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
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:
1.
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:
(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 C. Thus, the total charge on the spheres is
(b)before they are allowed to touch each other and
Answer:
on sphere B, and
.
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 C. Thus, the total charge on the spheres is
(c)after they have touched?
Answer:
on sphere B, and
.
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 C. Thus, the total charge on the spheres is
on sphere B, and
.
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
sphere C. Thus, the total charge on the spheres is
on
.
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,
Substituting this result into Equation 1 gives
,for
:
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
from the first sphere is
Thus, the first sphere carries a charge
, the amount of negative charge removed
, 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
from the first sphere is
Thus, the first sphere carries a charge
, the amount of negative charge removed
, 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
.
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.
Two tiny spheres have the same mass and carry charges of the same magnitude. The mass of each sphere
15.
is
. The gravitational force that each sphere exerts on the other is balanced by the electric force.
What
algebraic
signs can the charges have?
(a)
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
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:
given by Coulomb's law (Equation 18.1:
since the spheres have the same mass
we find
), while the magnitude of
is
). Therefore, we have
and carry charges of the same magnitude
. Solving for
,
(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:
given by Coulomb's law (Equation 18.1:
since the spheres have the same mass
we find
), while the magnitude of
is
). Therefore, we have
and carry charges of the same magnitude
. Solving for
,
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
). Therefore, we have
is given by Coulomb's law (Equation 18.1:
since the spheres have the same mass
16.
17.
and carry charges of the same magnitude
. Solving for
, we find
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?
Two particles, with identical positive charges and a separation of
Immediately after the release, particle 1 has an acceleration
particle 2 has an acceleration
whose magnitude is
Find
(a)the charge on each particle and
Answer:
, are released from rest.
whose magnitude is
, while
. Particle 1 has a mass of
.
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,
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
electron's centripetal acceleration.
Answer:
*20.
. Determine the magnitude of the
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
.
*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
charges.
Answer:
by the other two
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.
*23.
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.
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
Determine the kinetic energy of the electron.
Answer:
*24.
. The radius of the orbit is
.
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?
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
force
. The magnitudes of these forces can be expressed using Coulomb's law
distance between the charges and . Thus, we have
where we have used the fact that the distance between the charges
and
must cancel out the horizontal
, where is the
is the diagonal of the rectangle, which is
according to the Pythagorean theorem, and the fact that the distance between the charges
. The horizontal component of
is
, which must be equal to
, so that we have
The drawing in the REASONING, reveals that
and
is
. 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
(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:
**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
Solving for the ratio
by the expression for
gives
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
Review the important features of electric field lines discussed in Conceptual Example 13. Three point charges (
32.
, 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.
35.
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.
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
of the net electric field at
,
(a)
Answer:
7700 N/C
(b)
, and
Answer:
1300 N/C
(c)
.
is placed at the origin. Determine the magnitude
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
we find that the magnitude of the net torque is
38.
39.
for the magnitude of each torque,
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?
A tiny ball
carries a charge of
needed to cause the ball to float above the ground?
Answer:
.
. What electric field (magnitude and direction) is
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
axis. Determine the magnitude and direction of the electric field.
Answer:
in the direction of the
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
at
, and
. 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
separation is
Answer:
, and the plate
. How fast is the electron moving just before it reaches the positive plate?
**50.Two particles are in a uniform electric field that points in the
mass and charge of particle 1 are
and
direction and has a magnitude of 2500 N/C. The
, 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:
and
REASONING AND SOLUTION The net electric field at point in Figure 1 is the vector sum of the fields
, 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
Since
in Figure 1, both
and
are horizontal and point to the right. Again using Equation 18.3, we find
, 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
the magnitude of the charge on each plate?
. What is
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).
The drawing shows an edge-on view of two planar surfaces that intersect and are mutually perpendicular. Surface
55.
1 has an area of
, while surface 2 has an area of
. The electric field
magnitude of 250 N/C. Find the magnitude of the electric flux through
in the drawing is uniform and has a
(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
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:
,
*59.
. What is the angle (less
,
,
,
, and
. What is Q?
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
shell, which has a radius of 0.050 m. A
charge is spread uniformly over the inner
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
parallel to the x, z plane at
four faces is zero.
is
is
. The flux through the face
. The flux through each of the remaining
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
) direction to the electric field, and
. Therefore,
On the top face of the cube, the outward normal points parallel to the
electric flux is, therefore,
axis, in the opposite
axis, and
. The
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
) direction to the electric field, and
. Therefore,
On the top face of the cube, the outward normal points parallel to the
electric flux is, therefore,
axis, in the opposite
axis, and
. The
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
electric field, and
. Therefore,
On the top face of the cube, the outward normal points parallel to the
is, therefore,
axis, in the opposite direction to the
axis, and
. The electric flux
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
Answer:
moves from the outside to the inside?
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.
Multiple-Concept Example 3 employs some of the concepts that are needed here. An electric car accelerates for
5.
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.
Multiple-Concept Example 4 deals with the concepts that are important in this problem. As illustrated in Figure
7.
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
as follows:
from this expression. With this in mind, we express the fact that
This equation can be simplified further, since the particle travels horizontally, so that
Solving for
, with the result that
gives
According to Equation 19.4, the difference in electric potential energies
potential difference
:
Substituting this expression into the expression for
8.
(energy is conserved)
is related to the electric
gives
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,
potentials by Equation 19.4,
. The difference in potential energies is related to the difference in
. 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,
between them?
Answer:
and
, are separated by 1.20 m. What is the electric potential midway
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?
Two charges A and B are fixed in place, at different distances from a certain spot. At this spot the potentials due
15.
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
16.
, it follows that
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
placed at corner 2, its electric potential energy
is
. When the 20.0-
charge 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
,
,
placed at corner 3, its electric potential energy
, and
is
. When the third charge
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
the initial separation between the particles.
Answer:
**30.Review Conceptual Example 7 as background for this problem. A positive charge
particle is 125 m/s. Find
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.
32.
Two equipotential surfaces surround a
surface?
Answer:
1.1 m
point charge. How far is the 190-V surface from the 75.0-V
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
Answer:
. What is the magnitude of the electric field in the membrane?
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
35.
. Find .
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
magnitude of the potential difference
between the conductors?
Answer:
. What is the
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
distance between the two conductors. We can use this relation to find
.
SOLUTION The potential difference between the conductors is
is the
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
(b)
. Therefore,
.
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
(c)
. Therefore,
.
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
such a path, it follows that
*38.
. Therefore,
.
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
*39.
is perpendicular to the electric field and no work is done in moving a charge along
. What is the kinetic energy of the electron just as it reaches the positive plate?
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:
0V
(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
between them?
43.
of charge on its plates when a voltage of 1.5 V is applied
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
potential across its plates is
.
SOLUTION Therefore, solving Equation 19.11b for
44.
45.
46.
and
, we have
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.
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
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
Answer:
5.3
. What is the dielectric constant of the dielectric?
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
of
5.0.
and a thickness
. Assume that the membrane behaves like a parallel plate capacitor and has a dielectric constant of
(a)The potential on the outer surface of the membrane is
How much charge resides on the outer surface?
Answer:
greater than that on the inside surface.
(b)If the charge in part (a) is due to positive ions (charge
surface?
Answer:
), how many such ions are present on the outer
*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
*52.
, we have
A capacitor is constructed of two concentric conducting cylindrical shells. The radius of the inner cylindrical
shell is
charges of magnitude
, and that of the outer shell is
. When the cylinders carry equal and opposite
, 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?
A battery charger is connected to a dead battery and delivers a current of
for 5.0 hours, keeping the
3.
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
outlet. What is the current in the heating element?
5.
. This heating element is energized by a
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
Answer:
ions
, how many such ions flow in
.
?
ions
*6.A car battery has a rating of 220 ampere·hours (A·h). 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
Since
provided by the kinetic energy of the protons is
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
Since
provided by the kinetic energy of the protons is
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
Since
provided by the kinetic energy of the protons is
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
. Each piece of material is connected to a
(a)the resistance and
(b)the current in each case.
, and the unit of length in the drawing is
battery. Find
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:
and a radius of
. It carries a current of
, when a voltage of
12.A cylindrical wire has a length of
is applied across the ends of the wire. From what material in Table 20.1 is the wire made?
at
and
at
. What is the temperature coefficient of
13.A coil of wire has a resistance of
resistivity?
