let the variable d
represent the 0.100 m length of a side of the square, and let the variable
Q represent the 6.00 mC charge at each corner.

q
 . If we let x = ,

Q
T

)
QT2 x − x 2 , where 0 ≤ x ≤ 1 . A graph of f ( x ) = x − x2 between the
limits of 0 and 1 shows that the maximum occurs at x = 0.5 , or q = 0.5 QT . Both
charges are half of the total, and the actual maximized force is FE = 0.25
d2
Q2
/
F12
83.7 N , and will point 30o below the − x axis . Finally, the force on the lower right
Let one of the charges be q , and then the other charge is QT − q . The force
2
Q2
/
F13
F41 = k
k
QT2 .
2
r
If one of the charges has all of the charge, and the other has no charge, then the
(b)
force between
them will be 0, which is the minimum possible force.
F42 = k
Q2
→ F41x = k
d2
Q2
2d
Q2
2d
2
cos45 = k
2Q2
o
2
4d
2
, F42 y = k
12. Let the right be the positive direction on the line of charges. Use the fact that like charges
repel and unlike charges attract to determine the direction of the forces. In the following
expressions, k = 8.988 ×109 N ⋅ m2 C 2 .
( 75µ C)( 48µ C) ( 75µ C)( 85µ C)
+k
= −147.2 N ≈ −1.5 ×102 N
F+75 = −k
2
( 0.70 m )2
( 0.35m )
( 75µ C)( 48µ C) ( 48µ C)( 85µC )
+k
= 563.5 N ≈ 5.6 ×10 2 N
F+48 = k
( 0.35m )2
( 0.35m )2
( 85µ C )( 75µ C ) (85µ C )( 48µ C )
−k
= −416.3 N ≈ −4.2 ×102 N
F−85 = −k
( 0.70 m )2
( 0.35 m )2
Q3
2Q 2
4d 2
2
F4 x = F41x + F42 x + F43x = k
Q2
d
2
+k
2Q2
4d
2
+0= k
Q2 
2
1 +
 = F4 y
4 
d 
2
Q 
2
Q2 
1
F4 = F42x + F42y = k 2 1 +
 2=k 2  2+ 
d 
4 
d 
2
2
(
= 8.988 ×10 N ⋅ m C
15
9
2
2
)
( 6.00 ×10 C)
−3
( 0.100m )
2
2
/
F41
d
, F41y = 0
d2
→ F42 x = k
Q4
Q
→ F43 x = 0 , F43 y = k 2
d2
d
Add the x and y components together to find the total force, noting that F4 x = F4 y .
F43 = k
Q
2
Q1
Q2
Q2
/
F42
/
F43
1

7
 2 +  = 6.19 ×10 N
2

16
Physics Bio 178.
θ = tan −1
Physics Bio 178.
F4 y
= 45o above the x-direction.
F4 x
For each charge, the net force will be the magnitude determined above, and will lie along the
line from the center of the square out towards the charge.
F2Qy = k
15. Determine the force on the upper right charge, and then the symmetry of the configuration
says that the force on the lower left charge is the opposite of the force on the upper right
charge. Likewise, determine the force on the lower right charge, and then the symmetry of
the configuration says that the force on the upper left charge is the opposite of the force on
the lower right charge.
The force at the upper right corner of the square is the vector sum of the
forces due to the other three charges. Let the variable d represent the
Q1
0.100 m length of a side of the square, and let the variable Q represent
the 6.00 mC charge at each corner.
F42 = k
F43 = k
Q
d
2
→ F41 x = −k
2
Q
2
2d 2
→ F42 x = k
Q
d
( 2Q )( 3Q )
l
2
l
2
2
+k
/
F41
/
F42
F3Qy = − k
Q4
/
F43
F3Q =
2
, F41 y = 0
Q
Q2
2
2
2Q
cos45 = k
o
Q2
4d 2
, F42 y = k
2Q
Q2
d
2
= ( 8.988 ×10 N ⋅ m
9
θ = tan −1
F4 y
F4 x
Q2
d
2
+k
( 0.64645)
2Q 2
4d
2
+0= k
2=k
Q2
d
l
F2 x
= tan−
o
2
2
2
1
8.8284
4.8284
= 61
o
Q2 
3
kQ 2

