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Problems
Section 18.1 The Origin of Electricity Section 18.2 Charged
Objects and the Electric Force Section 18.3 Conductors and
Insulators Section 18.4 Charging by Contact and by Induction
1. Iron atoms have been detected in the sun's outer atmosphere,
some with many of their electrons stripped away.
What is the net electric charge (in coulombs) of an iron atom
with 26 protons and 7 electrons? Be sure to include the
algebraic sign ( or ) in your answer.
Answer:
REASONING The charge of a single proton is , and the charge of a
single electron is , where
. The net charge of the ionized atom is the sum of the charges
of its constituent protons and
electrons.
SOLUTION The ionized atom has 26 protons and 7 electrons, so its
net electric charge is
2. An object has a charge of . How many electrons must be
removed so that the charge becomes ?
3. Four identical metallic objects carry the following charges:
, , , and . The objects are
brought simultaneously into contact, so that each touches the
others. Then they are separated.
(a) What is the final charge on each object?
Answer:
(b) How many electrons (or protons) make up the final charge on
each object?
Answer:
4.
Four identical metal spheres have charges of , , , and
.
(a) Two of the spheres are brought together so they touch, and
then they are separated. Which spheres are they,
if the final charge on each one is ?
(b) In a similar manner, which three spheres are brought
together and then separated, if the final charge on each
of the three is ?
(c) The final charge on each of the three separated spheres in
part (b) is . How many electrons would
have to be added to one of these spheres to make it electrically
neutral?
5. Consider three identical metal spheres, A, B, and C. Sphere A
carries a charge of . Sphere B carries a
charge of . Sphere C carries no net charge. Spheres A and B are
touched together and then separated. Sphere C is
then touched to sphere A and separated from it. Last, sphere C
is touched to sphere B and separated from it.
(a) How much charge ends up on sphere C? What is the total
charge on the three spheres
Answer:
REASONING Identical conducting spheres equalize their charge
upon touching. When spheres A
and B touch, an amount of charge , flows from A and
instantaneously neutralizes the charge on B
leaving B momentarily neutral. Then, the remaining amount of
charge, equal to , is equally split
between A and B, leaving A and B each with equal amounts of
charge . Sphere C is initially neutral, so
when A and C touch, the on A splits equally to give on A and on
C. When B and C touch, the
on B and the on C combine to give a total charge of , which is
then equally divided between
the spheres B and C; thus, B and C are each left with an amount
of charge .
SOLUTION Taking note of the initial values given in the problem
statement, and summarizing the final
results determined in the REASONING above, we conclude the
following:
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(a) Sphere C ends up with an amount of charge equal to .
(b) The charges on the three spheres before they were touched,
are, according to the problem statement,
on sphere A, on sphere B, and zero charge on sphere C. Thus, the
total charge on the spheres
is .
(c) The charges on the spheres after they are touched are on
sphere A, on sphere B, and
on sphere C. Thus, the total charge on the spheres is .
(b) before they are allowed to touch each other and
Answer:
REASONING Identical conducting spheres equalize their charge
upon touching. When spheres A
and B touch, an amount of charge , flows from A and
instantaneously neutralizes the charge on B
leaving B momentarily neutral. Then, the remaining amount of
charge, equal to , is equally split
between A and B, leaving A and B each with equal amounts of
charge . Sphere C is initially neutral, so
when A and C touch, the on A splits equally to give on A and on
C. When B and C touch, the
on B and the on C combine to give a total charge of , which is
then equally divided between
the spheres B and C; thus, B and C are each left with an amount
of charge .
SOLUTION Taking note of the initial values given in the problem
statement, and summarizing the final
results determined in the REASONING above, we conclude the
following:
(a) Sphere C ends up with an amount of charge equal to .
(b) The charges on the three spheres before they were touched,
are, according to the problem statement,
on sphere A, on sphere B, and zero charge on sphere C. Thus, the
total charge on the spheres
is .
(c) The charges on the spheres after they are touched are on
sphere A, on sphere B, and
on sphere C. Thus, the total charge on the spheres is .
(c) after they have touched?
Answer:
REASONING Identical conducting spheres equalize their charge
upon touching. When spheres A
and B touch, an amount of charge , flows from A and
instantaneously neutralizes the charge on B
leaving B momentarily neutral. Then, the remaining amount of
charge, equal to , is equally split
between A and B, leaving A and B each with equal amounts of
charge . Sphere C is initially neutral, so
when A and C touch, the on A splits equally to give on A and on
C. When B and C touch, the
on B and the on C combine to give a total charge of , which is
then equally divided between
the spheres B and C; thus, B and C are each left with an amount
of charge .
SOLUTION Taking note of the initial values given in the problem
statement, and summarizing the final
results determined in the REASONING above, we conclude the
following:
(a) Sphere C ends up with an amount of charge equal to .
(b) The charges on the three spheres before they were touched,
are, according to the problem statement,
on sphere A, on sphere B, and zero charge on sphere C. Thus, the
total charge on the spheres
is .
(c) The charges on the spheres after they are touched are on
sphere A, on sphere B, and
on sphere C. Thus, the total charge on the spheres is .
REASONING Identical conducting spheres equalize their charge
upon touching. When spheres A and B
touch, an amount of charge , flows from A and instantaneously
neutralizes the charge on B leaving B
momentarily neutral. Then, the remaining amount of charge, equal
to , is equally split between A and B, leaving
A and B each with equal amounts of charge . Sphere C is
initially neutral, so when A and C touch, the on A
splits equally to give on A and on C. When B and C touch, the on
B and the on C combine to give a
total charge of , which is then equally divided between the
spheres B and C; thus, B and C are each left with an
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amount of charge .
SOLUTION Taking note of the initial values given in the problem
statement, and summarizing the final results
determined in the REASONING above, we conclude the
following:
(a) Sphere C ends up with an amount of charge equal to .
(b) The charges on the three spheres before they were touched,
are, according to the problem statement, on
sphere A, on sphere B, and zero charge on sphere C. Thus, the
total charge on the spheres is
.
(c) The charges on the spheres after they are touched are on
sphere A, on sphere B, and on
sphere C. Thus, the total charge on the spheres is .
6. A plate carries a charge of , while a rod carries a charge of
. How many electrons must be
transferred from the plate to the rod, so that both objects have
the same charge?
*7. Water has a mass per mole of 18.0 g/mol, and each water
molecule has 10 electrons.
(a)
How many electrons are there in one liter of water?
Answer:
(b) What is the net charge of all these electrons?
Answer:
Section
18.5 Coulomb's Law
8. In a vacuum, two particles have charges of and , where . They
are separated by a distance of
0.26 m, and particle 1 experiences an attractive force of 3.4 N.
What is (magnitude and sign)?
9. Two spherical objects are separated by a distance that is .
The objects are initially electrically
neutral and are very small compared to the distance between
them. Each object acquires the same negative charge due
to the addition of electrons. As a result, each object
experiences an electrostatic force that has a magnitude of
. How many electrons did it take to produce the charge on one of
the objects?
Answer: 8 electrons
REASONING The number of excess electrons on one of the objects
is equal to the charge on it divided by
the charge of an electron , or . Since the charge on the object
is negative, we can write
, where is the magnitude of the charge. The magnitude of the
charge can be found from Coulomb's law
(Equation 18.1), which states that the magnitude of the
electrostatic force exerted on each object is given by
, where is the distance between them.
SOLUTION The number of excess electrons on one of the objects
is
(1)
To find the magnitude of the charge, we solve Coulomb's law,
,for :
Substituting this result into Equation 1 gives
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10. Two tiny conducting spheres are identical and carry charges
of and . They are separated by a
distance of 2.50 cm.
(a) What is the magnitude of the force that each sphere
experiences, and is the force attractive or repulsive?
(b) The spheres are brought into contact and then separated to a
distance of 2.50 cm. Determine the magnitude
of the force that each sphere now experiences, and state whether
the force is attractive or repulsive.
11. Two very small spheres are initially neutral and separated
by a distance of 0.50 m. Suppose that
electrons are removed from one sphere and placed on the
other.
(a) What is the magnitude of the electrostatic force that acts
on each sphere?
Answer: 0.83 N
REASONING Initially, the two spheres are neutral. Since negative
charge is removed from the
sphere which loses electrons, it then carries a net positive
charge. Furthermore, the neutral sphere to which
the electrons are added is then negatively charged. Once the
charge is transferred, there exists an
electrostatic force on each of the two spheres, the magnitude of
which is given by Coulomb's law (Equation
18.1), .