Answer:
of insulation-coated wire coiled around it. When the electrician
14.A large spool in an electrician's workshop has
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?
Two wires have the same length and the same resistance. One is made from aluminum and the other from
15.
copper. Obtain the ratio of the cross-sectional area of the aluminum wire to the cross-sectional area of the copper wire.
Answer:
1.64
REASONING The resistance of a metal wire of length , cross-sectional area and resistivity is given by
Equation 20.3:
. Solving for , we have
. We can use this expression to find the ratio of the
cross-sectional area of the aluminum wire to that of the copper wire.
SOLUTION Forming the ratio of the areas and using resistivity values from Table 20.1, we have
16.High-voltage power lines are a familiar sight throughout the country. The aluminum wire used for some of these lines
has a cross-sectional area of
*17.
. What is the resistance of ten kilometers of this wire?
The temperature coefficient of resistivity for the metal gold is
, and for tungsten it is
. The resistance of a gold wire increases by 7.0% due to an increase in temperature. For the same
increase in temperature, what is the percentage increase in the resistance of a tungsten wire?
Answer:
9.3%
*18.
A tungsten wire has a radius of
and is heated from 20.0 to
. The temperature coefficient of
resistivity is
. When
is applied across the ends of the hot wire, a current of
is
produced. How long is the wire? Neglect any effects due to thermal expansion of the wire.
*19.
Two wires have the same cross-sectional area and are joined end to end to form a single wire. One is tungsten,
which has a temperature coefficient of resistivity of
. The other is carbon, for which
. The total resistance of the composite wire is the sum of the resistances of the pieces. The
total resistance of the composite does not change with temperature. What is the ratio of the lengths of the tungsten and
carbon sections? Ignore any changes in length due to thermal expansion.
Answer:
70
REASONING We will ignore any changes in length due to thermal expansion. Although the resistance of
each section changes with temperature, the total resistance of the composite does not change with temperature.
Therefore,
From Equation 20.5, we know that the temperature dependence of the resistance for a wire of resistance
temperature
Since
is given by
at
, where is the temperature coefficient of resistivity. Thus,
is the same for each wire, this simplifies to
(1)
This expression can be used to find the ratio of the resistances. Once this ratio is known, we can find the ratio of the
lengths of the sections with the aid of Equation 20.3
.
SOLUTION From Equation 1, the ratio of the resistances of the two sections of the wire is
Thus, using Equation 20.3, we find the ratio of the tungsten and carbon lengths to be
where we have used resistivity values from Table 20.1 and the fact that the two sections have the same cross-sectional
areas.
*20.
Two cylindrical rods, one copper and the other iron, are identical in lengths and cross-sectional areas. They are
joined end to end to form one long rod. A
battery is connected across the free ends of the copper–iron rod.
What is the voltage between the ends of the copper rod?
**21.
A digital thermometer employs a thermistor as the temperature-sensing element. A thermistor is a kind of
semiconductor and has a large negative temperature coefficient of resistivity . Suppose that
for the thermistor in a digital thermometer used to measure the temperature of a patient. The resistance of the
thermistor decreases to 85% of its value at the normal body temperature of
. What is the patient's
temperature?
Answer:
Section 20.4 Electric Power
outlet and consumes
of power. What is the resistance of the
22.An electric blanket is connected to a
heater wire in the blanket?
. The iron is plugged into a
outlet. What is the power
23.The heating element in an iron has a resistance of
delivered to the iron?
Answer:
. The current rating of the blow-dryer is
24.A blow-dryer and a vacuum cleaner each operate with a voltage of
, and that of the vacuum cleaner is
. Determine the power consumed by
(a)the blow-dryer and
(b)the vacuum cleaner.
(c)Determine the ratio of the energy used by the blow-dryer in 15 minutes to the energy used by the vacuum
cleaner in one-half hour.
25.There are approximately 110 million households that use TVs in the United States. Each TV uses, on average,
of power and is turned on for 6.0 hours a day. If electrical energy costs $0.12 per kWh, how much money is spent
every day in keeping 110 million TVs turned on?
Answer:
, and is using
of power. Find the current being supplied by
26.An MP3 player operates with a voltage of
the player's battery.
In doing a load of clothes, a clothes dryer uses
of current at
for 45 min. A personal computer, in
27.
contrast, uses
of current at
. With the energy used by the clothes dryer, how long (in hours) could you
use this computer to “surf” the Internet?
Answer:
8.9 h
REASONING According to Equation 6.10b, the energy used is
, where is the power and is
the time. According to Equation 20.6a, the power is
, where is the current and is the voltage. Thus,
, and we apply this result first to the dryer and then to the computer.
SOLUTION The energy used by the dryer is
For the computer, we have
Solving for we find
*28.
*29.
An electric heater used to boil small amounts of water consists of a
coil that is immersed directly in
the water. It operates from a
socket. How much time is required for this heater to raise the temperature of
0.50 kg of water from
to the normal boiling point?
The rear window of a van is coated with a layer of ice at
van turns on the rear-window defroster, which operates at
. The density of ice is
. The driver of the
and
. The defroster directly heats an area of
of the rear window. What is the maximum thickness of ice coating this area that the defroster can melt in 3.0
minutes?
Answer:
*30.
A piece of Nichrome wire has a radius of
. It is used in a laboratory to make a heater that
uses
of power when connected to a voltage source of
. Ignoring the effect of temperature
on resistance, estimate the necessary length of wire.
*31.
Tungsten has a temperature coefficient of resistivity of
. A tungsten wire is connected to
a source of constant voltage via a switch. At the instant the switch is closed, the temperature of the wire is
, and the initial power delivered to the wire is
. At what wire temperature will the power that is
delivered to the wire be decreased to
Answer:
?
REASONING AND SOLUTION As a function of temperature, the resistance of the wire is given by
Equation 20.5:
Equation 20.6c, we have
, where is the temperature coefficient of resistivity. From
. Combining these two equations, we have
where
, since the voltage is constant. But
, so we find
Solving for , we find
Section
20.5
Alternating
Current
32.According to Equation 20.7, an ac voltage V is given as a function of time t by
, where
is the
peak voltage and f is the frequency (in hertz). For a frequency of 60.0 Hz, what is the smallest value of the time at
which the voltage equals one-half of the peak value?
, and the resistance of the machine is
. What are
33.The rms current in a copy machine is
(a)the average power and
Answer:
(b)the peak power delivered to the machine?
Answer:
34.
The rms current in a
resistor is
. What is the peak value of the voltage across this resistor?
(rms)
35.A 550-W space heater is designed for operation in Germany, where household electrical outlets supply
service. What is the power output of the heater when plugged into a
(rms) electrical outlet in a house in the
United States? Ignore the effects of temperature on the heater's resistance.
Answer:
36.
Review Conceptual Example 7 as an aid in solving this problem. A portable electric heater uses
of
current. The manufacturer recommends that an extension cord attached to the heater receive no more than
of
power per meter of length. What is the smallest radius of copper wire that can be used in the extension cord? (Note:
An extension cord contains two wires.)
The average power used by a stereo speaker is
. Assuming that the speaker can be treated as a
37.
resistance, find the peak value of the ac voltage applied to the speaker.
Answer:
21 V
REASONING The average power is given by Equation 20.15c as
. In this expression the rms
voltage
appears. However, we seek the peak voltage
. The relation between the two types of voltage is given
by Equation 20.13 as
Equation 20.15c.
SOLUTION Substituting
, so we can obtain the peak voltage by using Equation 20.13 to substitute into
from Equation 20.13 into Equation 20.15c gives
Solving for the peak voltage
gives
*38.
The recovery time of a hot water heater is the time required to heat all the water in the unit to the desired
temperature. Suppose that a 52-gal
unit starts with cold water at
and delivers
hot water at
. The unit is electric and utilizes a resistance heater (120 V ac,
) to heat the water.
Assuming that no heat is lost to the environment, determine the recovery time (in hours) of the unit.
*39.
A light bulb is connected to a
wall socket. The current in the bulb depends on the time t according to
the relation
.
(a)What is the frequency f of the alternating current?