12
2
13.0607
+
=


l2 
4
l2

Q2 
3
kQ 2

6+
2  = −7.0607 2
2 
l 
4
l

θ 3Q = tan−1
l2
( 2 + 2 2 ) = 4.8284 kQl
( 6 + 2 2 ) = 8.8284 kQl
cos 45 = k
o
l
2
2
F2 y
sin 45 = − k
kQ 2
2
Q2
F3 y
F3 x
= tan
−7.0607
−1
13.0607
o
= 332
2
3Q F3Q
4Q
Q2 
2
Q2
 −1 +
 = −0.64645 k 2 = F4 y
4 
d 
d
2
17. The forces on each charge lie along a line connecting the charges. Let
the variable d represent the length of a side of the triangle. Since the
triangle is equilateral, each angle is 60o. First calculate the magnitude
of each individual force.
( 0.9142 )
2
2
7
( 0.100 m )
2Q
F2Q
l
( 6.00 ×10 C) ( 0.9142) = 2.96 ×10 N
C )
−3
2
2
Q
4d 2
Q2
F42x + F42y = k
2l 2
Q3
2
→ F43 x = 0 , F43 y = −k 2
d2
d
Add the x and y components together to find the total force, noting that F4 x = F4 y .
F4 =
( 3Q ) Q
( 3Q ) Q
F32Qx + F32Qy = 14.8
Q2
o
θ 2Q = tan−1
+k
2l
o
sin 45 = k
l2
−k
cos 45 = k
2
2
2d 2
F4 x = F41x + F42 x + F43 x = − k
kQ
2
l2
( 3Q )( 2Q )
l
2l
2
( 3Q )( 4Q )
2
2l
( 2Q )( 4Q )
2
3Q : F3Qx = k
(b)
( 2Q )( 4Q )
+k
F2Q = F2Qx + F2Qy = 10.1
d
F41 = k
( 2Q ) Q
2Q : F2Qx = k
(a)
F12 = k
2
= 225o from the x-direction, or exactly towards the center of the square.
For each charge, the net force will be the magnitude of 2.96 ×10 N and each net force will
lie along the line from the charge inwards towards the center of the square.
7
16.
Take the lower left hand corner of the square to be the origin of coordinates. Each charge
will have a
horizontal force on it due to one charge, a vertical force on it due to one charge, and a
diagonal force on it due to one charge. Find the components of each force, add the
components, find the magnitude of the net force, and the direction of the net force. At the
conclusion of the problem is a diagram showing the net force on each of the two charges.
17
Q1Q2
d2
(
= 8.988 ×109 N ⋅ m2 C2
)
= 0.1997 N = F21
F13 = k
Q1Q3
d2
(
= 8.988 ×109 N ⋅ m2 C2
= 0.1498 N = F31
F23 = k
Q2 Q3
d2
(
= 8.988 ×10 N ⋅ m C
9
2
2
)
)
( 4.0 × 10 C)( 8.0 × 10 C)
−6
−6
(1.20 m )2
F13
Q1
d
/
F23
F12
d
Q2
Q3
F21
F31
d
( 4.0 ×10 C )( 6.0 ×10 C )
−6
−6
(1.20 m )2
( 8.0 × 10 C)( 6.0 × 10 C) = 0.2996 N = F
−6
(1.20 m )2
F32
−6
32
Now calculate the net force on each charge and the direction of that net force, using
components.
18
Physics Bio 178.
Physics Bio 178.
F1x = F12 x + F13x = − ( 0.1997 N ) cos 60o + ( 0.1498 N ) cos 60o = −2.495 ×10−2 N
left: k
F1 y = F12 y + F13 y = − ( 0.1997 N) sin 60o − ( 0.1498 N) sin 60o = −3.027 × 10−1 N
θ1 = tan −1
F1 = F12x + F12y = 0.30 N
F1 y
= tan −1
F1x
−3.027 ×10−1 N
= 265o
−2
−2.495 ×10 N
F2 x = F21x + F23x = ( 0.1997 N ) cos 60o − ( 0.2996 N ) = −1.998 × 10−1 N
F2 y = F21 y + F23 y = ( 0.1997 N ) sin 60 + 0 = 1.729 × 10 N
θ 2 = tan −1
F2 y
F2 x
= tan −1
1.729 × 10−1 N
−1.998 ×10 N
F3 y
F3 x
= tan −1
1.297 ×10−1 N
2.247 ×10−1 N
= 30o
18.
Since the force is repulsive, both charges must be the same sign. Since the total charge is
positive,
both charges must be positive. Let the total charge be Q. Then if one charge is of magnitude
q, then the other charge must be of magnitude Q − q . Write a Coulomb’s law expression for
one of the charges.
q (Q − q)
Fr 2
2
→ q − Qq +
=0 →
F =k
2
r
k
Q ± Q2 −
q=
2
4 Fr
k
( 560 ×10 C) ± ( 560 × 10 C)
2
−6
=
= 5.54 × 10−4 C , 5.54 × 10−6 C
Q0Q
−6
2
−
(
x
2
=k
=k
=k
3Q02
l
3Q0Q
( l − x)
3Q02
l
2
2
3Q0Q
right: k
2
( l − x)
l
→ x=
3 +1
→ Q = 3Q0
x2
l2
2
3Q02
=k
→
l2
= 0.366l
= Q0
(
3
)
3 +1
2
= 0.402Q0
0.37 l from − Q0 towards − 3Q0 .
F3 y = F31 y + F32 y = ( 0.1498 N ) sin 60o + 0 = 1.297 × 10−1 N
θ 3 = tan −1
k
x2
x
2
Thus the charge should be of magnitude 0.40Q0 , and a distance
= 139o
−1
F3 x = F31x + F32 x = − ( 0.1498 N ) cos 60o + ( 0.2996 N ) = 2.247 ×10 −1 N
F3 = F32x + F32y = 0.26 N
Q0Q
−1
o
F2 = F22x + F22y = 0.26 N
k
Q0 Q
4 ( 22.8 N )(1.10 m )
F1 = F2
x=d
2
8.988 × 109 N ⋅ m2 C2
20. Assume that the negative charge is d = 18.5 cm to the right of the
Q1
Q2
Q
positive charge, on the x-axis. To experience no net force, the
4.7 µC
–3.5 µC
third charge Q must be closer to the smaller magnitude charge
–
+
d
x
(the negative charge). The third charge cannot be between the
charges, because it would experience a force from each charge in
the same direction, and so the net force could not be zero. And the third charge must be on
the line joining the other two charges, so that the two forces on the third charge are along the
same line. See the diagram. Equate the magnitudes of the two forces on the third charge,
and solve for x > 0.
)
2
Q − q = 5.54 × 10−6 C , 5.54 ×10−4 C
19. The negative charges will repel each other, and so the third charge
−Q0
Q
−3Q0
must put an opposite force on each of the original charges.
x
l–x
Consideration of the various possible configurations leads to the
conclusion that the third charge must be positive and must be between
l
the other two charges. See the diagram for the definition of variables.
For each negative charge, equate the magnitudes of the two forces on the charge. Also note
that 0 < x < l .
(
→ k
Q1 Q
(d + x)
Q2
Q1 −
Q2
)
2
Q2 Q
=k
x
= (18.5 cm )
(
→ x =d
Q2
Q2
)
4.7 ×10 −6 C − 3.5 × 10−6 C
)
2
(
Q1 −
3.5 ×10−6 C
= 116cm
21.
(a)
If the force is repulsive, both charges must be positive since the total charge is
positive. Call the
total charge Q.
kQ ( Q − Q )
kQ Q
Fd 2
=0
Q1 + Q2 = Q
F = 12 2 = 1 2 1 → Q12 − QQ1 +
d
d
k
Q ± Q2 − 4
Q1 =
2
Fd 2
k
Q ± Q2 − 4
=
2
(90.0 ×10 C ) ± (90.0 ×10 C )
−6
=
Fd 2
k
−6
2
−4
(12.0N )(1.06 m )
(8.988 × 10
9
2
N ⋅ m C2
2
)
2
= 69.9 × 10−6 C , 22.1 ×10 −6 C
(b)
If the force is attractive, then the charges are of opposite sign. The value used for
F must then
be negative. Other than that, the solution method is the same as for part (a).
19
20
Physics Bio 178.
Physics Bio 178.
Q1 + Q2 = Q
F=
Q ± Q2 − 4
Q1 =
Fd
k
kQ1Q2
d
2
=
kQ1 ( Q − Q1 )
d
2
2
Q ± Q2 − 4
=
2
Fd
k
→ Q1 − QQ1 +
2
Fd 2
k
=0
27.
Assuming the electric force is the only force on the electron, then Newton’s 2nd law may
be used to
find the acceleration.
1.602 × 10−19 C
q
Fnet = ma = qE → a = E =
750 N C = 1.32 ×1014 m s 2
m
9.11×10−31 kg
2
(
(
2
(90.0 ×10 C ) ± (90.0 ×10 C )
−6
−6
=
2
( −12.0N )(1.06 m )
−4
(8.988 × 10
9
2
N ⋅ m2 C 2
)
Since the charge is negative, the direction of the acceleration is opposite to the field .
2
−6
28. The electric field due to the negative charge will point Q > 0
1
toward the negative charge, and the electric field due
to the positive charge will point away from the
positive charge. Thus both fields point in the same
d 2
direction, towards the negative charge, and so can be
added.
Q
Q
Q1
Q2
4k
E = E1 + E2 = k 21 + k 22 = k
+k
=
( Q1 + Q2 )
2
r1
r2
( d / 2)
( d / 2 )2 d 2
−6
= 104.4 × 10 C , − 14.4 × 10 C
22.
The spheres can be treated as point charges since they are spherical, and so Coulomb’s
law may be
used to relate the amount of charge to the force of attraction. Each sphere will have a
magnitude Q
of charge, since that amount was removed from one sphere and added to the other, being
initially uncharged.
F =k
Q1Q2
r
2
=k
Q
r
2
2
→ Q=r
F
k
= ( 0.12 m)
−2
1.7 ×10 N
8.988 ×10 N ⋅ m C
9
2
)
)
=
2
 1 electron  = 1.0 ×1012 electrons