SOLUTION
(a) Since each electron carries a charge of , the amount of
negative charge removed
from the first sphere is
Thus, the first sphere carries a charge , while the second
sphere carries a charge
. The magnitude of the electrostatic force that acts on each
sphere is, therefore,
(b) Since the spheres carry charges of opposite sign, the force
is .
(b) Is the force attractive or repulsive? Why?
Answer: attractive
REASONING Initially, the two spheres are neutral. Since negative
charge is removed from the
sphere which loses electrons, it then carries a net positive
charge. Furthermore, the neutral sphere to which
the electrons are added is then negatively charged. Once the
charge is transferred, there exists an
electrostatic force on each of the two spheres, the magnitude of
which is given by Coulomb's law (Equation
18.1), .
SOLUTION
(a) Since each electron carries a charge of , the amount of
negative charge removed
from the first sphere is
Thus, the first sphere carries a charge , while the second
sphere carries a charge
. The magnitude of the electrostatic force that acts on each
sphere is, therefore,
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(b) Since the spheres carry charges of opposite sign, the force
is .
REASONING Initially, the two spheres are neutral. Since negative
charge is removed from the sphere which
loses electrons, it then carries a net positive charge.
Furthermore, the neutral sphere to which the electrons are added
is
then negatively charged. Once the charge is transferred, there
exists an electrostatic force on each of the two spheres,
the magnitude of which is given by Coulomb's law (Equation
18.1), .
SOLUTION
(a) Since each electron carries a charge of , the amount of
negative charge removed from the
first sphere is
Thus, the first sphere carries a charge , while the second
sphere carries a charge
. The magnitude of the electrostatic force that acts on each
sphere is, therefore,
(b) Since the spheres carry charges of opposite sign, the force
is .
12. Two charges attract each other with a force of 1.5 N. What
will be the force if the distance between them is reduced to
one-ninth of its original value?
13. Two point charges are fixed on the y axis: a negative point
charge at and a positive
point charge at . A third point charge is fixed at the origin.
The net electrostatic
force exerted on the charge q by the other two charges has a
magnitude of 27 N and points in the direction. Determine the
magnitude of .
Answer:
14.
The drawings show three charges that have the same magnitude but
may have different signs. In all cases the
distance d between the charges is the same. The magnitude of the
charges is , and the distance between
them is . Determine the magnitude of the net force on charge 2
for each of the three drawings.
15. Two tiny spheres have the same mass and carry charges of the
same magnitude. The mass of each sphere
is . The gravitational force that each sphere exerts on the
other is balanced by the electric force.
(a) What algebraic signs can the charges have?
Answer: the same algebraic signs, both positive or both
negative
REASONING AND SOLUTION
(a) Since the gravitational force between the spheres is one of
attraction and the electrostatic force must
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balance it, the electric force must be one of repulsion.
Therefore, the charges must have
.
(b) There are two forces that act on each sphere; they are the
gravitational attraction of one sphere for
the other, and the repulsive electric force of one sphere on the
other. From the problem statement,
we know that these two forces balance each other, so that . The
magnitude of is given
by Newton's law of gravitation (Equation 4.3: ), while the
magnitude of is
given by Coulomb's law (Equation 18.1: ). Therefore, we have
since the spheres have the same mass and carry charges of the
same magnitude . Solving for ,
we find
(b) Determine the charge magnitude.
Answer:
REASONING AND SOLUTION
(a) Since the gravitational force between the spheres is one of
attraction and the electrostatic force must
balance it, the electric force must be one of repulsion.
Therefore, the charges must have
.
(b) There are two forces that act on each sphere; they are the
gravitational attraction of one sphere for
the other, and the repulsive electric force of one sphere on the
other. From the problem statement,
we know that these two forces balance each other, so that . The
magnitude of is given
by Newton's law of gravitation (Equation 4.3: ), while the
magnitude of is
given by Coulomb's law (Equation 18.1: ). Therefore, we have
since the spheres have the same mass and carry charges of the
same magnitude . Solving for ,
we find
REASONING AND SOLUTION
(a) Since the gravitational force between the spheres is one of
attraction and the electrostatic force must balance it,
the electric force must be one of repulsion. Therefore, the
charges must have
.
(b) There are two forces that act on each sphere; they are the
gravitational attraction of one sphere for the other,
and the repulsive electric force of one sphere on the other.
From the problem statement, we know that these
two forces balance each other, so that . The magnitude of is
given by Newton's law of gravitation
(Equation 4.3: ), while the magnitude of is given by Coulomb's
law (Equation 18.1:
). Therefore, we have
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since the spheres have the same mass and carry charges of the
same magnitude . Solving for , we find
16. A charge is located at the origin, while an identical charge
is located on the x axis at . A third
charge of is located on the x axis at such a place that the net
electrostatic force on the charge at the origin
doubles, its direction remaining unchanged. Where should the
third charge be located?
17. Two particles, with identical positive charges and a
separation of , are released from rest.
Immediately after the release, particle 1 has an acceleration
whose magnitude is , while
particle 2 has an acceleration whose magnitude is . Particle 1
has a mass of .
Find
(a) the charge on each particle and
Answer:
REASONING Each particle will experience an electrostatic force
due to the presence of the other
charge. According to Coulomb's law (Equation 18.1), the
magnitude of the force felt by each particle can be
calculated from , where are the respective charges on particles
1 and 2 and is
the distance between them. According to Newton's second law, the
magnitude of the force experienced by
each particle is given by , where is the acceleration of the
particle and we have assumed that the
electrostatic force is the only force acting.
SOLUTION
(a
) Since the two particles have identical positive charges, , and
we have, using the data
for particle 1,
Solving for , we find that
(b
)
Since each particle experiences a force of the same magnitude
(From Newton's third law), we can write
, or . Solving this expression for the mass of particle 2, we
have
(b) the mass of particle 2.
Answer:
REASONING Each particle will experience an electrostatic force
due to the presence of the other
charge. According to Coulomb's law (Equation 18.1), the
magnitude of the force felt by each particle can be
calculated from , where are the respective charges on particles
1 and 2 and is
the distance between them. According to Newton's second law, the
magnitude of the force experienced by
each particle is given by , where is the acceleration of the
particle and we have assumed that the
electrostatic force is the only force acting.
SOLUTION
(a
) Since the two particles have identical positive charges, , and
we have, using the data
for particle 1,
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Solving for , we find that
(b
)
Since each particle experiences a force of the same magnitude
(From Newton's third law), we can write
, or . Solving this expression for the mass of particle 2, we
have
REASONING Each particle will experience an electrostatic force
due to the presence of the other charge.
According to Coulomb's law (Equation 18.1), the magnitude of the
force felt by each particle can be calculated from
, where are the respective charges on particles 1 and 2 and is
the distance between
them. According to Newton's second law, the magnitude of the
force experienced by each particle is given by
, where is the acceleration of the particle and we have assumed
that the electrostatic force is the only force acting.
SOLUTION
(a) Since the two particles have identical positive charges, ,
and we have, using the data for
particle 1,
Solving for , we find that
(b) Since each particle experiences a force of the same
magnitude (From Newton's third law), we can write
, or . Solving this expression for the mass of particle 2, we
have
18. A charge of is fixed at the center of a compass. Two
additional charges are fixed on the circle of the
compass, which has a radius of 0.100 m. The charges on the
circle are at the position due north and
at the position due east. What are the magnitude and direction
of the net electrostatic force acting on the
charge at the center? Specify the direction relative to due
east.
*19. Multiple-Concept Example 3 provides some pertinent
background for this problem. Suppose a single electron orbits
about a nucleus containing two protons , as would be the case
for a helium atom from which one of the two
naturally occurring electrons is removed. The radius of the
orbit is . Determine the magnitude of the
electron's centripetal acceleration.
Answer:
*20.
The drawing shows an equilateral triangle, each side of which
has a length of 2.00 cm. Point charges are fixed
to each corner, as shown. The charge experiences a net force due
to the charges and . This net force
points vertically downward and has a magnitude of 405 N.
Determine the magnitudes and algebraic signs of the
charges and .
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*21. The drawing shows three point charges fixed in place. The
charge at the coordinate origin has a value of
; the other two charges have identical magnitudes, but opposite
signs: and
.
(a) Determine the net force (magnitude and direction) exerted on
by the other two
charges.
Answer:
0.166 N directed along the
(b) If had a mass of 1.50 g and it were free to move, what would
be its acceleration?