Answer:
50.0 Hz
REASONING
(a) We can obtain the frequency of the alternating current by comparing this specific expression for the
current with the more general one in Equation 20.8.
(b)The resistance of the light bulb is, according to Equation 20.14, equal to the rms-voltage divided by
the rms-current. The rms-voltage is given, and we can obtain the rms-current by dividing the peak
current by
, as expressed by Equation 20.12.
(c) The average power is given by Equation 20.15a as the product of the rms-current and the rms-voltage.
SOLUTION
(a) By comparing
with the general expression (see Equation 20.8) for
the current in an ac circuit,
, we see that
(b)The resistance is equal to
, where the rms-current is related to the peak current
by
. Thus, the resistance of the light bulb is
(20.14)
(c) The average power is the product of the rms-current and rms-voltage:
(20.15a)
(b)Determine the resistance of the bulb's filament.
Answer:
REASONING
(a) We can obtain the frequency of the alternating current by comparing this specific expression for the
current with the more general one in Equation 20.8.
(b)The resistance of the light bulb is, according to Equation 20.14, equal to the rms-voltage divided by
the rms-current. The rms-voltage is given, and we can obtain the rms-current by dividing the peak
current by
, as expressed by Equation 20.12.
(c) The average power is given by Equation 20.15a as the product of the rms-current and the rms-voltage.
SOLUTION
(a) By comparing
with the general expression (see Equation 20.8) for
the current in an ac circuit,
, we see that
(b)The resistance is equal to
, where the rms-current is related to the peak current
by
. Thus, the resistance of the light bulb is
(20.14)
(c) The average power is the product of the rms-current and rms-voltage:
(20.15a)
(c)What is the average power delivered to the light bulb?
Answer:
REASONING
(a) We can obtain the frequency of the alternating current by comparing this specific expression for the
current with the more general one in Equation 20.8.
(b)The resistance of the light bulb is, according to Equation 20.14, equal to the rms-voltage divided by
the rms-current. The rms-voltage is given, and we can obtain the rms-current by dividing the peak
current by
, as expressed by Equation 20.12.
(c) The average power is given by Equation 20.15a as the product of the rms-current and the rms-voltage.
SOLUTION
(a) By comparing
with the general expression (see Equation 20.8) for
the current in an ac circuit,
, we see that
(b)The resistance is equal to
, where the rms-current is related to the peak current
by
. Thus, the resistance of the light bulb is
(20.14)
(c) The average power is the product of the rms-current and rms-voltage:
(20.15a)
REASONING
(a) We can obtain the frequency of the alternating current by comparing this specific expression for the current
with the more general one in Equation 20.8.
(b)The resistance of the light bulb is, according to Equation 20.14, equal to the rms-voltage divided by the rmscurrent. The rms-voltage is given, and we can obtain the rms-current by dividing the peak current by
, as
expressed by Equation 20.12.
(c) The average power is given by Equation 20.15a as the product of the rms-current and the rms-voltage.
SOLUTION
(a) By comparing
with the general expression (see Equation 20.8) for the current
in an ac circuit,
, we see that
(b)The resistance is equal to
, where the rms-current is related to the peak current
by
. Thus, the resistance of the light bulb is
(20.14)
(c) The average power is the product of the rms-current and rms-voltage:
(20.15a)
**40.To save on heating costs, the owner of a greenhouse keeps 660 kg of water around in barrels. During a winter
day, the water is heated by the sun to
. During the night the water freezes into ice at
in nine
hours. What is the minimum ampere rating of an electric heating system (240 V) that would provide the same
heating effect as the water does?
Section
20.6
Series
Wiring
41.
Three resistors, 25, 45, and
, are connected in series, and a 0.51-A current passes through them. What are
(a)the equivalent resistance and
Answer:
REASONING The equivalent series resistance is the sum of the resistances of the three resistors.
The potential difference can be determined from Ohm's law according to
.
SOLUTION
(a) The equivalent resistance is
(b)The potential difference across the three resistors is
(b)the potential difference across the three resistors?
Answer:
74 V
REASONING The equivalent series resistance is the sum of the resistances of the three resistors.
The potential difference can be determined from Ohm's law according to
.
SOLUTION
(a) The equivalent resistance is
(b)The potential difference across the three resistors is
REASONING The equivalent series resistance is the sum of the resistances of the three resistors. The
potential difference can be determined from Ohm's law according to
.
SOLUTION
(a) The equivalent resistance is
(b)The potential difference across the three resistors is
42.
A 60.0-W lamp is placed in series with a resistor and a
what is the resistance R of the resistor?
The current in a series circuit is
. When an additional
43.
drops to
. What is the resistance in the original circuit?
Answer:
32 V
source. If the voltage across the lamp is
,
resistor is inserted in series, the current
REASONING Using Ohm's law (Equation 20.2) we can write an expression for the voltage across the
original circuit as
. When the additional resistor is inserted in series, assuming that the battery remains the
same, the voltage across the new combination is given by
write
. This expression can be solved for
SOLUTION Solving for , we have
. Since
is the same in both cases, we can
.
Therefore, we find that
44.
Multiple-Concept Example 9 discusses the physics principles used in this problem. Three resistors, 2.0, 4.0,
and
, are connected in series across a
battery. Find the power delivered to each resistor.
resistor is
. This resistor is in series with a
resistor, and the series
45.The current in a
combination is connected across a battery. What is the battery voltage?
Answer:
9.0 V
*46.
Multiple-Concept Example 9 reviews the concepts that are important to this problem. A light bulb is wired in
series with a
resistor, and they are connected across a
source. The power delivered to the light
bulb is
. What is the resistance of the light bulb? Note that there are two possible answers.
*47.
Three resistors are connected in series across a battery. The value of each resistance and its maximum
power rating are as follows:
and
,
and
, and
and
.
(a)What is the greatest voltage that the battery can have without one of the resistors burning up?
Answer:
15.5 V
REASONING
(a) The greatest voltage for the battery is the voltage that generates the maximum current that the circuit
can tolerate. Once this maximum current is known, the voltage can be calculated according to Ohm's
law, as the current times the equivalent circuit resistance for the three resistors in series. To determine
the maximum current we note that the power dissipated in each resistance is
according to
Equation 20.6b. Since the power rating and resistance are known for each resistor, the maximum
current that can be tolerated by a resistor is
. By examining this maximum current for
each resistor, we will be able to identify the maximum current that the circuit can tolerate.
(b)The battery delivers power to the circuit that is given by the battery voltage times the current,
according to Equation 20.6a.
SOLUTION
(a) Solving Equation 20.6b for the current, we find that the maximum current for each resistor is as
follows:
The smallest of these three values is 0.913 A and is the maximum current that the circuit can tolerate.
Since the resistors are connected in series, the equivalent resistance of the circuit is
Using Ohm's law with this equivalent resistance and the maximum current of 0.913 A reveals that the
maximum battery voltage is
(b)The power delivered by the battery in part (a) is given by Equation 20.6a as
(b)How much power does the battery deliver to the circuit in (a)?
Answer:
REASONING
(a) The greatest voltage for the battery is the voltage that generates the maximum current that the circuit
can tolerate. Once this maximum current is known, the voltage can be calculated according to Ohm's
law, as the current times the equivalent circuit resistance for the three resistors in series. To determine
the maximum current we note that the power dissipated in each resistance is
according to
Equation 20.6b. Since the power rating and resistance are known for each resistor, the maximum
current that can be tolerated by a resistor is
. By examining this maximum current for
each resistor, we will be able to identify the maximum current that the circuit can tolerate.
(b)The battery delivers power to the circuit that is given by the battery voltage times the current,
according to Equation 20.6a.
SOLUTION
(a) Solving Equation 20.6b for the current, we find that the maximum current for each resistor is as
follows:
The smallest of these three values is 0.913 A and is the maximum current that the circuit can tolerate.
Since the resistors are connected in series, the equivalent resistance of the circuit is
Using Ohm's law with this equivalent resistance and the maximum current of 0.913 A reveals that the
maximum battery voltage is
(b)The power delivered by the battery in part (a) is given by Equation 20.6a as
REASONING
The
greatest voltage for the battery is the voltage that generates the maximum current that the circuit can
(a)
tolerate. Once this maximum current is known, the voltage can be calculated according to Ohm's law, as the
current times the equivalent circuit resistance for the three resistors in series. To determine the maximum
current we note that the power dissipated in each resistance is
according to Equation 20.6b. Since
the power rating and resistance are known for each resistor, the maximum current that can be tolerated by a
resistor is
. By examining this maximum current for each resistor, we will be able to identify the
maximum current that the circuit can tolerate.