−19
 1.602 × 10 C 
(
4 8.988 ×109 N ⋅ m2 C2
(8.0 ×10 m)
−2
2
)
( 8.0 ×10
−6
E1
Q2 > 0
E2
)
C + 7.0 × 10−6 C = 8.4 ×107 N C
The direction is towards the negative charge .
= 1.650 × 10−7 C 
29.
23.
Use Eq. 16–3 to calculate the force.
F
E=
→ F = qE = ( −1.602 ×10−19 C) ( 2360 N C east ) = 3.78 ×10−16 N west
q
24.
Use Eq. 16–3 to calculate the electric field.
F 3.75 ×10−14 Nsouth
E= =
= 2.34 × 105 N C south
q
1.602 × 10−19 C
25.
Use Eq. 16–3 to calculate the electric field.
F 8.4 N down
E= =
= 9.5 ×105 N C up
−6
q −8.8 × 10 C
26.
Use Eq. 16–4a to calculate the electric field due to a point charge.
E=k
Q
r2
(
= 8.988 × 109 N ⋅ m2 C2
)
33.0 × 10−6 C
( 2.00 ×10 m )
−1
2
30.
Assuming the electric force is the only force on the electron, then Newton’s 2nd law may
be used to
find the electric field strength.
Fnet = ma = qE → E =
= 7.42 ×106 N C up
ma
q
(1.67 × 10
−27
=
)(
kg 1×106
(1.602 × 10
−19
)( 9.80 m s ) = 0.102 N C ≈ 0.1N C
C)
2
31.
Since the electron accelerates from rest towards the north, the net force on it must be to
the north.
Assuming the electric force is the only force on the electron, then Newton’s 2nd law may be
used to
find the electric field.
Note that the electric field points away from the positive charge.
21
22
Physics Bio 178.
Physics Bio 178.
(
−31
)
m
9.11×10 kg
Fnet = ma = qE → E = a =
115 m s2 north = 6.54 ×10−10 N C south
q
−1.602 ×10−19 C
(
)
(
)
32.
The field due to the negative charge will point
towards
the negative charge, and the field due to the positive
charge will point towards the negative charge. Thus
the magnitudes of the two fields can be added together
to find the charges.
Enet = 2 EQ = 2k
Q
( d / 2)
2
=
8kQ
d
2
→ Q=
Ed 2
8k
=
( 745 N C) (1.60 × 10−1 m)
(
Q1
2
d 2
+k
Q2
2
d 2
= (8.988 ×109 N ⋅ m2 C2 )
34.
of
=k
)
Q2
d
Q2 = −27.0 µ C
E1
E2
Q2
E2 = k
E3 = k
Q
d2
Q
2d 2
Q
→ E1x = k
Q
d2
E3
→ E2 x = k
2d 2
cos45o = k
4d 2
, E2 y = k
E2
E1
Q3
Q2
2Q
−6
(1.00 m )
2