Answer:
directed along the
*22. An electrically neutral model airplane is flying in a
horizontal circle on a 3.0-m guideline, which is nearly
parallel
to the ground. The line breaks when the kinetic energy of the
plane is 50.0 J. Reconsider the same situation, except
that now there is a point charge of on the plane and a point
charge of at the other end of the guideline. In this
case, the line breaks when the kinetic energy of the plane is
51.8 J. Find the magnitude of the charges.
*23. Multiple-Concept Example 3 illustrates several of the
concepts that come into play in this problem. A single
electron orbits a lithium nucleus that contains three protons .
The radius of the orbit is .
Determine the kinetic energy of the electron.
Answer:
*24.
An unstrained horizontal spring has a length of 0.32 m and a
spring constant of 220 N/m. Two small charged
objects are attached to this spring, one at each end. The
charges on the objects have equal magnitudes. Because of
these charges, the spring stretches by 0.020 m relative to its
unstrained length. Determine
(a) the possible algebraic signs and
(b) the magnitude of the charges.
*25. In the rectangle in the drawing, a charge is to be placed
at the empty corner to make the net force on the charge
at corner A point along the vertical direction. What charge
(magnitude and algebraic sign) must be placed at the empty
corner?
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Answer:
REASONING Consider the drawing at the right. It is given that
the charges , , and are each positive.
Therefore, the charges and each exert a repulsive force on the
charge . As the drawing shows, these forces
have magnitudes (vertically downward) and (horizontally to the
left). The unknown charge placed at the
empty corner of the rectangle is , and it exerts a force on that
has a magnitude . In order that the net force
acting on point in the vertical direction, the horizontal
component of must cancel out the horizontal force
. Therefore, must point as shown in the drawing, which means
that it is an attractive force and must be
negative, since is positive.
SOLUTION The basis for our solution is the fact that the
horizontal component of must cancel out the horizontal
force . The magnitudes of these forces can be expressed using
Coulomb's law , where is the
distance between the charges and . Thus, we have
where we have used the fact that the distance between the
charges and is the diagonal of the rectangle, which is
according to the Pythagorean theorem, and the fact that the
distance between the charges and is
. The horizontal component of is , which must be equal to , so
that we have
The drawing in the REASONING, reveals that . Therefore, we find
that
As discussed in the REASONING, the algebraic sign of the charge
is .
**26. There are four charges, each with a magnitude of . Two are
positive and two are negative. The charges are fixed
to the corners of a 0.30-m square, one to a corner, in such a
way that the net force on any charge is directed toward the
center of the square. Find the magnitude of the net
electrostatic force experienced by any charge.
**27. A small spherical insulator of mass and charge is hung by
a thread of negligible
mass. A charge of is held 0.150 m away from the sphere and
directly to the right of it, so the thread makes
an angle with the vertical (see the drawing). Find
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(a) the angle and
Answer:
REASONING The charged insulator experiences an electric force
due to the
presence of the charged sphere shown in the drawing in the text.
The forces acting on the
insulator are the downward force of gravity (i.e., its weight,
), the electrostatic
force (see Coulomb's law, Equation 18.1) pulling to the right,
and the
tension in the thread pulling up and to the left at an angle
with respect to the vertical,
as shown in the drawing in the problem statement. We can analyze
the forces to determine
the desired quantities and .
SOLUTION
(a
)
We can see from the diagram given with the problem statement
that
and
Dividing the first equation by the second yields
Solving for , we find that
(b
) Since , the tension can be obtained as follows:
(b) the tension in the thread.
Answer: 0.813 N
REASONING The charged insulator experiences an electric force
due to the presence of the charged
sphere shown in the drawing in the text. The forces acting on
the insulator are the downward force of gravity
(i.e., its weight, ), the electrostatic force (see Coulomb's
law, Equation 18.1)
pulling to the right, and the tension in the thread pulling up
and to the left at an angle with respect to the
vertical, as shown in the drawing in the problem statement. We
can analyze the forces to determine the desired
quantities and .
-
SOLUTION
(a) We can see from the diagram given with the problem statement
that
and
Dividing the first equation by the second yields
Solving for , we find that
(b
) Since , the tension can be obtained as follows:
REASONING The charged insulator experiences an electric force
due to the presence of the charged sphere
shown in the drawing in the text. The forces acting on the
insulator are the downward force of gravity (i.e., its weight,
), the electrostatic force (see Coulomb's law, Equation 18.1)
pulling to the right, and the
tension in the thread pulling up and to the left at an angle
with respect to the vertical, as shown in the drawing in the
problem statement. We can analyze the forces to determine the
desired quantities and .
SOLUTION
(a) We can see from the diagram given with the problem statement
that
and
Dividing the first equation by the second yields
Solving for , we find that
(b) Since , the tension can be obtained as follows:
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**28. Two objects carry initial charges that are and ,
respectively, where . They are located 0.200 m
apart and behave like point charges. They attract each other
with a force that has a magnitude of 1.20 N. The
objects are then brought into contact, so the net charge is
shared equally, and then they are returned to their initial
positions. Now it is found that the objects repel one another
with a force whose magnitude is equal to the
magnitude of the initial attractive force. What are the
magnitudes of the initial charges on the objects?
Section 18.6 The Electric Field
Section 18.7 Electric Field Lines Section 18.8 The Electric
Field Inside a Conductor: Shielding
29. At a distance from a point charge, the magnitude of the
electric field created by the charge is 248 N/C. At a
distance from the charge, the field has a magnitude of 132 N/C.
Find the ratio / .
Answer: 1.37
REASONING The electric field created by a point charge is
inversely proportional to the square of the
distance from the charge, according to Equation 18.3. Therefore,
we expect the distance to be greater than the
distance , since the field is smaller at than it is at . The
ratio , then, should be greater than one.
SOLUTION Applying Equation 18.3 to each position relative to the
charge, we have
Dividing the expression for by the expression for gives
Solving for the ratio gives
As expected, this ratio is greater than one.
30. Suppose you want to determine the electric field in a
certain region of space. You have a small object of known
charge and an instrument that measures the magnitude and
direction of the force exerted on the object by the electric
field.
(a) The object has a charge of and the instrument indicates that
the electric force exerted on it is
, due east. What are the magnitude and direction of the electric
field?
(b) What are the magnitude and direction of the electric field
if the object has a charge of and the
instrument indicates that the force is , due west?
31. An electric field of 260 000 N/C points due west at a
certain spot. What are the magnitude and direction of the force
that acts on a charge of at this spot?
Answer: 1.8 N due east
32. Review the important features of electric field lines
discussed in Conceptual Example 13. Three point charges (
, and ) are at the corners of an equilateral triangle. Sketch in
six electric field lines between the three
charges.
33. Four point charges have the same magnitude of and are fixed
to the corners of a square that is 4.0 cm
on a side. Three of the charges are positive and one is
negative. Determine the magnitude of the net electric field
that
-
exists at the center of the square.
Answer: 54 N/C
34. The drawing shows two situations in which charges are placed
on the x and y axes. They are all located at the
same distance of 6.1 cm from the origin O. For each of the
situations in the drawing, determine the magnitude of the
net electric field at the origin.
35. A uniform electric field exists everywhere in the x, y
plane. This electric field has a magnitude of 4500 N/C and
is directed in the positive x direction. A point charge is
placed at the origin. Determine the magnitude of the net electric
field at
(a) ,
Answer: 7700 N/C
(b) , and
Answer: 1300 N/C
(c) .
Answer: 5500 N/C
36.
The membrane surrounding a living cell consists of aninner and
an outer wall that are separated by a small
space. Assume that the membrane acts like a parallel plate
capacitor in which the effective charge density on the inner
and outer walls has a magnitude of .
(a) What is the magnitude of the electric field within the cell
membrane?
(b) Find the magnitude of the electric force that would be
exerted on a potassium ion
placed inside the membrane.
37. A long, thin rod lies along the x axis, with its midpoint at
the origin. In a vacuum, a
point charge is fixed to one end of the rod, and a point charge
is fixed to the other end. Everywhere in the x,
y plane there is a constant external electric field that is
perpendicular to the rod. With respect to the z axis, find the
magnitude of the net torque applied to the rod.