(b)The battery delivers power to the circuit that is given by the battery voltage times the current, according to
Equation 20.6a.
SOLUTION
(a) Solving Equation 20.6b for the current, we find that the maximum current for each resistor is as follows:
The smallest of these three values is 0.913 A and is the maximum current that the circuit can tolerate. Since the
resistors are connected in series, the equivalent resistance of the circuit is
Using Ohm's law with this equivalent resistance and the maximum current of 0.913 A reveals that the
maximum battery voltage is
(b)The power delivered by the battery in part (a) is given by Equation 20.6a as
*48.
One heater uses
of power when connected by itself to a battery. Another heater uses
of power
when connected by itself to the same battery. How much total power do the heaters use when they are both connected
in series across the battery?
**49.Two resistances, and , are connected in series across a
battery. The current increases by
when is removed, leaving connected across the battery. However, the current increases by just
when
is removed, leaving connected across the battery. Find
(a) and
Answer:
(b) .
Answer:
Section
20.7
Parallel
Wiring
50.A coffee-maker
and a toaster
are connected in parallel to the same
much total power is supplied to the two appliances when both are turned on?
outlet in a kitchen. How
51.For the three-way bulb
discussed in Conceptual Example 11, find the resistance of each of
the two filaments. Assume that the wattage ratings are not limited by significant figures, and ignore any heating
effects on the resistances.
Answer:
(50.0-W filament) and
(100.0-W filament)
52.
53.
The drawing shows three different resistors in two different circuits. The battery has a voltage of
and the resistors have values of
,
, and
.
(a)For the circuit on the left, determine the current through and the voltage across each resistor.
(b)Repeat part (a) for the circuit on the right.
,
The drawing shows a circuit that contains a battery, two resistors, and a switch. What is the equivalent resistance
of the circuit when the switch is
(a)open and
Answer:
REASONING When the switch is open, no current goes to the resistor . Current exists only in
, so it is the equivalent resistance. When the switch is closed, current is sent to both resistors. Since they are
wired in parallel, we can use Equation 20.17 to find the equivalent resistance. Whether the switch is open or
closed, the power delivered to the circuit can be found from the relation
(Equation 20.6c),
where is the battery voltage and is the equivalent resistance.
SOLUTION
(a) When the switch is open, there is current only in resistor . Thus, the equivalent resistance is
.
(b)When the switch is closed, there is current in both resistors and, furthermore, they are wired in
parallel. The equivalent resistance is
(20.17)
(c) When the switch is open, the power delivered to the circuit by the battery is given by
since the only resistance in the circuit is
,
. Thus, the power is
(20.6)
(d)When the switch is closed, the power delivered to the circuit is
equivalent resistance of the two resistors wired in parallel:
, where
is the
(20.6)
(b)closed? What is the total power delivered to the resistors when the switch is
Answer:
REASONING When the switch is open, no current goes to the resistor . Current exists only in
, so it is the equivalent resistance. When the switch is closed, current is sent to both resistors. Since they are
wired in parallel, we can use Equation 20.17 to find the equivalent resistance. Whether the switch is open or
closed, the power delivered to the circuit can be found from the relation
(Equation 20.6c),
where is the battery voltage and is the equivalent resistance.
SOLUTION
(a) When the switch is open, there is current only in resistor . Thus, the equivalent resistance is
.
When
the
switch
is closed, there is current in both resistors and, furthermore, they are wired in
(b)
parallel. The equivalent resistance is
(20.17)
(c) When the switch is open, the power delivered to the circuit by the battery is given by
since the only resistance in the circuit is
,
. Thus, the power is
(20.6)
(d)When the switch is closed, the power delivered to the circuit is
equivalent resistance of the two resistors wired in parallel:
, where
is the
(20.6)
(c)open and
Answer:
REASONING When the switch is open, no current goes to the resistor . Current exists only in
, so it is the equivalent resistance. When the switch is closed, current is sent to both resistors. Since they are
wired in parallel, we can use Equation 20.17 to find the equivalent resistance. Whether the switch is open or
closed, the power delivered to the circuit can be found from the relation
(Equation 20.6c),
where is the battery voltage and is the equivalent resistance.
SOLUTION
(a) When the switch is open, there is current only in resistor . Thus, the equivalent resistance is
.
(b)When the switch is closed, there is current in both resistors and, furthermore, they are wired in
parallel. The equivalent resistance is
(20.17)
(c) When the switch is open, the power delivered to the circuit by the battery is given by
since the only resistance in the circuit is
,
. Thus, the power is
(20.6)
(d)When the switch is closed, the power delivered to the circuit is
equivalent resistance of the two resistors wired in parallel:
, where
is the
(20.6)
(d)closed?
Answer:
REASONING When the switch is open, no current goes to the resistor . Current exists only in
, so it is the equivalent resistance. When the switch is closed, current is sent to both resistors. Since they are
wired in parallel, we can use Equation 20.17 to find the equivalent resistance. Whether the switch is open or
closed, the power delivered to the circuit can be found from the relation
(Equation 20.6c),
where is the battery voltage and is the equivalent resistance.
SOLUTION
(a) When the switch is open, there is current only in resistor . Thus, the equivalent resistance is
.
When
the
switch
is closed, there is current in both resistors and, furthermore, they are wired in
(b)
parallel. The equivalent resistance is
(20.17)
(c) When the switch is open, the power delivered to the circuit by the battery is given by
since the only resistance in the circuit is
,
. Thus, the power is
(20.6)
(d)When the switch is closed, the power delivered to the circuit is
equivalent resistance of the two resistors wired in parallel:
, where
is the
(20.6)
REASONING When the switch is open, no current goes to the resistor . Current exists only in , so it is
the equivalent resistance. When the switch is closed, current is sent to both resistors. Since they are wired in parallel,
we can use Equation 20.17 to find the equivalent resistance. Whether the switch is open or closed, the power
delivered to the circuit can be found from the relation
is the equivalent resistance.
SOLUTION
(a) When the switch is open, there is current only in resistor
(Equation 20.6c), where
is the battery voltage and
. Thus, the equivalent resistance is
.
(b)When the switch is closed, there is current in both resistors and, furthermore, they are wired in parallel. The
equivalent resistance is
(20.17)
(c) When the switch is open, the power delivered to the circuit by the battery is given by
only resistance in the circuit is
, since the
. Thus, the power is
(20.6)
(d)When the switch is closed, the power delivered to the circuit is
resistance of the two resistors wired in parallel:
, where
is the equivalent
(20.6)
loudspeaker, an
loudspeaker, and a
loudspeaker are connected in parallel across the
54.A
terminals of an amplifier. Determine the equivalent resistance of the three speakers, assuming that they all behave as
resistors.
Two resistors, 42.0 and
, are connected in parallel. The current through the
resistor is
55.
.
(a)Determine the current in the other resistor.
Answer:
REASONING Since the resistors are connected in parallel, the voltage across each one is the same
and can be calculated from Ohm's Law (Equation 20.2:
). Once the voltage across each resistor is
known, Ohm's law can again be used to find the current in the second resistor. The total power consumed by
the parallel combination can be found calculating the power consumed by each resistor from Equation
20.6b:
. Then, the total power consumed is the sum of the power consumed by each resistor.
SOLUTION Using data for the second resistor, the voltage across the resistors is equal to
(a) The current through the 42.0-
resistor is
(b)The power consumed by the 42.0-
resistor is
while the power consumed by the 64.0-
resistor is
Therefore the total power consumed by the two resistors is
(b)What is the total power supplied to the two resistors?
Answer:
.
REASONING Since the resistors are connected in parallel, the voltage across each one is the same
and can be calculated from Ohm's Law (Equation 20.2:
). Once the voltage across each resistor is
known, Ohm's law can again be used to find the current in the second resistor. The total power consumed by
the parallel combination can be found calculating the power consumed by each resistor from Equation
20.6b:
. Then, the total power consumed is the sum of the power consumed by each resistor.