4
 2 +  = 3.87 ×10 N C
2

1
from the x-direction.
2Q
Q
( x + a)
2
−k
 1

1
−4kQxa
−
=
 ( x + a )2 ( x − a ) 2 
2
2 2

 (x −a )
Q
( x − a)
= kQ 
2
36.
For the net field to be zero at point P, the magnitudes of the fields created by Q1 and Q2
must be
equal. Also, the distance x will be taken as positive to the left of Q1 . That is the only
region where the total field due to the two charges can be zero. Let the variable d represent
the 12 cm distance, and note that Q1 = 12 Q2 .
4.70 ×106 N C at 45o
, E1y = 0
Q
o
( 2.25 × 10 C) 
The negative sign means the field points to the left .
d2 2
the fields due to the other three charges. Let the variable d represent Q1
the 1.0 m length of a side of the square, and let the variable Q
represent the charge at each of the three occupied corners.
d
E1 = k
= 45
)
35.
Choose the rightward direction to be positive. Then the field due to +Q will be positive,
and the
field due to –Q will be negative.
E=k
Q1 = +45.0 µ C
The field at the upper right corner of the square is the vector sum
Ex
Q
2
1 +
 = Ey
4 
d2 
= 2.65 ×10−10 C
Q1 + Q2
( 47.0 + 27.0 ) ×10−6 C
=
( 0.525m ) 2 2
Ey
+0= k
2
Q
1
1 +
 2 =k 2  2+ 
d 
4 
d 
2
(
θ = tan −1
2Q
4d 2
2
= 8.988 × 109 N ⋅ m2 C2
E −Q
d 2
8 8.988 ×109 N ⋅ m2 C2
−Q
+k
d2
Q
E = Ex2 + Ey2 = k
EQ
Q
33. The field at the center due to the two negative charges on opposite
corners (lower right and upper left in the diagram) will cancel
each other, and so only the positive charge and the opposite
negative charge need to be considered. The field due to the
negative charge will point directly toward it, and the field due to
the positive charge will point directly away from it. Accordingly,
the two fields are in the same direction and can be added
algebraically.
E = E1 + E2 = k
Q
Ex = E1 x + E2 x + E3 x = k
E1 = E2
x =d
(
→ k
Q1
Q1
Q2 −
x
2
=k
=d
Q1
Q2
→
( x + d )2
) (
1
2
Q2 −
Q2
1
2
Q2
=
) (
d
)
2 −1
=
12 cm
2 −1
= 29cm
37. (a) The field due to the charge at A will point straight downward, and the
field due to the charge at B will point along the line from A to the
origin, 30o below the negative x axis.
A
+Q
l
4d 2
+Q
l
Q
→ E3 x = 0 , E1 y = k 2
d2
d
Add the x and y components together to find the total electric field, noting that Ex = Ey .
23
EB
B
l
EA
24
Physics Bio 178.
EA = k
EB = k
Physics Bio 178.
Q
l
Q
→ EBx = −k
l2
EBy = −k
Ex = EAx + EBx = −k
E = Ex2 + Ey2 =
θ = tan
−1
Q
→ EAx = 0 , EAx = −k
2
Ey
Ex
= tan
Q
l2
Q
3Q
4l
4
EA = 2
cos30 = −k
3Q
o
sin 30 = −k
o
2l 2
Q
+
9k 2 Q 2
4l
12k 2 Q2
=
4
4l
4
3Q
2l 2
3kQ
=
l
−k
d right =
EA = k
EB = k
l
2
Q
l2
Ex = EAx + EBx = k
E= E +E =
2
x
θ = tan
−1
2
y
Ey
Ex
3Q
3k 2Q 2
4l
4
−1
−k
+
Q
l
cos30 = k
o
3Q
2l 2
, EBy = k
k 2Q 2
4l
4
4k 2Q 2
=
4l
4
=
Q
l2
sin 30 = k
o
Q
2l 2
d=
( 5.0cm ) + (10.0cm )
2
2
2

)( 7.0 ×10 C) 
−6
)(
d right
d left
cos45o
 ( 0.0707 m )
)
EB = Ex2 + Ey2 = 1.2 × 107 N C
θ right
θ left
+Q
2
−