Answer:
REASONING The drawing at the right shows the set-up. Here, the
electric field E points along the axis
and applies a force of to the charge and a force of to the
charge, where denotes the
magnitude of each charge. Each force has the same magnitude of ,
according to Equation 18.2. The torque is
measured as discussed in Section 9.1. According to Equation 9.1,
the torque produced by each force has a magnitude
-
given by the magnitude of the force times the lever arm, which
is the perpendicular distance between the point of
application of the force and the axis of rotation. In the
drawing the axis is the axis of rotation and is midway between
the ends of the rod. Thus, the lever arm for each force is half
the length of the rod or , and the magnitude of the
torque produced by each force is .
SOLUTION The and the force each cause the rod to rotate in the
same sense about the axis. Therefore, the
torques from these forces reinforce one another. Using the
expression for the magnitude of each torque,
we find that the magnitude of the net torque is
38. A point charge is placed in an external uniform electric
field that has a magnitude of .
At what distance from the charge is the net electric field
zero?
39. A tiny ball carries a charge of . What electric field
(magnitude and direction) is
needed to cause the ball to float above the ground?
Answer:
directed downward
REASONING Two forces act on the charged ball (charge ); they are
the downward force of gravity and
the electric force F due to the presence of the charge in the
electric field E. In order for the ball to float, these two
forces must be equal in magnitude and opposite in direction, so
that the net force on the ball is zero (Newton's second
law). Therefore, F must point upward, which we will take as the
positive direction. According to Equation 18.2,
. Since the charge is negative, the electric field E must point
downward, as the product in the expression
must be positive, since the force F points upward. The
magnitudes of the two forces must be equal, so that
. This expression can be solved for .
SOLUTION The magnitude of the electric field E is
As discussed in the reasoning, this electric field points .
40. A proton and an electron are moving due east in a constant
electric field that also points due east. The electric
field has a magnitude of . Determine the magnitude of the
acceleration of the proton and the
electron.
41. Review Conceptual Example 12 before attempting to work this
problem. The magnitude of each of the charges in
Figure 18.21 is . The lengths of the sides of the rectangles are
3.00 cm and 5.00 cm. Find the
magnitude of the electric field at the center of the rectangle
in Figures 18.21a and b.
Answer:
-
in Figure 18.21a and in Figure 18.21b
42. Two charges are placed between the plates of a parallel
plate capacitor. One charge is and the other is
. The charge per unit area on each of the plates has a magnitude
of . The
magnitude of the force on due to equals the magnitude of the
force on due to the electric field of the parallel
plate capacitor. What is the distance r between the two
charges?
*43. A small object has a mass of and a charge of . It is placed
at a certain spot where there
is an electric field. When released, the object experiences an
acceleration of in the direction of the
axis. Determine the magnitude and direction of the electric
field.
Answer:
directed along the
*44. A spring with an unstrained length of 0.074 m and a spring
constant of 2.4 N/m hangs vertically downward
from the ceiling. A uniform electric field directed vertically
upward fills the region containing the spring. A sphere
with a mass of and a net charge of is attached to the lower end
of the spring. The spring is
released slowly, until it reaches equilibrium. The equilibrium
length of the spring is 0.059 m. What is the magnitude
of the external electric field?
*45. Two point charges are located along the x axis: at , and
at
. Two other charges are located on the y axis: at , and
at . Find the net electric field (magnitude and direction) at
the origin.
Answer:
directed along the
REASONING The two charges lying on the axis produce no net
electric field at the coordinate origin. This is
because they have identical charges, are located the same
distance from the origin, and produce electric fields that
point
in opposite directions. The electric field produced by at the
origin points away from the charge, or along the
direction. The electric field produced by at the origin points
toward the charge, or along the direction. The net
electric field is, then, , where and can be determined by using
Equation 18.3.
SOLUTION The net electric field at the origin is
The plus sign indicates that .
*46. The total electric field consists of the vector sum of two
parts. One part has a magnitude of
and points at an angle above the axis. The other part has a
magnitude of and points
at an angle above the axis. Find the magnitude and direction of
the total field. Specify the directional
angle relative to the axis.
*47. In Multiple-Concept Example 9 you can see the concepts that
are important in this problem. A particle of
charge and mass is released from rest in a region where there is
a constant electric field of
. What is the displacement of the particle after a time of ?
Answer:
-
*48. The drawing shows a positive point charge , a second point
charge that may be positive or negative, and a
spot labeled P, all on the same straight line. The distance d
between the two charges is the same as the distance
between and the spot P. With present, the magnitude of the net
electric field at P is twice what it is when is
present alone. Given that , determine when it is
(a) positive and
(b) negative.
*49. Multiple-Concept Example 9 illustrates the concepts in this
problem. An electron is released from rest at the negative
plate of a parallel plate capacitor. The charge per unit area on
each plate is , and the plate
separation is . How fast is the electron moving just before it
reaches the positive plate?
Answer:
**50. Two particles are in a uniform electric field that points
in the direction and has a magnitude of 2500 N/C. The
mass and charge of particle 1 are and , while the corresponding
values for
particle 2 are and . Initially the particles are at rest. The
particles are both
located on the same electric field line but are separated from
each other by a distance d. Particle 1 is located to the left
of particle 2. When released, they accelerate but always remain
at this same distance from each other. Find d. **51. Two point
charges of the same magnitude but opposite signs are fixed to
either end of the base of an isosceles
triangle, as the drawing shows. The electric field at the
midpoint M between the charges has a magnitude . The
field directly above the midpoint at point P has a magnitude .
The ratio of these two field magnitudes is
. Find the angle in the drawing.
Answer:
REASONING AND SOLUTION The net electric field at point in Figure
1 is the vector sum of the fields
and , which are due, respectively, to the charges and . These
fields are shown in Figure 2.
Figure 1
-
Figure 2
According to Equation 18.3, the magnitudes of the fields and are
the same, since the triangle is an isosceles
triangle with equal sides of length . Therefore, . The vertical
components of these two fields
cancel, while the horizontal components reinforce, leading to a
total field at point that is horizontal and has a
magnitude of
At point in Figure 1, both and are horizontal and point to the
right. Again using Equation 18.3, we find
Since , we have
But from Figure 1, we can see that . Thus, it follows that
The value for is, then, .
**52. The drawing shows an electron entering the lower left side
of a parallel plate capacitor and exiting at the upper right
side. The initial speed of the electron is . The capacitor is
2.00 cm long, and its plates are separated
by 0.150 cm. Assume that the electric field between the plates
is uniform everywhere and find its magnitude.
**53. A small plastic ball with a mass of and with a charge of
is suspended from an
insulating thread and hangs between the plates of a capacitor
(see the drawing). The ball is in equilibrium, with
the thread making an angle of with respect to the vertical. The
area of each plate is . What is
the magnitude of the charge on each plate?
-
Answer:
Section
18.9 Gauss' Law
54. A spherical surface completely surrounds a collection of
charges. Find the electric flux through the surface if the
collection consists of
(a) a single charge,
(b) a single charge, and
(c) both of the charges in (a) and (b).
55. The drawing shows an edge-on view of two planar surfaces
that intersect and are mutually perpendicular. Surface
1 has an area of , while surface 2 has an area of . The electric
field in the drawing is uniform and has a
magnitude of 250 N/C. Find the magnitude of the electric flux
through
(a) surface 1 and
Answer:
REASONING As discussed in Section 18.9, the magnitude of the
electric flux through a surface is equal to the magnitude of the
component of
the electric field that is normal to the surface multiplied by
the area of the
surface, ,where is the magnitude of the component of E that
is
normal to the surface of area . We can use this expression and
the figure in the
text to determine the desired quantities.
SOLUTION
(a
)
The magnitude of the flux through surface 1 is
(b
)
Similarly, the magnitude of the flux through surface 2 is
(b) surface 2.
Answer:
-
REASONING As discussed in Section 18.9, the magnitude of the
electric flux through a surface is
equal to the magnitude of the component of the electric field
that is normal to the surface multiplied by the
area of the surface, ,where is the magnitude of the component of
E that is normal to the
surface of area . We can use this expression and the figure in
the text to determine the desired quantities.
SOLUTION
(a) The magnitude of the flux through surface 1 is
(b) Similarly, the magnitude of the flux through surface 2
is
REASONING As discussed in Section 18.9, the magnitude of the
electric flux through a surface is equal to
the magnitude of the component of the electric field that is
normal to the surface multiplied by the area of the surface,
,where is the magnitude of the component of E that is normal to
the surface of area . We can use this
expression and the figure in the text to determine the desired
quantities.