SOLUTION Using data for the second resistor, the voltage across the resistors is equal to
(a) The current through the 42.0-
resistor is
(b)The power consumed by the 42.0-
resistor is
while the power consumed by the 64.0-
resistor is
Therefore the total power consumed by the two resistors is
.
REASONING Since the resistors are connected in parallel, the voltage across each one is the same and can be
calculated from Ohm's Law (Equation 20.2:
). Once the voltage across each resistor is known, Ohm's law can
again be used to find the current in the second resistor. The total power consumed by the parallel combination can be
found calculating the power consumed by each resistor from Equation 20.6b:
. Then, the total power
consumed is the sum of the power consumed by each resistor.
SOLUTION Using data for the second resistor, the voltage across the resistors is equal to
(a) The current through the 42.0-
resistor is
(b)The power consumed by the 42.0-
resistor is
while the power consumed by the 64.0-
resistor is
Therefore the total power consumed by the two resistors is
.
56.
Two identical resistors are connected in parallel across a
battery, which supplies them with a total
power of
. While the battery is still connected, one of the resistors is heated so that its resistance doubles. The
resistance of the other resistor remains unchanged. Find
(a)the initial resistance of each resistor and
(b)the total power delivered to the resistors after one resistor has been heated.
A coffee cup heater and a lamp are connected in parallel to the same
outlet. Together, they use a total
57.
of
of power. The resistance of the heater is
Answer:
. Find the resistance of the lamp.
*58.Two resistors have resistances and . When the resistors are connected in series to a
battery, the current
from the battery is
. When the resistors are connected in parallel to the battery, the total current from the
battery is
. Determine and .
*59.
A cylindrical aluminum pipe of length
has an inner radius of
and an outer radius of
. The interior of the pipe is completely filled with copper. What is the resistance of this unit?
(Hint: Imagine that the pipe is connected between the terminals of a battery and decide whether the aluminum and
copper parts of the pipe are in series or in parallel.)
Answer:
REASONING AND SOLUTION The aluminum and copper portions may be viewed a being connected in
parallel since the same voltage appears across them. Using a and b to denote the inner and outer radii, respectively,
and using Equation 20.3 to express the resistance for each portion, we find for the equivalent resistance that
We have taken resistivity values for copper and aluminum from Table 20.1
*60.
The drawing shows two circuits, and the same battery is used in each. The two resistances
in circuit A are the
same, and the two resistances
in circuit B are the same. Knowing that the same total power is delivered in each
circuit, find the ratio
for the circuits.
**61.
The rear window defogger of a car consists of thirteen thin wires
glass. The wires are connected in parallel to the
battery, and each has a length of
embedded in the
. The defogger can
melt
of ice at
into water at
in two minutes. Assume that all the power delivered to the
wires is used immediately to melt the ice. Find the cross-sectional area of each wire.
Answer:
Section 20.8 Circuits Wired Partially in Series and Partially in Parallel
62.A
with a
resistor is connected in parallel with a
resistor. This parallel group is connected in series
resistor. The total combination is connected across a
battery. Find
(a)the current and
resistor.
(b)the power delivered to the
A
coffee
maker
and
a
frying
pan are connected in series across a
source of voltage.
63.
A
bread maker is also connected across the
source and is in parallel with the series combination.
Find the total current supplied by the source of voltage.
Answer:
REASONING To find the current, we will use Ohm's law, together with the proper equivalent resistance. The
coffee maker and frying pan are in series, so their equivalent resistance is given by Equation 20.16 as
. This total resistance is in parallel with the resistance of the bread maker, so the equivalent resistance of the parallel
combination can be obtained from Equation 20.17 as
SOLUTION Using Ohm's law and the expression developed above for
64.
Find the equivalent resistance between points A and B in the drawing.
.
, we find
65.
Determine the equivalent resistance between the points A and B for the group of resistors in the drawing.
Answer:
REASONING When two or more resistors are in series, the equivalent resistance is given by Equation 20.16:
. Likewise, when resistors are in parallel, the expression to be solved to find the equivalent
resistance is given by Equation 20.17:
. We will successively apply these to the
individual resistors in the figure in the text beginning with the resistors on the right side of the figure.
SOLUTION Since the 4.0- and the 6.0- resistors are in series, the equivalent resistance of the combination of
those two resistors is
. The 9.0- and 8.0- resistors are in parallel; their equivalent resistance is
. The
equivalent resistances of the parallel combination (
and
) and the series combination (
and the
)
are in parallel; therefore, their equivalent resistance is
. The 2.98- combination is in series with the 3.0resistor, so that equivalent resistance is
. Finally, the 5.98- combination and the 20.0- resistor are in
parallel, so the equivalent resistance between the points and is
66.
.
The circuit in the drawing contains three identical resistors. Each resistor has a value of
equivalent resistance between the points a and b, b and c, and a and c.
67.Find the equivalent resistance between the points A and B in the drawing.
Answer:
. Determine the
*68.
Each resistor in the three circuits in the drawing has the same resistance R, and the batteries have the same
voltage V. The values for R and V are
and
, respectively. Determine the total power delivered by the
battery in each of the three circuits.
*69.Eight different values of resistance can be obtained by connecting together three resistors (1.00, 2.00, and
all possible ways. What are the values?
Answer:
,
,
,
,
,
,
,
*70.
Determine the power supplied to each of the resistors in the drawing.
*71.
The circuit in the drawing contains five identical resistors. The
circuit. What is the resistance R of each resistor?
battery delivers
) in
of power to the
Answer:
REASONING The power delivered to the circuit is, according to Equation 20.6c,
, where
is the voltage of the battery and
is the equivalent resistance of the five-resistor circuit. The voltage and power
are known, so that the equivalent resistance can be calculated. We will use our knowledge of resistors wired in series
and parallel to evaluate
in terms of the resistance of each resistor. In this manner we will find the value for .
SOLUTION First we note that all the resistors are equal, so
. We can find the
equivalent resistance
as follows. The resistors and are in series, so the equivalent resistance
of these
two is
. The resistors ,
, and are in parallel, and the reciprocal of the equivalent
resistance
is
so
. The resistor is in series with
equivalent resistance of the circuit. Thus, we have
, and the equivalent resistance of this combination is the
The power delivered to the circuit is
Solving for the resistance , we find that
**72.The circuit shown in the drawing is constructed with six identical resistors and an ideal battery. When the
resistor is removed from the circuit, the current in the battery decreases by
. Determine the resistance
of each resistor.
Section
20.9
Internal
Resistance
73.
A battery has an internal resistance of
. A number of identical light bulbs, each with a resistance of
, are connected in parallel across the battery terminals. The terminal voltage of the battery is observed to be one-half
the emf of the battery. How many bulbs are connected?
Answer:
30 bulbs
REASONING The terminal voltage of the battery is given by
, where is the internal
resistance of the battery. Since the terminal voltage is observed to be one-half of the emf of the battery, we have
and
. From Ohm's law, the equivalent resistance of the circuit is
.
We can also find the equivalent resistance of the circuit by considering that the identical bulbs are in parallel across
the battery terminals, so that the equivalent resistance of the bulbs is found from
This equivalent resistance is in series with the battery, so we find that the equivalent resistance of the circuit is
This expression can be solved for .
SOLUTION Solving the above expression for
, we have
resistor is connected across a
battery. The voltage between the terminals of the battery is
74.A
observed to be only
. Find the internal resistance of the battery.
of power to the bulb. A
75.When a light bulb is connected across the terminals of a battery, the battery delivers
voltage of
exists between the terminals of the battery, which has an internal resistance of
. What is the
emf of the battery?
Answer:
12.0 V
and an emf of
. What is the maximum current that can be
76.A battery has an internal resistance of
drawn from the battery without the terminal voltage dropping below
?
*77.
A battery delivering a current of
to a circuit has a terminal voltage of
. The electric power being
dissipated by the internal resistance of the battery is
. Find the emf of the battery.
Answer:
24.0 V
*78.
When a “dry-cell” flashlight battery with an internal resistance of
is connected to a
light
bulb, the bulb shines dimly. However, when a lead-acid “wet-cell” battery with an internal resistance of
is
connected, the bulb is noticeably brighter. Both batteries have the same emf. Find the ratio
of the power
delivered to the bulb by the wet-cell battery to the power delivered by the dry-cell battery.