2
+
sin45o
 ( 0.0707 m )
θB = tan −1
cos18.4o 
6
 = 6.51 ×10 N C
( 0.1581m ) 2 
sin18.4o 
6
 = 9.69 ×10 N C
( 0.1581m ) 
2
Ey
= 56o
Ex
The results are consistent with Figure 16-31b. In the figure, the field at Point A points
straight up, matching the calculations. The field at Point B should be to the right and
vertical, matching the calculations. Finally, the field lines are closer together at Point B than
at Point A, indicating that the field is stronger there, matching the calculations.
2
39. Both charges must be of the same sign so that the electric fields created by the two charges
oppose each other, and so can add to zero. The magnitudes of the two electric fields must be
equal.
( )
)
E1 = E2 → k
A
d
+Q
Eleft
Eright
+Q
θ A = 90o
= 0.1581m
2l 2
2l 2
Point A: From the symmetry of the geometry, in
calculating the electric field at point A only the vertical
components of the fields need to be considered. The
horizontal components will cancel each other.
5.0
θ = tan −1
= 26.6o
10.0
sin 26.6o = 4.5 ×106 N C
= 0.0707 m
2
( 5.0 cm ) + (15.0cm )
(
Q
2l 2 = tan −1 1 = 330o
3Q
− 3
(
( 0.1118 m )2
= 8.988 ×109 N ⋅ m2 C2 7.0 ×10 −6 C 
Q
kQ
l
7.0 × 10−6 C
Q
Q
Ey = ( Eleft ) y + ( Eright ) y = k 2 sinθleft + k 2 sin θright
dleft
d right
2
38. In each case, find the vector sum of the field caused by the charge on the left Eleft and the
Eright
field caused by the charge on the right Eright
Eleft
2
2
(
Ey = EAy + EBy = −k
2l 2
k
= tan
Q
l2
( 5.0 cm ) + ( 5.0cm )
= 8.988 × 109 N ⋅ m2 C2
→ EAx = 0 , EAx = −k
→ EBx = k
sinθ = 2 (8.988 ×109 N ⋅ m 2 C 2 )
Q
Q
Ex = ( Eleft )x + ( Eright ) x = k 2 cosθleft − k 2 cos θright
dleft
d right
2l 2
(b) Now reverse the direction of EA
Q
d
2
dleft =
2
3Q
2l 2 = tan −1 −3 = tan −1 3 = 240o
− 3
3Q
−k
kQ
Point B: Now the point is not symmetrically placed, and
so horizontal and vertical components of each individual
field need to be calculated to find the resultant electric
field.
5.0
5.0
o
o
θ left = tan −1
= 45
θ left = tan −1
= 18.4
5.0
15.0
2l 2
Ey = EAy + EBy = −k
2l 2
3k 2Q2
−1
l2
l2
θ
d
θ
+Q
Q1
( l 3)
2
=k
Q2
( 2l 3)
2
→ 9Q1 =
9Q2
4
→
Q1
Q2
=
1
4
40. From the diagram, we see that the x components of the two fields will cancel each other at
the point P. Thus the net electric field will be in
+Q
the negative y-direction, and will be twice the ycomponent of either electric field vector.
a
x
= 0.1118m
a
25
−Q
θ
EQ
E− Q
26
Physics Bio 178.
Physics Bio 178.
kQ
Enet = 2 E sin θ = 2
x +a
a
2kQ
= 2
x + a 2 x 2 + a 2 1/ 2
2
(
=
(x
+ a2
43.
sinθ
)
2kQa
2
2
)
3/ 2
44. (a) Φ E = E⊥ A = Eπ r 2 = ( 5.8 ×102 N C) π (1.8 ×10−1 m ) = 59 N ⋅ m2 C
2
(b) Φ E = E⊥ A = ( E cos 45o ) π r 2 = ( 5.8 ×102 N C )( cos 45o ) π (1.8 ×10−1 m ) = 42 N ⋅ m 2 C
2
in the negative y direction
(c) Φ E = E⊥ A = ( E cos 90o ) π r 2 = 0
41.
We assume that gravity can be ignored, which is proven in part (b).
(a)
The electron will accelerate to the right. The magnitude of the acceleration can
be found from
setting the net force equal to the electric force on the electron. The acceleration is
constant, so constant acceleration relationships can be used.
Fnet = ma = q E → a =
(b) Since there is no charge enclosed by surface A2, ΦE = 0 .
m
qE
m
46. (a) Assuming that there is no charge contained within the cube, then the net flux through the
cube is
0 . All of the field lines that enter the cube also leave the cube.
(b)
There are four faces that have no flux through
them,
because none of the field lines pass through those faces.
In the diagram shown, the left face has a positive flux
and the right face has the opposite amount of negative
flux.
∆x
(1.602 ×10 C)(1.45 ×10
( 9.11×10 kg)
−19
(b)
45. (a) Use Gauss’s law to determine the electric flux.
Q
−1.0 ×10−6 C
ΦE = encl =
= −1.1×105 N ⋅ m2 C
8.85 ×10−12 C2 N ⋅ m2
εo
qE
v 2 = v02 + 2a ∆x → v = 2a∆x = 2
= 2
4
N C)
−31
(1.10 ×10 m ) = 7.49 ×10
−2
6
ms
The value of the acceleration caused by the electric field is compared to g.
−19
4
q E (1.602 × 10 C)(1.45 ×10 N C )
a=
=
= 2.55 ×1015 m s 2
m
( 9.11× 10−31 kg)
a
=
2.55 ×1015 m s 2
2
Φleft = EA = El 2 = ( 6.50 ×103 N C) l 2
Φright = − ( 6.50 ×103 N C ) l 2
= 2.60 ×1014
g
9.80m s
The acceleration due to gravity can be ignored compared to the acceleration
47.
caused by the
electric field.
F = qE = ma → a =
E=
2q∆x
qE
m
v 2 = v02 + 2a∆x = v02 + 2
qE
m
−31
−19
−2
Q A
ε0
(
)
→ Q = ε0 EA = 8.85 ×10 −12 C2 N ⋅ m2 (130 N C )(1.0m ) = 1.15 ×10−9 C
2
The electric field can be calculated by Eq. 16-4a, and that can be solved for the charge.
E=k
Q
r2
→ Q=
Er 2
k
=
( 2.75 ×10
2
)(
−2
N C 3.50 ×10 m
8.988 ×109 N ⋅ m2 C2
)
2
= 3.75 ×10−11 C
This corresponds to about 2 ×108 electrons. Since the field points toward the ball, the
charge must be negative .
49.
2
6
Φother = 0
∆x →
( 9.11×10 kg )(3.0 ×10 m s ) = 6.4 ×10
=
=−
2q∆x
2 ( −1.602 ×10 C)( 4.0 ×10 m)
−mv02
48.
l
Equation 16-10 applies.
E=
42.
(a)
The electron will experience a force in the opposite direction to the electric field.
Since the
electron is to be brought to rest, the electric field must be in the same direction as the
initial velocity of the electron, and so is to the right .
(b) Since the field is uniform, the electron will experience a constant force, and therefore
have a constant acceleration. Use constant acceleration relationships to find the field
strength.
m ( v 2 − v02 )
Use Gauss’s law to determine the enclosed charge.
Q
3
2
−12
2
2
−8
ΦE = encl → Qencl = Φ Eε o = (1.45 × 10 N ⋅ m C)( 8.85× 10 C N ⋅ m ) = 1.28 ×10 C
εo
2
NC
27
See Example 16-11 for a detailed discussion related to this problem.
(a) Inside a solid metal sphere the electric field is 0 .
(b)
Inside a solid metal sphere the electric field is 0 .
28
Physics Bio 178.
Physics Bio 178.
(c)
Outside a solid metal sphere the electric field is the same as if all the charge were
concentrated
at the center as a point charge.
−6
3.50 × 10 C
Q
= 3.27 ×103 N C
E = k 2 = 8.988 ×109 N ⋅ m2 C2
r
( 3.10m)2
(d)
Same reasoning as in part (c).
(
E =k
(e)
50.
Q
r2
)
(
= 8.988 ×109 N ⋅ m2 C2
)
(
= (.02004 )
9
( 3.50 × 10 C) = 8.74 ×10
(b)
2
2
(a) Inside the shell, the field is that of the point charge, E = k
r
2
O – N repulsion:
FON = k
H – N attraction:
FHN = k
N – N repulsion:
FNN = k
(2 of these)
2
2
 2.90 Ao 