SOLUTION
(a) The magnitude of the flux through surface 1 is
(b) Similarly, the magnitude of the flux through surface 2
is
56. A surface completely surrounds a charge. Find the electric
flux through this surface when the surface
is
(a) a sphere with a radius of 0.50 m,
(b) a sphere with a radius of 0.25 m, and
(c) a cube with edges that are 0.25 m long.
57. A circular surface with a radius of 0.057 m is exposed to a
uniform external electric field of magnitude
. The magnitude of the electric flux through the surface is .
What is the angle (less
than 90) between the direction of the electric field and the
normal to the surface?
Answer:
58.
A charge Q is located inside a rectangular box. The electric
flux through each of the six surfaces of the box is:
, , , ,
, and . What is Q? *59. A solid nonconducting sphere has a
positive charge q spread uniformly throughout its volume. The
charge
density or charge per unit volume, therefore, is . Use Gauss'
law to show that the electric field at a point
within the sphere at a radius r has a magnitude of . (Hint: For
a Gaussian surface, use a sphere of radius r
centered within the solid sphere of radius. Note that the net
charge within any volume is the charge density times the
volume.)
Answer: The answer is a proof.
*60. Two spherical shells have a common center. A charge is
spread uniformly over the inner
shell, which has a radius of 0.050 m. A charge is spread
uniformly over the outer shell, which has a
-
radius of 0.15 m. Find the magnitude and direction of the
electric field at a distance (measured from the common
center) of
(a) 0.20 m,
(b) 0.10 m, and
(c) 0.025 m.
*61. A cube is located with one corner situated at the origin of
an x, y, z coordinate system. One of the cube's faces lies
in the x, y plane, another in the y, z plane, and another in the
x, z plane. In other words, the cube is in the first octant of the
coordinate system. The edges of the cube are 0.20 m long. A uniform
electric field is parallel to the x, y plane and
points in the direction of the axis. The magnitude of the field
is 1500 N/C.
(a) Using the outward normal for each face of the cube, find the
electric flux through each of the six faces.
Answer:
The flux through the face in the x, z plane at is . The flux
through the face
parallel to the x, z plane at is . The flux through each of the
remaining four faces is zero.
REASONING The electric flux through each face of the cube is
given by (see
Section 18.9) where is the magnitude of the electric field at
the face, is the area of the face, and is the
angle between the electric field and the outward normal of that
face. We can use this expression to calculate
the electric flux through each of the six faces of the cube.
SOLUTION
(a
) On the bottom face of the cube, the outward normal points
parallel to the axis, in the opposite
direction to the electric field, and . Therefore,
On the top face of the cube, the outward normal points parallel
to the axis, and . The
electric flux is, therefore,
On each of the other four faces, the outward normals are
perpendicular to the direction of the electric
field, so . So for each of the four side faces,
(b
)
The total flux through the cube is
Therefore,
(b) Add the six values obtained in part (a) to show that the
electric flux through the cubical surface is zero, as
Gauss' law predicts, since there is no net charge within the
cube.
Answer:
REASONING The electric flux through each face of the cube is
given by (see
Section 18.9) where is the magnitude of the electric field at
the face, is the area of the face, and is the
-
angle between the electric field and the outward normal of that
face. We can use this expression to calculate
the electric flux through each of the six faces of the cube.
SOLUTION
(a
) On the bottom face of the cube, the outward normal points
parallel to the axis, in the opposite
direction to the electric field, and . Therefore,
On the top face of the cube, the outward normal points parallel
to the axis, and . The
electric flux is, therefore,
On each of the other four faces, the outward normals are
perpendicular to the direction of the electric
field, so . So for each of the four side faces,
(b
)
The total flux through the cube is
Therefore,
REASONING The electric flux through each face of the cube is
given by (see Section
18.9) where is the magnitude of the electric field at the face,
is the area of the face, and is the angle between the
electric field and the outward normal of that face. We can use
this expression to calculate the electric flux through
each of the six faces of the cube.
SOLUTION
(a) On the bottom face of the cube, the outward normal points
parallel to the axis, in the opposite direction to the
electric field, and . Therefore,
On the top face of the cube, the outward normal points parallel
to the axis, and . The electric flux
is, therefore,
-
On each of the other four faces, the outward normals are
perpendicular to the direction of the electric field, so
. So for each of the four side faces,
(b
)
The total flux through the cube is
Therefore,
**62. A long, thin, straight wire of length L has a positive
charge Q distributed uniformly along it. Use Gauss' law to show
that the electric field created by this wire at a radial
distance r has a magnitude of , where
. (Hint: For a Gaussian surface, use a cylinder aligned with its
axis along the wire and note that the cylinder has a flat surface
at either end, as well as a curved surface.)
Copyright 2012 John Wiley & Sons, Inc. All rights
reserved.
Problems
Section 19.1 Potential Energy Section 19.2 The Electric
Potential Difference
1. During a particular thunderstorm, the electric potential
difference between a cloud and the ground is
, with the cloud being at the higher potential. What is the
change in an electron's
electric potential energy when the electron moves from the
ground to the cloud?
Answer:
2.
A particle with a charge of and a mass of is released from rest
at point A and accelerates
toward point B, arriving there with a speed of 42 m/s. The only
force acting on the particle is the electric force.
(a) Which point is at the higher potential? Give your
reasoning.
(b) What is the potential difference between A and B?
3.
Suppose that the electric potential outside a living cell is
higher than that inside the cell by 0.070 V. How
much work is done by the electric force when a sodium ion moves
from the outside to the inside?
Answer:
REASONING AND SOLUTION Combining Equations 19.1 and 19.3, we
have
4. A particle has a charge of and moves from point A to point B,
a distance of 0.20 m. The particle
experiences a constant electric force, and its motion is along
the line of action of the force. The difference between the
particle's electric potential energy at A and at B is .
(a) Find the magnitude and direction of the electric force that
acts on the particle.
(b) Find the magnitude and direction of the electric field that
the particle experiences.
5. Multiple-Concept Example 3 employs some of the concepts that
are needed here. An electric car accelerates for
7.0 s by drawing energy from its 290-V battery pack. During this
time, 1200 C of charge passes through the battery
pack. Find the minimum horsepower rating of the car.
Answer: 67 hp
-
6. Review Multiple-Concept Example 4 to see the concepts that
are pertinent here. In a television picture tube,
electrons strike the screen after being accelerated from rest
through a potential difference of 25 000 V. The speeds of
the electrons are quite large, and for accurate calculations of
the speeds, the effects of special relativity must be taken
into account. Ignoring such effects, find the electron speed
just before the electron strikes the screen.
7. Multiple-Concept Example 4 deals with the concepts that are
important in this problem. As illustrated in Figure
19.5b, a negatively charged particle is released from rest at
point B and accelerates until it reaches point A. The mass
and charge of the particle are and , respectively. Only the
gravitational force and the
electrostatic force act on the particle, which moves on a
horizontal straight line without rotating. The electric potential
at
A is 36 V greater than that at B; in other words, . What is the
translational speed of the particle at point A?
Answer: 19 m/s
REASONING The translational speed of the particle is related to
the particle's translational kinetic energy,
which forms one part of the total mechanical energy that the
particle has. The total mechanical energy is conserved,
because only the gravitational force and an electrostatic force,
both of which are conservative forces, act on the particle
(see Section 6.5). Thus, we will determine the speed at point by
utilizing the principle of conservation of mechanical
energy.
SOLUTION The particle's total mechanical energy is
Since the particle does not rotate the angular speed is always
zero, and since there is no elastic force we may omit the
terms and from this expression. With this in mind, we express
the fact that (energy is conserved)
as follows:
This equation can be simplified further, since the particle
travels horizontally, so that , with the result that
Solving for gives
According to Equation 19.4, the difference in electric potential
energies is related to the electric
potential difference :
Substituting this expression into the expression for gives
8. An electron and a proton, starting from rest, are accelerated
through an electric potential difference of the same
magnitude. In the process, the electron acquires a speed , while
the proton acquires a speed . Find the ratio
.
*9. The potential at location A is 452 V. A positively charged
particle is released there from rest and arrives at
location B with a speed . The potential at location C is 791 V,
and when released from rest from this spot, the
particle arrives at B with twice the speed it previously had, or
. Find the potential at B.
-
Answer: 339 V
REASONING The only force acting on the moving charge is the
conservative electric force. Therefore, the
total energy of the charge remains constant. Applying the
principle of conservation of energy between locations A and
B, we obtain
Since the charged particle starts from rest, . The difference in
potential energies is related to the difference in
potentials by Equation 19.4, . Thus, we have
(1)
Similarly, applying the conservation of energy between locations
C and B gives
(2)
Dividing Equation 1 by Equation 2 yields
This expression can be solved for .