Section 20.10 Kirchhoff's Rules
79.
Consider the circuit in the drawing. Determine
(a)the magnitude of the current in the circuit and
Answer:
REASONING The current can be found by using Kirchhoff's loop rule.
Once the current is known, the voltage between points and can be determined.
SOLUTION
(a) We assume that the current is directed clockwise around the circuit. Starting
at the upper left corner and going clockwise around the circuit, we set the
potential drops equal to the potential rises:
Solving for the current gives
(b)The voltage between points and is
.
(c)
is at the higher potential.
(b)the magnitude of the voltage between the points labeled A and B.
Answer:
REASONING The current can be found by using Kirchhoff's loop rule. Once the current is known,
the voltage between points and can be determined.
SOLUTION
(a) We assume that the current is directed clockwise around the circuit. Starting at the upper left corner
and going clockwise around the circuit, we set the potential drops equal to the potential rises:
Solving for the current gives
(b)The voltage between points and is
.
(c)
is at the higher potential.
(c)State which point, A or B, is at the higher potential.
Answer:
Point B is at the higher potential.
REASONING The current can be found by using Kirchhoff's loop rule. Once the current is known,
the voltage between points and can be determined.
SOLUTION
(a) We assume that the current is directed clockwise around the circuit. Starting at the upper left corner
and going clockwise around the circuit, we set the potential drops equal to the potential rises:
Solving for the current gives
(b)The voltage between points and is
(c)
.
is at the higher potential.
REASONING The current can be found by using Kirchhoff's loop rule. Once the current is known, the
voltage between points and can be determined.
SOLUTION
(a) We assume that the current is directed clockwise around the circuit. Starting at the upper left corner and going
clockwise around the circuit, we set the potential drops equal to the potential rises:
Solving for the current gives
(b)The voltage between points and is
(c)
80.
81.
.
is at the higher potential.
The drawing shows a portion of a larger circuit. Current flows left to right in each resistor. What is the current in
the resistor R?
Find the magnitude and the direction of the current in the
resistor in the drawing.
Answer:
to the left
82.
Using Kirchhoff's loop rule, find the value of the current I in part c of the drawing, where
. (Note:
Parts a and b of the drawing are used in the online tutorial help that is provided for this problem in the WileyPLUS
homework management program.)
83.Determine the current (both magnitude and direction) in the 8.0- and
Answer:
(left to right) in the
*84.
resistor,
Determine the voltage across the
potential?
*85.
(left to right) in the
resistor
resistor in the drawing. Which end of the resistor is at the higher
Find the current in the
Answer:
(downward) in the
resistors in the drawing.
resistor in the drawing. Specify the direction of the current.
resistor
REASONING In preparation for applying Kirchhoff's rules, we now choose the currents in each
resistor. The directions of the currents are arbitrary, and should they be incorrect, the currents will turn out to
be negative quantities. Having chosen the currents, we also mark the ends of the resistors with the plus and
minus signs that indicate that the currents are directed from higher
toward lower
plus and minus signs will guide us when we apply Kirchhoff's loop rule.
potential. These
SOLUTION Applying the junction rule to junction B, we find
(1)
Applying the loop rule to loop ABCD (going clockwise around the loop), we obtain
(2)
Applying the loop rule to loop BEFC (going clockwise around the loop), we obtain
(3)
Substituting
from Equation 1 into Equation 3 gives
(4)
Solving Equation 2 for
gives
This result may be substituted into Equation 4 to show that
The minus sign indicates that
, rather than
upward as selected arbitrarily in the drawing.
**86.None of the resistors in the circuit shown in the drawing is connected in series or in parallel with one another.
Find
(a)the current
(b)
(c)
and the resistances
and
.
Section 20.11
The
Measurement
of Current
and Voltage
87.
The coil of a galvanometer has a resistance of
, and its meter deflects full scale when a current of
passes through it. To make the galvanometer into a nondigital ammeter, a 24.8-mΩ shunt resistor is added to
it. What is the maximum current that this ammeter can read?
Answer:
REASONING As discussed in Section 20.11, some of the current (6.20 mA) goes directly through the
galvanometer and the remainder goes through the shunt resistor. Since the resistance of the coil
and the shunt
resistor are in parallel, the voltage across each is the same. We will use this fact to determine how much current goes
through the shunt resistor. This value, plus the 6.20 mA that goes through the galvanometer, is the maximum current
that this ammeter can read.
SOLUTION The voltage across the coil resistance is equal to the voltage across the shunt resistor, so
So
. The maximum current is
.
88.The coil of wire in a galvanometer has a resistance of
. The galvanometer exhibits a full-scale deflection
when the current through it is
. A resistor is connected in series with this combination so as to produce a
nondigital voltmeter. The voltmeter is to have a full-scale deflection when it measures a potential difference of
. What is the resistance of this resistor?
89.
Nondigital voltmeter A has an equivalent resistance of
and a full-scale voltage of
. Nondigital
voltmeter B, using the same galvanometer as voltmeter A, has an equivalent resistance of
. What is its
full-scale voltage?
Answer:
30.0 V
is used with a shunt resistor to make a nondigital ammeter that has an
90.A galvanometer with a coil resistance of
equivalent resistance of
. The current in the shunt resistor is
when the galvanometer reads full scale.
Find the full-scale current of the galvanometer.
*91.
Two scales on a nondigital voltmeter measure voltages up to 20.0 and
, respectively. The resistance
connected in series with the galvanometer is
for the
scale and
for the
scale.
Determine the coil resistance and the full-scale current of the galvanometer that is used in the voltmeter.
Answer:
and
REASONING AND SOLUTION For the 20.0 V scale
For the 30.0 V scale
Subtracting and rearranging yields
Substituting this value into either of the equations for
or
gives
.
**92.In measuring a voltage, a voltmeter uses some current from the circuit. Consequently, the voltage measured is only an
approximation to the voltage present when the voltmeter is not connected. Consider a circuit consisting of two
resistors connected in series across a
battery.
(a)Find the voltage across one of the resistors.
and uses a galvanometer with a full-scale
(b)A nondigital voltmeter has a full-scale voltage of
deflection of
. Determine the voltage that this voltmeter registers when it is connected across the
resistor used in part (a).
Section 20.12 Capacitors in Series and in Parallel
93.Two capacitors are connected in parallel across the terminals of a battery. One has a capacitance of
and the
other a capacitance of
. These two capacitors together store
of charge. What is the voltage of the
battery?
Answer:
9.0 V
94.Three parallel plate capacitors are connected in series. These capacitors have identical geometries. However, they are
filled with three different materials. The dielectric constants of these materials are 3.30, 5.40, and 6.70. It is desired to
replace this series combination with a single parallel plate capacitor. Assuming that this single capacitor has the same
geometry as each of the other three capacitors, determine the dielectric constant of the material with which it is filled.
95.
Three capacitors are connected in series. The equivalent capacitance of this combination is 3.00
. Two of the
individual capacitances are 6.00
and 9.00
. What is the third capacitance (in
)?
Answer:
REASONING The equivalent capacitance
of a set of three capacitors connected in series is given by
(Equation 20.19). In this case, we know that the equivalent capacitance is
and the capacitances of two of the individual capacitors in this series combination are
and
.We will use Equation 20.19 to determine the remaining capacitance
.
SOLUTION Solving Equation 20.19 for
we obtain
Therefore, the third capacitance is
96.
Two capacitors are connected to a battery. The battery voltage is
, and the capacitances are
and
. Determine the total energy stored by the two capacitors when they are wired
(a)in parallel and
(b)in series.
,
97.Determine the equivalent capacitance between A and B for the group of capacitors in the drawing.
Answer:
98.
A
and a
capacitor are connected to a
battery. What is the total charge
supplied to the capacitors when they are wired
(a)in parallel and
(b)in series with each other?
99.Suppose that two capacitors (
and
) are connected in series. Show that the sum of the energies stored in these
capacitors is equal to the energy stored in the equivalent capacitor. [Hint: The energy stored in a capacitor can be
expressed as
.]
Answer:
The answer is a proof.
*100.A
and a
capacitor are connected in series across a
battery. A
capacitor
is then connected in parallel across the
capacitor. Determine the voltage across the
capacitor.
*101.