( 0.2e )(0.2e )
2
.
r
(d)
The shell does not affect the field due to Q alone, except in the shell material,
where the field is
0. The charge Q does affect the shell – it polarizes it. There will be an induced charge
of –Q uniformly distributed over the inside surface of the shell, and an induced charge
of +Q uniformly distributed over the outside surface of the shell.
o
1.90 A





( 0.4e )( 0.2e )
.
Q
= 4.623 × 10−10 N ≈ 4.6 × 10−10 N
2
these)
There is no field inside the conducting material: E = 0 .
(c) Outside the shell, the field is that of the point charge, E = k
)
2
1.90 Ao 




NC
The answers would be no different for a thin metal shell.
Q
(
)
The net force between the cytosine and guanine is due to the following forces.
( 0.4e )( 0.2e ) 0.08ke2
FOH = k
=
O – H attraction:
(2 of
2
2
−6
( 6.00 m)
)(
N ⋅ m2 C2 1.602 × 10−19 C
1.0 ×10−10 m
)
See Figure 16-33 in the text for additional insight into this problem.
(b)
(8.988 ×10
o
 2.00 A





( 0.2e )( 0.2e )
2
 3.00Ao 




=
=
=
0.08ke 2
o
 2.90 A





2
0.04ke2
o
 2.00 A





2
0.04ke2
2
o
 3.00A





2
51.
(a)
The net force between the thymine and adenine is due to the following forces.
( 0.4e )( 0.2e ) 0.08ke2
FOH = k
=
O – H attraction:
2
2
1.80Ao 




O – N repulsion:
FON = k
( 0.4e )( 0.2e )
2
 2.80 Ao 




o
1.80 A





=
( 0.2e )( 0.2e ) =
=k
N – N repulsion:
FNN
H – N attraction:
FHN = k
2
 3.00Ao 




( 0.2e )(0.2e )
2
o
 2.00 A





=
= (.03085)
0.08ke 2
9
)(
)
N ⋅ m2 C2 1.602 ×10−19 C
(1.0 ×10
−10
m
)
2
2
−10
−10
= 7.116 × 10 N ≈ 7.1× 10 N
(c) For 10 pairs of molecules, we assume that half are A-T pairs and half are C-G pairs. We
average the above results and multiply by 105 .
2
5
5
−5
−10
−10
)
−5
= 5.850 ×10 N ≈ 6 ×10 N
2
52.
Set the magnitude of the electric force equal to the magnitude of the force of gravity and
solve for
the distance.
2
e
FE = FG → k 2 = mg →
r
2
o
 2.00 A





(
Fnet = 12 10 ( FA-T + FC-G ) = 10 4.623 ×10 N + 7.116 ×10 N
2
o
 3.00A





0.04ke
( 8.988 ×10
5
o
 2.80 A





0.04ke
2
0.08 0.04 0.04 
1
 0.08
 ke
−2
−
+
 2
2
2
2
2 
−10
2.90 3.00 2.00  1.0 ×10 m  d
 1.90
FC-G = 2FOH − 2 FON − FNN + FHN =  2
2
2
2
1
 0.08 − 0.08 − 0.04 + 0.04 
 ke
 2
2
2
2
2 
−10
 1.80 2.80 3.00 2.00  1.0 ×10 m  d
FA-T = FOH − FON − FNN + FHN = 
r =e
29
k
mg
(
= 1.602 ×10 −19 C
)
( 8.988× 10
9
( 9.11×10
−31
N ⋅ m2 C2
)(
)
kg 9.80 m s 2
)
= 5.08 m
30
Physics Bio 178.
Physics Bio 178.
53.
Calculate the total charge on all electrons in 3.0 g of copper, and then compare the 38µC
to that
value.
 1 mole   6.02 ×1023 atoms   29 e   1.602 ×10−19C 