SOLUTION Solving for , we find that
*10. A moving particle encounters an external electric field
that decreases its kinetic energy from 9520 eV to 7060
eV as the particle moves from position A to position B. The
electric potential at A is , and the electric
potential at B is . Determine the charge of the particle.
Include the algebraic sign ( or ) with your
answer.
*11. During a lightning flash, there exists a potential
difference of between
a cloud and the ground. As a result, a charge of is transferred
from the ground to the cloud.
(a) How much work is done on the charge by the electric
force?
Answer:
(b) If the work done by the electric force were used to
accelerate a 1100-kg automobile from rest, what
would be its final speed?
Answer:
(c) If the work done by the electric force were converted into
heat, how many kilograms of water at
could be heated to ?
Answer:
**12. A particle is uncharged and is thrown vertically upward
from ground level with a speed of 25.0 m/s. As a result,
it attains a maximum height h. The particle is then given a
positive charge and reaches the same maximum
height h when thrown vertically upward with a speed of 30.0 m/s.
The electric potential at the height h exceeds the electric
potential at ground level. Finally, the particle is given a
negative charge . Ignoring air
resistance, determine the speed with which the negatively
charged particle must be thrown vertically upward,
so that it attains exactly the maximum height h. In all three
situations, be sure to include the effect of gravity. Section
19.3 The Electric
-
Potential Difference Created by Point Charges
13. Two point charges, and , are separated by 1.20 m. What is
the electric potential midway
between them?
Answer:
14. An electron and a proton are initially very far apart
(effectively an infinite distance apart). They are then brought
together to form a hydrogen atom, in which the electron orbits
the proton at an average distance of .
What is , which is the change in the electric potential
energy?
15. Two charges A and B are fixed in place, at different
distances from a certain spot. At this spot the potentials due
to the two charges are equal. Charge A is 0.18 m from the spot,
while charge B is 0.43 m from it. Find the ratio
of the charges.
Answer: 2.4
REASONING The potential of each charge at a distance away is
given by Equation 19.6 as . By
applying this expression to each charge, we will be able to find
the desired ratio, because the distances are given for
each charge.
SOLUTION According to Equation 19.6, the potentials of each
charge are
Since we know that , it follows that
16. The drawing shows a square, each side of which has a length
of . On two corners of the square are
fixed different positive charges, and . Find the electric
potential energy of a third charge
placed at corner A and then at corner B.
17. The drawing shows four point charges. The value of q is ,
and the distance d is 0.96 m. Find the total
potential at the location P. Assume that the potential of a
point charge is zero at infinity.
-
Probolem 17
Answer:
REASONING The electric potential at a distance from a point
charge is given by Equation 19.6 as
. The total electric potential at location due to the four point
charges is the algebraic sum of the
individual potentials.
SOLUTION The total electric potential at is (see the
drawing)
Substituting in the numbers gives
18. A charge of is fixed at the center of a square that is 0.64
m on a side. How much work is done by the
electric force as a charge of is moved from one corner of the
square to any other empty corner? Explain.
19. The drawing shows six point charges arranged in a rectangle.
The value of q is , and the distance d is 0.13 m.
Find the total electric potential at location P, which is at the
center of the rectangle.
-
Answer:
20. Location A is 3.00 m to the right of a point charge q.
Location B lies on the same line and is 4.00 m to the right of
the
charge. The potential difference between the two locations is .
What are the magnitude and sign
of the charge?
21. Identical charges are fixed to adjacent corners of a square.
What charge (magnitude and algebraic sign)
should be fixed to one of the empty corners, so that the total
electric potential at the remaining empty corner is 0 V?
Answer:
22.
Charges of and are fixed in place, with a distance of 2.00 m
between them. A dashed line is drawn
through the negative charge, perpendicular to the line between
the charges. On the dashed line, at a distance L from the negative
charge, there is at least one spot where the total potential is
zero. Find L.
*23. Determine the electric potential energy for the array of
three charges in the drawing, relative to its value when
the charges are infinitely far away and infinitely far
apart.
Probolem 23
Answer:
REASONING Initially, the three charges are infinitely far apart.
We will proceed as in Example 8 by adding
charges to the triangle, one at a time, and determining the
electric potential energy at each step. According to Equation
19.3, the electric potential energy EPE is the product of the
charge and the electric potential at the spot where the
charge is placed, . The total electric potential energy of the
group is the sum of the energies of each step in
assembling the group.
SOLUTION Let the corners of the triangle be numbered clockwise
as 1, 2 and 3, starting with the top corner. When
the first charge is placed at a corner 1, the charge has no
electric potential energy, . This
is because the electric potential produced by the other two
charges at corner 1 is zero, since they are infinitely far
away.
Once the 8.00- charge is in place, the electric potential that
it creates at corner 2 is
-
where is the distance between corners 1 and 2, and . When the
20.0- charge is
placed at corner 2, its electric potential energy is
The electric potential at the remaining empty corner is the sum
of the potentials due to the two charges that are
already in place on corners 1 and 2:
where , , , and . When the third charge is
placed at corner 3, its electric potential energy is
The electric potential energy of the entire array is given
by
*24.
Two identical point charges are fixed at diagonally opposite
corners of a square with
sides of length 0.480 m. A test charge , with a mass of , is
released from
rest at one of the empty corners of the square. Determine the
speed of the test charge when it reaches the center of the
square.
*25. Two protons are moving directly toward one another. When
they are very far apart, their initial speeds are
. What is the distance of closest approach?
Answer:
*26.
Four identical charges ( each) are brought from infinity and
fixed to a straight line. The charges are
located 0.40 m apart. Determine the electric potential energy of
this group.
*27. A charge of is fixed in place. From a horizontal distance
of 0.0450 m, a particle of mass
and charge is fired with an initial speed of 65.0 m/s directly
toward the fixed charge. How
far does the particle travel before its speed is zero?
Answer: 0.0342 m
REASONING The only force acting on the moving charge is the
conservative electric force. Therefore, the
sum of the kinetic energy KE and the electric potential energy
EPE is the same at points A and B:
Since the particle comes to rest at B, . Combining Equations
19.3 and 19.6, we have
-
and
where is the initial distance between the fixed charge and the
moving charged particle, and is the distance between
the charged particles after the moving charge has stopped.
Therefore, the expression for the conservation of energy
becomes
This expression can be solved for . Once is known, the distance
that the charged particle moves can be determined.
SOLUTION Solving the expression above for gives
Therefore, the charge moves a distance of .
*28. Identical point charges of are fixed to diagonally opposite
corners of a square. A third charge is then
fixed at the center of the square, such that it causes the
potentials at the empty corners to change signs without
changing magnitudes. Find the sign and magnitude of the third
charge.
**29. One particle has a mass of and a charge of . A second
particle has a mass of
and the same charge. The two particles are initially held in
place and then released. The particles fly
apart, and when the separation between them is 0.100 m, the
speed of the particle is 125 m/s. Find
the initial separation between the particles.
Answer:
**30. Review Conceptual Example 7 as background for this
problem. A positive charge is located to the left of a
negative charge . On a line passing through the two charges,
there are two places where the total potential is zero.
The first place is between the charges and is 4.00 cm to the
left of the negative charge. The second place is 7.00 cm to
the right of the negative charge.
(a) What is the distance between the charges?
(b) Find , the ratio of the magnitudes of the charges.
Section 19.4 Equipotential Surfaces andTheir Relation to the
Electric Field 31.
Two equipotential surfaces surround a point charge. How far is
the 190-V surface from the 75.0-V
surface?
Answer: 1.1 m
32. An equipotential surface that surrounds a point charge q has
a potential of 490 V and an area of . Determine q.
33.
The inner and outer surfaces of a cell membrane carry a negative
and a positive charge, respectively.
-
Because of these charges, a potential difference of about 0.070
V exists across the membrane. The thickness of the
cell membrane is . What is the magnitude of the electric field
in the membrane?
Answer:
REASONING AND SOLUTION From Equation 19.7a we know that , where
is the potential
difference between the two surfaces of the membrane, and is the
distance between them. If is a point on the
positive surface and is a point on the negative surface, then .
The electric field
between the surfaces is
34.
A positive point charge is surrounded by an equipotential
surface A, which has a radius
of . A positive test charge moves from surface A to another
equipotential
surface B, which has a radius . The work done as the test charge
moves from surface A to surface B is
. Find .