A
and a
capacitor are connected in series across a
battery. What voltage is
required to charge a parallel combination of the two capacitors to the same total energy?
Answer:
11 V
REASONING When two or more capacitors are in series, the equivalent capacitance of the combination
can be obtained from Equation 20.19,
. Equation 20.18 gives the equivalent
capacitance for two or more capacitors in parallel:
is given by
where
. The energy stored in a capacitor
, according to Equation 19.11. Thus, the energy stored in the series combination is
Similarly, the energy stored in the parallel combination is
,
where
The voltage required to charge the parallel combination of the two capacitors to the same total energy as the
series combination can be found by equating the two energy expressions and solving for
.
SOLUTION Equating the two expressions for the energy, we have
Solving for
, we obtain the result
**102.The drawing shows two capacitors that are fully charged
. The switch is closed, and charge flows until
equilibrium is reestablished (i.e., until both capacitors have the same voltage across their plates). Find the
resulting voltage across either capacitor.
Section
20.13
RC
Circuits
103.
In a heart pacemaker, a pulse is delivered to the heart 81 times per minute. The capacitor that controls this
pulsing rate discharges through a resistance of
. One pulse is delivered every time the fully charged
capacitor loses 63.2% of its original charge. What is the capacitance of the capacitor?
Answer:
REASONING The charge on a discharging capacitor in a RC circuit is given by Equation 20.22:
, where is the original charge at time
. Once (time for one pulse) and the ratio
known, this expression can be solved for .
SOLUTION Since the pacemaker delivers 81 pulses per minute, the time for one pulse is
are
Since one pulse is delivered every time the fully-charged capacitor loses 63.2% of its original charge, the charge
remaining is 36.8% of the original charge. Thus, we have
From Equation 20.22, we have
, or
.
Taking the natural logarithm of both sides, we have,
Solving for
, we find
104.A circuit contains a resistor in series with a capacitor, the series combination being connected across the terminals of a
battery, as in Figure 20.37a. The time constant for charging the capacitor is 1.5 s when the resistor has a resistance of
. What would the time constant be if the resistance had a value of
105.
?
The circuit in the drawing contains two resistors and two capacitors that are connected to a battery via a switch.
When the switch is closed, the capacitors begin to charge up. What is the time constant for the charging process?
Problem 105
Answer:
*106.
How many time constants must elapse before a capacitor in a series RC circuit is charged to 80.0% of its
equilibrium charge?
*107.
Four identical capacitors are connected with a resistor in two different ways. When they are connected as in
part a of the drawing, the time constant to charge up this circuit is 0.72 s. What is the time constant when they are
connected with the same resistor, as in part b?
Answer:
0.29 s
Copyright © 2012 John Wiley & Sons, Inc. All rights reserved.
Additional
Problems
72.In a certain
region, the
earth's magneti
field has a
magnitude of
and is directed
north at an
angle of
below the
horizontal. An
electrically
charged bullet
fired north and
above the
horizontal, with
a speed of
. The
magnetic force
on the bullet is
, directed due
east. Determine
the bullet's
electric charge
including its
algebraic sign
or
).
73.
An electron is moving through a magnetic field whose magnitude is
only a magnetic force and has an acceleration of magnitude
. Determine the angle (less than
. The electron experiences
. At a certain instant, it has a speed of
) between the electron's velocity and the magnetic field.
Answer:
REASONING The angle between the electron's velocity and the magnetic field can be found from Equation
21.1,
According to Newton's second law, the magnitude
the magnitude of its acceleration,
.
SOLUTION The angle is
74.
75.
of the force is equal to the product of the electron's mass
and
A very long, straight wire carries a current of
. This wire is tangent to a single-turn, circular wire loop that
also carries a current. The directions of the currents are such that the net magnetic field at the center of the loop is
zero. Both wires are insulated and have diameters that can be neglected. How much current is there in the loop?
The maximum torque experienced by a coil in a 0.75-T magnetic field is
. The coil is circular
and consists of only one turn. The current in the coil is
. What is the length of the wire from which the coil is
made?
Answer:
REASONING According to Equation 21.4, the maximum torque is
, where
is the number of
turns in the coil, is the current,
is the area of the circular coil, and is the magnitude of the magnetic field.
Since the coil contains only one turn, the length of the wire is the circumference of the circle, so that
or
. Since
, , and are known we can solve for .
SOLUTION According to Equation 21.4 and the fact that
, we have
Solving this result for gives
76.Multiple-Concept Example 7 discusses how problems like this one can be solved. A
with a speed of
charge is moving
parallel to a very long, straight wire. The wire is 5.00 cm from the charge and carries
a current of
on the charge.
77.
78.
in a direction opposite to that of the moving charge. Find the magnitude and direction of the force
The x, y, and z components of a magnetic field are
wire is oriented along the z axis and carries a current of
on this wire?
Answer:
0.19 N
,
, and
. A 25-cm
. What is the magnitude of the magnetic force that acts
In a lightning bolt, a large amount of charge flows during a time of
. Assume that the bolt can be treated
as a long, straight line of current. At a perpendicular distance of
from the bolt, a magnetic field of
is measured. How much charge has flowed during the lightning bolt? Ignore the earth's magnetic field.
79.
80.
A charge is moving perpendicular to a magnetic field and experiences a force whose magnitude is
. If
this same charge were to move at the same speed and the angle between its velocity and the same magnetic field were
, what would be the magnitude of the magnetic force that the charge would experience?
Answer:
The drawing shows four insulated wires overlapping one another, forming a square with 0.050-m sides. All four
wires are much longer than the sides of the square. The net magnetic field at the center of the square is
.
Calculate the current I.
Problem 80
*81.
A particle of charge
and mass
drawing shows. The speed of the particle is
is traveling perpendicular to a 1.6-T magnetic field, as the
.
(a)What is the value of the angle , such that the particle's subsequent path
will intersect the y axis at the greatest possible value of y?
Answer:
REASONING From the discussion in Section 21.3, we know that
when a charged particle moves perpendicular to a magnetic field, the
trajectory of the particle is a circle. The drawing at the right shows a
particle moving in the plane of the paper (the magnetic field is
perpendicular to the paper). If the particle is moving initially through the
coordinate origin and to the right (along the
axis), the subsequent
circular path of the particle will intersect the axis at the greatest possible
value, which is equal to twice the radius of the circle.
SOLUTION
(a From the drawing above, it can be seen that the largest value of is
)
equal to the diameter
of the circle. When the particle passes
through the coordinate origin its velocity must be parallel to the
axis. Thus, the angle is
.
(b The maximum value of is twice the radius of the circle.
) According to Equation 21.2, the radius of the circular path is
. The maximum value
is, therefore,
(b)Determine this value of y.
Answer:
REASONING From the discussion in Section 21.3, we know that when a charged particle moves
perpendicular to a magnetic field, the trajectory of the particle is a circle. The drawing at the right shows a
particle moving in the plane of the paper (the magnetic field is perpendicular to the paper). If the particle is
moving initially through the coordinate origin and to the right (along the
axis), the subsequent circular
path of the particle will intersect the axis at the greatest possible value, which is equal to twice the radius of
the circle.
SOLUTION
(a) From the drawing above, it can be seen that the largest value of is equal to the diameter
of the
circle. When the particle passes through the coordinate origin its velocity must be parallel to the
axis. Thus, the angle is
.
(b)The maximum value of is twice the radius of the circle. According to Equation 21.2, the radius of
the circular path is
. The maximum value
is, therefore,
REASONING From the discussion in Section 21.3, we know that when a charged particle moves perpendicular
to a magnetic field, the trajectory of the particle is a circle. The drawing at the right shows a particle moving in the
plane of the paper (the magnetic field is perpendicular to the paper). If the particle is moving initially through the
coordinate origin and to the right (along the
axis), the subsequent circular path of the particle will intersect the
axis at the greatest possible value, which is equal to twice the radius of the circle.
SOLUTION
(a) From the drawing above, it can be seen that the largest value of is equal to the diameter
When the particle passes through the coordinate origin its velocity must be parallel to the
of the circle.
axis. Thus, the
angle is
.
(b)The maximum value of is twice the radius of the circle. According to Equation 21.2, the radius of the circular
path is
*82.