  atoms  

mole
1e

 63.5g 


Total electron charge = 3.0 g 
= 1.32 ×105 C
38 ×10−6 C
Fraction lost =
= 2.9 ×10−10
1.32 ×105 C
55.
mg
q
(1.67 ×10 kg )( 9.80m s ) = 1.02 ×10
(1.602 ×10 C)
−27
=
Q
→ Q=
Er 2
=
2
−7
−19
N C, up
n=
(150 N C) ( 6.38 ×106 m)
4π r 3 ρ g
= 6.8 ×105 C
59.
orbit.
m
3
3
3
2
6
−19
≈ 1.0 × 107 electrons
k ( 0.40e )( 0.20e ) 
(1× 10 m)
−9
2
1
1
1
1 
+
+
−
−
2
2
2
2
 ( 0.30 ) ( 0.40 ) (0.18 ) ( 0.28 ) 
FE = Fradial → k
Q2
mv2
Q2
2
orbit
r
=
mv2
rorbit
→
= (8.988 ×109 N ⋅ m2 C2 )
(1.602 ×10 C)
( 9.11×10 kg )(1.1×10
−19
−31
2
6
m s)
2
= 2.1× 10−10 m
−19
=
13
−31
m s 2 ≈ 2.6 ×10 13 m s 2 , up
60.
Set the Coulomb electrical force equal to the Newtonian gravitational force on one of the
bodies (the
Moon).
FE = FG → k
FE = eE = ma →
eE
) (1.00 ×10 kg m )( 9.80 m s ) = 9.96 ×10
3 (1.602 ×10 C) (150 N C)
The electric force must be a radial force in order for the electron to move in a circular
rorbit = k
(1.602 ×10 C) (150 N C) = 2.638 ×10
( 9.11×10 kg )
(
4π 1.8 ×10−5 m
= 2.445 ×10−10 N ≈ 2.4 ×10−10 N
(b)
A proton in the field would experience a downwards force of magnitude
FE = eE . The force of
gravity on the proton will be negligible compared to the electric force.
a=
3eE
=
2
FE = eE = ma →
m
≈ 1.5 ×109
58.
There are four forces to calculate. Call the rightward direction the positive direction.
The value of k
is 8.988 ×10 9 N ⋅ m 2 C 2 and the value of e is 1.602 ×10−19 C .
56.
(a)
From problem 55, we know that the electric field is pointed towards the Earth’s
center. Thus an
electron in such a field would experience an upwards force of magnitude FE = eE .
The force of gravity on the electron will be negligible compared to the electric force.
eE
9.80 m s2
Fnet = FCH + FCN + FOH + FON =
r2
k
8.988 ×109 N ⋅ m2 C2
Since the electric field is pointing towards the Earth’s center, the charge must be
negative .
a=
g
1.439 ×1010 m s 2
57. For the droplet to remain stationary, the magnitude of the electric force on the droplet must
be the same as the weight of the droplet. The mass of the droplet is found from its volume
times the density of water. Let n be the number of excess electrons on the water droplet.
Use Eq. 16-4a to calculate the magnitude of the electric charge on the Earth.
E=k
=
FE = q E = mg → neE = 43 π r 3 ρ g →
54.
Since the gravity force is downward, the electric force must be upward. Since the charge
is positive,
the electric field must also be upward. Equate the magnitudes of the two forces and solve
for the electric field.
FE = FG → qE = mg → E =
a
For the proton:
(1.602 ×10 C) (150 N C) = 1.439 ×10
(1.67 × 10 kg)
Q2
2
rorbit
=G
−19
=
(c) For the electron:
10
−27
a
g
=
2.638 ×1013 m s 2
9.80 m s 2
m s 2 ≈ 1.4 ×1010 m s 2 , down
Q=
≈ 2.7 ×1012
GM Moon M Earth
61.
(a)
Thus the
31
k
M Moon M Earth
2
rorbit
( 6.67 ×10
−11
=
→
)(
( 8.988 ×10
)(
)
Nim2 kg 2 7.35 ×1022 kg 5.98 ×1024 kg
9
N ⋅ m2 C2
) = 5.71×10
13
C
The electron will experience a force in the opposite direction to the electric field.
32
Physics Bio 178.
Physics Bio 178.
acceleration is in the opposite direction to the initial velocity. The force is constant, and
so constant acceleration equations apply. To find the stopping distance, set the final
velocity to 0.
eE
F = eE = ma → a =
v2 = v02 + 2a∆x →
m
∆x =
(b)
to return.
v2 − v02
2a
=−
mv02
2eE
2
6
−19
3
FE = k
To return to the starting point, the velocity will reverse. Use that to find the time
v = v0 + at →
t=
(9.11×10 kg )( 21.5 ×10 m s ) = 0.115m
2 ( −1.602 ×10 C )(11.4 ×10 N C)
−31
=−
the spheres, write the net force in both the horizontal and vertical directions and solve for the
electric force. Then write the electric force by Coulomb’s law, and equate the two
expressions for the electric force to find the charge.
mg
∑ Fy = FT cos θ − mg = 0 → FT = cosθ
mg
∑ Fx = FT sin θ − FE = 0 → FE = FT sin θ = cos θ sinθ = mg tan θ
v − v0
a
=
−v0 − v0
a
=−
2mv0
qE
=−
(
)(
)(
2 9.11 ×10 kg 21.5 ×10 m s
(
−31
6
) = 2.14 ×10
)
−1.602 ×10 −19 C 11.4 ×103 N C
Q2
Q1 − Q2
d=
5.0 ×10−6 C
−5
−6
2.5 × 10 C − 5.0 ×10 C
( 2.0m) =
−8
E=k
=
64.
Q
=
(
)(
[0.150 − 0.050cos (12.5t )]
1.08 ×107
[3.00 − cos (12.5t )]
2
2
m2
)=
9
2
o
= 6.064 ×10−6 C ≈ 6.1×10−6 C
2
2.6m from Q1
2
Er 2
( 3×10
6
=
)(
−3
N C 3.75 ×10 m
)
2
= 5 ×10−9 C
)
(
)(
)
(
)(
)
= 0.45 N, right
67.
the
NC
FE
33
Since the electric field exerts a force on the charge in
same direction as the electric field, the charge is
positive. Use the free-body diagram to write the
equilibrium equations for both the horizontal and vertical
directions, and use those equations to find the magnitude
of the charge.
N C, upwards
The wires form two sides of an equilateral triangle, and so the two charges are
separated by a distance d = 78 cm and are directly horizontal from each other. Thus
the electric force on each charge is horizontal. From the free-body diagram for one of
→ Q=
(
4
[ 0.150 − 0.050cos (12.5t )]
Q
66.
There will be a rightward force on Q1 due to Q2 , given by Coulomb’s law. There will be
a leftward
force on Q1 due to the electric field created by the parallel plates. Let right be the positive
direction.
QQ
∑ F = k x1 2 2 − Q1 E
6.7 ×10−6 C 1.8 ×10−6 C
= 8.988 ×109 N ⋅ m2 C2
− 6.7 ×10−6 C 7.3× 104 N C
( 0.34m)2
1.6 m from Q2 ,
2.70 ×10
( 24 ×10 kg )( 9.80m s ) tan 30
(8.988 ×10 N ⋅ m C )
2
r2
k
8.988 ×109 N ⋅ m2 C2
This corresponds to about 3 billion electrons.