35. A spark plug in an automobile engine consists of two metal
conductors that are separated by a distance of 0.75
mm. When an electric spark jumps between them, the magnitude of
the electric field is . What is the
magnitude of the potential difference between the
conductors?
Answer:
REASONING The magnitude of the electric field is given by
Equation 19.7a (without the minus sign) as
, where is the potential difference between the two metal
conductors of the spark plug, and is the
distance between the two conductors. We can use this relation to
find .
SOLUTION The potential difference between the conductors is
36. The drawing that accompanies Problem 60 shows a graph of a
set of equipotential surfaces in cross section. The grid
lines are 2.0 cm apart. Determine the magnitude and direction of
the electric field at position D. Specify whether the electric
field points toward the top or the bottom of the drawing.
*37. An electric field has a constant value of and is directed
downward. The field is the same
everywhere. The potential at a point P within this region is 155
V. Find the potential at the following points:
(a) directly above P,
Answer: 179 V
REASONING The drawing shows the electric field and the three
points, , , and , in the
vicinity of point , which we take as the origin. We choose the
upward direction as being positive. Thus,
, since the electric field points straight down. The electric
potential at points and
can be determined from Equation 19.7a as , since and are known.
Since the path from to
is perpendicular to the electric field, no work is done in
moving a charge along such a path. Thus, the
potential difference between these two points is zero.
-
SOLUTION
(a) The potential difference between points and is . The
potential at is
(b) The potential difference between points and is . The
potential at is
(c) Since the path from to is perpendicular to the electric
field and no work is done in moving a
charge along such a path, it follows that . Therefore, .
(b) directly below P,
Answer: 143 V
REASONING The drawing shows the electric field and the three
points, , , and , in the
vicinity of point , which we take as the origin. We choose the
upward direction as being positive. Thus,
, since the electric field points straight down. The electric
potential at points and
can be determined from Equation 19.7a as , since and are known.
Since the path from to
is perpendicular to the electric field, no work is done in
moving a charge along such a path. Thus, the
potential difference between these two points is zero.
SOLUTION
(a) The potential difference between points and is . The
potential at is
(b) The potential difference between points and is . The
potential at is
-
(c) Since the path from to is perpendicular to the electric
field and no work is done in moving a
charge along such a path, it follows that . Therefore, .
(c) directly to the right of P.
Answer: 155 V
REASONING The drawing shows the electric field and the three
points, , , and , in the
vicinity of point , which we take as the origin. We choose the
upward direction as being positive. Thus,
, since the electric field points straight down. The electric
potential at points and
can be determined from Equation 19.7a as , since and are known.
Since the path from to
is perpendicular to the electric field, no work is done in
moving a charge along such a path. Thus, the
potential difference between these two points is zero.
SOLUTION
(a) The potential difference between points and is . The
potential at is
(b) The potential difference between points and is . The
potential at is
(c) Since the path from to is perpendicular to the electric
field and no work is done in moving a
charge along such a path, it follows that . Therefore, .
REASONING The drawing shows the electric field and the three
points, , , and , in the vicinity of
point , which we take as the origin. We choose the upward
direction as being positive. Thus,
, since the electric field points straight down. The electric
potential at points and can be
determined from Equation 19.7a as , since and are known. Since
the path from to is
perpendicular to the electric field, no work is done in moving a
charge along such a path. Thus, the potential
difference between these two points is zero.
-
SOLUTION
(a) The potential difference between points and is . The
potential at is
(b) The potential difference between points and is . The
potential at is
(c) Since the path from to is perpendicular to the electric
field and no work is done in moving a charge along
such a path, it follows that . Therefore, .
*38. An electron is released from rest at the negative plate of
a parallel plate capacitor and accelerates to the positive
plate (see the drawing). The plates are separated by a distance
of 1.2 cm, and the electric field within the capacitor has
a magnitude of . What is the kinetic energy of the electron just
as it reaches the positive plate?
*39. The drawing shows the electric potential as a function of
distance along the x axis. Determine the magnitude of
the electric field in the region
-
(a) A to B,
Answer: 0 V/m
(b) B to C, and
Answer:
(c) C to D.
Answer: 5.0 V/m
*40. At a distance of 1.60 m from a point charge of , there is
an equipotential surface. At greater distances
there are additional equipotential surfaces. The potential
difference between any two successive surfaces is
. Starting at a distance of 1.60 m and moving radially outward,
how many of the additional
equipotential surfaces are crossed by the time the electric
field has shrunk to one-half of its initial value? Do not
include the starting surface.
*41. The drawing shows a uniform electric field that points in
the negative y direction; the magnitude of the
field is 3600 N/C. Determine the electric potential
difference
Probolem 41
(a) between points A and B,
Answer:
-
0 V
(b) between points B and C, and
Answer:
(c) between points C and A.
Answer:
Section
19.5 Capacitors and Dielectrics
42. What is the capacitance of a capacitor that stores of charge
on its plates when a voltage of 1.5 V is applied
between them?
43.
The electric potential energy stored in the capacitor of a
defibrillator is 73 J, and the capacitance is
. What is the potential difference that exists across the
capacitor plates?
Answer:
REASONING According to Equation 19.11b, the energy stored in a
capacitor with capacitance and
potential across its plates is .
SOLUTION Therefore, solving Equation 19.11b for , we have
44. Two identical capacitors store different amounts of energy:
capacitor A stores , and capacitor B stores
. The voltage across the plates of capacitor B is 12 V. Find the
voltage across the plates of capacitor A.
45. The electronic flash attachment for a camera contains a
capacitor for storing the energy used to produce the
flash. In one such unit, the potential difference between the
plates of an capacitor is 280 V.
(a) Determine the energy that is used to produce the flash in
this unit.
Answer: 33 J
(b) Assuming that the flash lasts for , find the effective power
or wattage of the flash. Answer: 8500 W
46. The same voltage is applied between the plates of two
different capacitors. When used with capacitor A, this
voltage causes the capacitor to store of charge and of energy.
When used with capacitor B, which
has a capacitance of , this voltage causes the capacitor to
store a charge that has a magnitude of . Determine
.
47. A parallel plate capacitor has a capacitance of when filled
with a dielectric. The area of each plate is
and the separation between the plates is . What is the
dielectric constant of the dielectric?
Answer: 5.3
REASONING AND SOLUTION Equation 19.10 gives the capacitance for
a parallel plate capacitor filled
with a dielectric of constant : . Solving for , we have
-
48. Two capacitors are identical, except that one is empty and
the other is filled with a dielectric . The
empty capacitor is connected to a 12.0-V battery. What must be
the potential difference across the plates of the
capacitor filled with a dielectric so that it stores the same
amount of electrical energy as the empty capacitor?
49.
The membrane that surrounds a certain type of living cell has a
surface area of and a thickness
of . Assume that the membrane behaves like a parallel plate
capacitor and has a dielectric constant of
5.0.
(a) The potential on the outer surface of the membrane is
greater than that on the inside surface.
How much charge resides on the outer surface?
Answer:
(b) If the charge in part (a) is due to positive ions (charge ),
how many such ions are present on the outer
surface?
Answer:
*50.
Capacitor A and capacitor B both have the same voltage across
their plates. However, the energy of capacitor A
can melt m kilograms of ice at , while the energy of capacitor B
can boil away the same amount of water at
. The capacitance of capacitor A is . What is the capacitance of
capacitor B?
*51. What is the potential difference between the plates of a
3.3-F capacitor that stores sufficient energy to
operate a 75-W light bulb for one minute?
Answer: 52 V
REASONING According to Equation 19.11b, the energy stored in a
capacitor with a capacitance and
potential across its plates is . Once we determine how much
energy is required to operate a 75-W
light bulb for one minute, we can then use the expression for
the energy to solve for .
SOLUTION The energy stored in the capacitor, which is equal to
the energy required to operate a 75-W bulb for one
minute , is
Therefore, solving Equation 19.11b for , we have
*52. A capacitor is constructed of two concentric conducting
cylindrical shells. The radius of the inner cylindrical
shell is , and that of the outer shell is . When the cylinders
carry equal and opposite
charges of magnitude , the electric field between the plates has
an average magnitude of
and is directed radially outward from the inner shell to the
outer shell. Determine
(a) the magnitude of the potential difference between the
cylindrical shells and
(b) the capacitance of this capacitor.
*53. Review Conceptual Example 11 before attempting this
problem. An empty capacitor is connected to a 12.0-V battery
and charged up. The capacitor is then disconnected from the
battery, and a slab of dielectric material is
inserted between the plates. Find the amount by which the
potential difference across the plates changes. Specify
whether the change is an increase or a decrease.
-
Answer: 7.7 V decrease
*54. An empty parallel plate capacitor is connected between the
terminals of a 9.0-V battery and charged up. The
capacitor is then disconnected from the battery, and the spacing
between the capacitor plates is doubled. As a result of
this change, what is the new voltage between the plates of the
capacitor?
**55. The potential difference between the plates of a capacitor
is 175 V. Midway between the plates, a proton and an
electron are released. The electron is released from rest. The
proton is projected perpendicularly toward the negative
plate with an initial speed. The proton strikes the negative
plate at the same instant that the electron strikes the
positive
plate. Ignore the attraction between the two particles, and find
the initial speed of the proton.
Answer:
REASONING If we assume that the motion of the proton and the
electron is horizontal in the direction,
the motion of the proton is determined by Equation 2.8, , where
is the distance traveled by the
proton, is its initial speed, and is its acceleration. If the
distance between the capacitor places is , then this
relation becomes , or
(1)
We can solve Equation 1 for the initial speed of the proton,
but, first, we must determine the time and the
acceleration of the proton . Since the proton strikes the
negative plate at the same instant the electron strikes the
positive plate, we can use the motion of the electron to
determine the time .
For the electron, , where we have taken into account the fact
that the electron is released from rest.
Solving this expression for we have . Substituting this
expression into Equation 1, we have
(2)
The accelerations can be found by noting that the magnitudes of
the forces on the electron and proton are equal, since
these particles have the same magnitude of charge. The force on
the electron is , and the
acceleration of the electron is, therefore,
(3)
Newton's second law requires that , so that
(4)
Combining Equations 2, 3 and 4 leads to the following expression
for , the initial speed of the proton:
SOLUTION Substituting values into the expression above, we
find
**56. If the electric field inside a capacitor exceeds the
dielectric strength of the dielectric between its plates, the
dielectric
will break down, discharging and ruining the capacitor. Thus,
the dielectric strength is the maximum magnitude that
the electric field can have without breakdown occurring. The
dielectric strength of air is , and that of
neoprene rubber is . A certain air-gap, parallel plate capacitor
can store no more than 0.075 J of
electrical energy before breaking down. How much energy can this
capacitor store without breaking down after the
-
gap between its plates is filled with neoprene rubber?
Copyright 2012 John Wiley & Sons, Inc. All rights
reserved.
Problems
Section 20.1 Electromotive Force and Current Section 20.2 Ohm's
Law
1.
A defibrillator is used during a heart attack to restore the
heart to its normal beating pattern (see Section 19.5).
A defibrillator passes of current through the torso of a person
in 2.0 ms.
(a) How much charge moves during this time?
Answer:
(b) How many electrons pass through the wires connected to the
patient?
Answer:
electrons
2. An especially violent lightning bolt has an average current
of lasting 0.138 s. How much charge is
delivered to the ground by the lightning bolt?
3. A battery charger is connected to a dead battery and delivers
a current of for 5.0 hours, keeping the
voltage across the battery terminals at in the process. How much
energy is delivered to the battery?
Answer:
REASONING AND SOLUTION First determine the total charge
delivered to the battery using Equation
20.1:
To find the energy delivered to the battery, multiply this
charge by the energy per unit charge (i.e., the voltage) to get
4. A coffee-maker contains a heating element that has a
resistance of . This heating element is energized by a
outlet. What is the current in the heating element?
5.
Suppose that the resistance between the walls of a biological
cell is .
(a) What is the current when the potential difference between
the walls is 75 mV?
Answer:
(b)
If the current is composed of ions , how many such ions flow in
?
Answer:
ions
*6. A car battery has a rating of 220 amperehours (Ah). This
rating is one indication of the total charge that the battery can
provide to a circuit before failing.
(a) What is the total charge (in coulombs) that this battery can
provide?
(b) Determine the maximum current that the battery can provide
for 38 minutes.
*7. A resistor is connected across the terminals of a battery,
which delivers of energy to the
resistor in six hours. What is the resistance of the
resistor?
Answer:
-
*8. The resistance of a bagel toaster is . To prepare a bagel,
the toaster is operated for one minute from a
outlet. How much energy is delivered to the toaster?
**9. A beam of protons is moving toward a target in a particle
accelerator. This beam constitutes a current
whose value is .
(a) How many protons strike the target in ?
Answer:
protons
REASONING The number of protons that strike the target is equal
to the amount of
electric charge striking the target divided by the charge of a
proton, . From
Equation 20.1, the amount of charge is equal to the product of
the current and the time . We can
combine these two relations to find the number of protons that
strike the target in 15 seconds.
The heat that must be supplied to change the temperature of the
aluminum sample of mass by an
amount is given by Equation 12.4 as , where is the specific heat
capacity of
aluminum. The heat is provided by the kinetic energy of the
protons and is equal to the number of
protons that strike the target times the kinetic energy per
proton. Using this reasoning, we can find
the change in temperature of the block for the 15 second-time
interval.
SOLUTION
(a) The number of protons that strike the target is
(b) The amount of heat provided by the kinetic energy of the
protons is
Since and since Table 12.2 gives the specific heat of aluminum
as
, the change in temperature of the block is
(b) Each proton has a kinetic energy of . Suppose the target is
a 15-gram block of
aluminum, and all the kinetic energy of the protons goes into
heating it up. What is the change in
temperature of the block that results from the 15-s bombardment
of protons?
Answer:
REASONING The number of protons that strike the target is equal
to the amount of
electric charge striking the target divided by the charge of a
proton, . From
Equation 20.1, the amount of charge is equal to the product of
the current and the time . We can
combine these two relations to find the number of protons that
strike the target in 15 seconds.
The heat that must be supplied to change the temperature of the
aluminum sample of mass by an
amount is given by Equation 12.4 as , where is the specific heat
capacity of
aluminum. The heat is provided by the kinetic energy of the
protons and is equal to the number of
protons that strike the target times the kinetic energy per
proton. Using this reasoning, we can find
the change in temperature of the block for the 15 second-time
interval.
SOLUTION
(a) The number of protons that strike the target is
-
(b) The amount of heat provided by the kinetic energy of the
protons is
Since and since Table 12.2 gives the specific heat of aluminum
as
, the change in temperature of the block is
REASONING The number of protons that strike the target is equal
to the amount of electric charge
striking the target divided by the charge of a proton, . From
Equation 20.1, the amount of
charge is equal to the product of the current and the time . We
can combine these two relations to find the
number of protons that strike the target in 15 seconds.
The heat that must be supplied to change the temperature of the
aluminum sample of mass by an amount
is given by Equation 12.4 as , where is the specific heat
capacity of aluminum. The heat is
provided by the kinetic energy of the protons and is equal to
the number of protons that strike the target times
the kinetic energy per proton. Using this reasoning, we can find
the change in temperature of the block for the
15 second-time interval.
SOLUTION
(a) The number of protons that strike the target is
(b) The amount of heat provided by the kinetic energy of the
protons is
Since and since Table 12.2 gives the specific heat of aluminum
as
, the change in temperature of the block is
Section 20.3 Resistance and Resistivity
10. The resistance and the magnitude of the current depend on
the path that the current takes. The drawing shows
three situations in which the current takes different paths
through a piece of material. Each of the rectangular pieces is
made from a material whose resistivity is , and the unit of
length in the drawing is
. Each piece of material is connected to a battery. Find
(a) the resistance and
(b) the current in each case.
-
11. Two wires are identical, except that one is aluminum and one
is copper. The aluminum wire has a resistance of
. What is the resistance of the copper wire?
Answer:
12. A cylindrical wire has a length of and a radius of . It
carries a current of , when a voltage of
is applied across the ends of the wire. From what material in
Table 20.1 is the wire made?
13. A coil of wire has a resistance of at and at . What is the
temperature coefficient of
resistivity?
Answer:
14. A large spool in an electrician's workshop has of
insulation-coated wire coiled around it. When the electrician
connects a battery to the ends of the spooled wire, the
resulting current is . Some weeks later, after cutting off
various lengths of wire for use in repairs, the electrician
finds that the spooled wire carries a 3.1-A current when the
same battery is connected to it. What is the length of wire
remaining on the spool?
15. Two wires have the same length and the same resistance. One
is made from aluminum and the other from
c