this distance and the charge to give the electric field value at the tabletop. That electric field
will point upwards at all times, towards the negative sphere.
r2
)
k
−3
65.
The electric field at the surface of the pea is given by Equation (16-4a). Solve that
equation for the
charge.
The distance of the sphere from the table is given by r = [ 0.150 − 0.050cos (12.5t )] m . Use
E=k
mg tan θ
= mg tan θ → Q = 2d
s
angular frequency of the sphere is given by ω = k m = 126 N m 0.800 kg = 12.5 rad s .
3.00 ×10−6 C
2
(
63.
The sphere will oscillate sinusoidally about the equilibrium point, with an amplitude of
5.0 cm. The
8.988 × 109 N ⋅ m2 C2
d
2
= 2 7.8 × 10−1 m
62.
Because of the inverse square nature of the electric
Q1
Q2
field,
any location where the field is zero must be closer to the
d
l
weaker charge ( Q2 ) . Also, in between the two charges,
the fields due to the two charges are parallel to each other and cannot cancel. Thus the only
places where the field can be zero are closer to the weaker charge, but not between them. In
the diagram, this means that x must be positive.
Q
Q1
2
E = −k 22 + k
= 0 → Q2 ( l + d ) = Q1l 2 →
l
( l + d )2
l=
(Q 2)
FT
θ
FT
mg
FE
43cm
θ
L = 55cm
θ
mg
34
Physics Bio 178.
Physics Bio 178.
θ = cos −1
55
= FE − FT sin θ = 0 → FE = FT sin θ = QE
∑F
= FT cos θ − mg = 0 → FT =
y
Q=
mg tan θ
E
71.
On the x-axis, the electric field can only be zero at a location closer to the smaller
magnitude charge.
Thus the field will never be zero to the left of the midpoint between the two charges. Also,
in between the two charges, the field due to both charges will point to the left, and so the
total field cannot be zero. Thus the only place on the x-axis where the field can be zero is to
the right of the negative charge, and so x must be positive. Calculate the field at point P and
set it equal to zero.
= 38.6o
∑F
x
68.
than
43
mg
cos θ
(1.0 ×10 kg)( 9.80m s ) tan 38.6
(1.2 ×10 N C)
−3
=
→ QE = mg tan θ
2
o
= 6.5 ×10−7 C
4
E=k
The weight of the mass is only about 2 N. Since the tension in the string is more
that, there must be a downward electric force on the positive charge, which means that
the electric field must be pointed down . Use the free-body diagram to write an
expression for the magnitude of the electric field.
∑ F = FT − mg − FE = 0 → FE = QE = FT − mg →
E=
FT − mg
Q
=
(
5.67 N − ( 0.210 kg ) 9.80 m s
3.40 ×10−7C
2
) = 1.06 ×10
7
= −7.0 × 10 C
8
The net charge of the bar is 0C , since there are equal numbers of protons and electrons.
(a)
The force of sphere B on sphere A is given by Coulomb’s law.
FAB =
kQ
(b)
shared between
the two spheres, and so the charge on B is reduced to Q 2 . Again use Coulomb’s law.
QQ 2
=
kQ2
, away from B
R
2 R2
The result of touching sphere A to sphere C is that the charge on the two spheres
2
(c)
is shared, and
so the charge on A is reduced to 3Q 4 . Again use Coulomb’s law.
FAB = k
( 3 Q 4 )( Q 2 ) =
R2
3kQ
Q
= 0 → 2x = ( x + d )
2
( x + d )2
2
→ x=
d
2 −1
≈ 2.41d
73.
FE
F43 = k
Q
→ F42 x = k
2d 2
Q2
d
Q
2
2d 2
cos45o = k
→ F43 x = 0 , F43 y = k
2
2Q
4d 2
2
, F42 y = k
2Q
Q
E
A negative charge must be placed at the center of the square. Let
Q = 8.0 µC be the charge at each corner, let -q be the magnitude of
negative charge in the center, and let d = 9.2cm be the side length of
the square. By the symmetry of the problem, if we make the net force
on one of the corner charges be zero, the net force on each other
corner charge will also be zero.
Q2
Q2
F41 = k 2 → F41x = k 2 , F41 y = 0
d
d
2
FE
−Q
Q1
d
F4 q
Q3
Q2
2
4d 2
Q2
d2
2qQ
2qQ
qQ
→ F4 qx = − k 2 cos 45o = − k
= F4 qy
d2 2
d
d2
The net force in each direction should be zero.
x
=k
Q2
d
2
+k
2Q2
4d 2
+ 0−k
2qQ
d2
, away from B
35
F41
−q
F4 q = k
∑F
F42
F43
Q4
 1 1
−6
+  = 7.66 ×10 C
 2 4
= 0 → q = Q
So the charge to be placed is −q = −7.66 ×10−6 C .
This is an unstable equilibrium . If the center charge were slightly displaced, say towards
the right, then it would be closer to the right charges than the left, and would be attracted
more to the right. Likewise the positive charges on the right side of the square would be
2
8R 2
+k
The electric field will put a force of magnitude FE = QE on
each charge. The distance of each charge from the pivot point is
QEL
L 2 , and so the torque caused by each force is τ = FE r⊥ =
.
2
Both torques will tend to make the rod rotate counterclockwise
 QEL  = QEL .
in the diagram, and so the net torque is τ net = 2 

 2 
F42 = k
2
, away from B
R2
The result of touching sphere B to uncharged sphere C is that the charge on B is
FAB = k
72.
NC
69.
To find the number of electrons, convert the mass to moles, the moles to atoms, and then
multiply by
the number of electrons in an atom to find the total electrons. Then convert to charge.
 1 mole Al   6.02 × 1023 atoms   13 electrons   −1.602 × 10−19 C 
15 kg Al = (15 kg Al ) 




−2
1 mole
electron
 2.7 × 10 kg 
  1 molecule  

70.
FE
x2
The field cannot be zero at any points off the x-axis. For any point off the x-axis, the electric
fields due to the two charges will not be along the same line, and so they can never combine
to give 0.
FT
mg
( − Q 2)
36
Physics Bio 178.
closer to it and would be attracted more to it, moving from their corner positions. The
system would not have a tendency to return to the symmetric shape, but rather would have a
tendency to move away from it if disturbed.
37

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