. The maximum value
A particle has a charge of
electric field of
is, therefore,
and is located at the coordinate origin. As the drawing shows, an
exists along the
axis. A magnetic field also exists, and its x and y components
are
and
. Calculate the force (magnitude and direction) exerted on the particle by
each of the three fields when it is
(a)stationary,
(b)moving along the
axis at a speed of
, and
(c)moving along the
axis at a speed of
.
*83.
Two parallel rods are each
in length. They are attached at their centers to either end of a spring
that is initially neither stretched nor compressed. When
of current is in each
rod in the same direction, the spring is observed to be compressed by 2.0 cm. Treat the rods as long, straight wires and
find the separation between them when the current is present.
Answer:
**84.A solenoid is formed by winding
of insulated silver wire around a hollow cylinder. The turns are wound as
closely as possible without overlapping, and the insulating coat on the wire is negligibly thin. When the solenoid is
connected to an ideal (no internal resistance) 3.00-V battery, the magnitude of the magnetic field inside the solenoid is
found to be
. Determine the radius of the wire.
(Hint: Because the solenoid is closely coiled, the number of turns per unit length depends on the radius of the wire.)
**85.
A charge of
is placed on a small conducting sphere that is located at the end of a thin insulating
rod whose length is
. The rod rotates with an angular speed of
about an axis that passes
perpendicularly through its other end. Find the magnetic moment of the rotating charge.
(Hint: The charge travels around a circle in a time equal to the period of the motion.)
Answer:
REASONING The magnetic moment of the rotating charge can be found from the expression
, as discussed in Section 21.6. For this situation,
. Thus, we need to find the
current and the area for the rotating charge. This can be done by resorting to first principles.
SOLUTION The current for the rotating charge is, by definition (see Equation 20.1),
, where
is the
amount of charge that passes by a given point during a time interval . Since the charge passes by once per
revolution, we can find the current by dividing the total rotating charge by the period of revolution.
The area of the rotating charge is
Therefore, the magnetic moment is
Copyright © 2012 John Wiley & Sons, Inc. All rights reserved.
Additional
Problems
70.In each of two
coils the rate o
change of the
magnetic flux
in a single loop
is the same. Th
emf induced in
coil 1, which
has 184 loops,
is 2.82 V. The
emf induced in
coil 2 is 4.23 V
How many
loops does coi
2 have?
71.
When its coil rotates at a frequency of 280 Hz, a certain generator has a peak emf of 75 V.
(a)What is the peak emf of the generator when its coil rotates at a frequency of 45 Hz?
Answer:
12 V
REASONING The peak emf of a generator is found from
(Equation 22.4), where
is the number of turns in the generator coil, is the coil's cross-sectional area, is the magnitude of the
uniform magnetic field in the generator, and
is the angular frequency of rotation of the coil. In terms
of the frequency (in Hz), the angular frequency is given by
(Equation 10.6). Substituting
Equation 10.6 into Equation 22.4, we obtain
(1)
When the rotational frequency of the coil changes, the peak emf also changes. The quantities , , and
remain constant, however, because they depend on how the generator is constructed, not on how rapidly
the coil rotates. We know the peak emf of the generator at one frequency, so we will use Equation 1 to
determine the peak emf for a different frequency in part (a), and the frequency needed for a different peak
emf in part (b).
SOLUTION
(a) Solving Equation 1 for the quantities that do not change with frequency, we find that
(2)
The peak emf is
when the frequency is
when the frequency is
. We wish to find the peak emf
. From Equation 2, we have that
(3)
Solving Equation 3 for
, we obtain
(b)Letting
, Equation 2 yields
(4)
Solving Equation 4 for
, we find that
(b)Determine the frequency of the coil's rotation when the peak emf of the generator is 180 V.
Answer:
670 Hz
REASONING The peak emf of a generator is found from
(Equation 22.4), where
is the number of turns in the generator coil, is the coil's cross-sectional area, is the magnitude of the
uniform magnetic field in the generator, and
is the angular frequency of rotation of the coil. In terms
of the frequency (in Hz), the angular frequency is given by
(Equation 10.6). Substituting
Equation 10.6 into Equation 22.4, we obtain
(1)
When the rotational frequency of the coil changes, the peak emf also changes. The quantities , , and
remain constant, however, because they depend on how the generator is constructed, not on how rapidly
the coil rotates. We know the peak emf of the generator at one frequency, so we will use Equation 1 to
determine the peak emf for a different frequency in part (a), and the frequency needed for a different peak
emf in part (b).
SOLUTION
(a) Solving Equation 1 for the quantities that do not change with frequency, we find that
(2)
The peak emf is
when the frequency is
when the frequency is
. We wish to find the peak emf
. From Equation 2, we have that
(3)
Solving Equation 3 for
(b)Letting
, we obtain
, Equation 2 yields
(4)
Solving Equation 4 for
, we find that
REASONING The peak emf of a generator is found from
(Equation 22.4), where is the
number of turns in the generator coil, is the coil's cross-sectional area, is the magnitude of the uniform magnetic
field in the generator, and
is the angular frequency of rotation of the coil. In terms of the frequency (in Hz),
the angular frequency is given by
(Equation 10.6). Substituting Equation 10.6 into Equation 22.4, we obtain
(1)
When the rotational frequency of the coil changes, the peak emf also changes. The quantities , , and remain
constant, however, because they depend on how the generator is constructed, not on how rapidly the coil rotates. We
know the peak emf of the generator at one frequency, so we will use Equation 1 to determine the peak emf for a
different frequency in part (a), and the frequency needed for a different peak emf in part (b).
SOLUTION
(a) Solving Equation 1 for the quantities that do not change with frequency, we find that
(2)
The peak emf is
the frequency is
when the frequency is
. From Equation 2, we have that
. We wish to find the peak emf
when
(3)
Solving Equation 3 for
(b)Letting
, we obtain
, Equation 2 yields
(4)
Solving Equation 4 for
72.
73.
, we find that
A planar coil of wire has a single turn. The normal to this coil is parallel to a uniform and constant (in time)
magnetic field of 1.7 T. An emf that has a magnitude of 2.6 V is induced in this coil because the coil's area A is
shrinking. What is the magnitude of
, which is the rate (in
) at which the area changes?
Review Conceptual Example 9 as an aid in understanding this problem. A long, straight wire lies on a table and
carries a current I. As the drawing shows, a small circular loop of wire is pushed across the top of the table from
position 1 to position 2. Determine the direction of the induced current, clockwise or counterclockwise, as the loop
moves past
(a)position 1 and
Answer:
clockwise
REASONING In solving this problem, we apply Lenz's law,
which essentially says that the change in magnetic flux must be
opposed by the induced magnetic field.
SOLUTION
(a) The magnetic field due to the wire in the vicinity of position 1
is directed out of the paper. The coil is moving closer to the
wire into a region of higher magnetic field, so the flux
through the coil is increasing. Lenz's law demands that the
induced field counteract this increase. The direction of the
induced field, therefore, must be into the paper. The current in
the coil must be
.
(b)At position 2 the magnetic field is directed into the paper and
is decreasing as the coil moves away from the wire. The
induced magnetic field, therefore, must be directed into the
paper, so the current in the coil must be
.
(b)position 2. Justify your answers.
Answer:
clockwise
REASONING In solving this problem, we apply Lenz's law, which essentially says that the change
in magnetic flux must be opposed by the induced magnetic field.
SOLUTION
(a) The magnetic field due to the wire in the vicinity of position 1 is directed out of the paper. The coil is
moving closer to the wire into a region of higher magnetic field, so the flux through the coil is
increasing. Lenz's law demands that the induced field counteract this increase. The direction of the
induced field, therefore, must be into the paper. The current in the coil must be
.
(b)At position 2 the magnetic field is directed into the paper and is decreasing as the coil moves away
from the wire. The induced magnetic field, therefore, must be directed into the paper, so the current in
the coil must be
.
REASONING In solving this problem, we apply Lenz's law, which essentially says that the change in
magnetic flux must be opposed by the induced magnetic field.
SOLUTION
(a) The magnetic field due to the wire in the vicinity of position 1 is directed out of the paper. The coil is moving
closer to the wire into a region of higher magnetic field, so the flux through the coil is increasing. Lenz's law
demands that the induced field counteract this increase. The direction of the induced field, therefore, must be
into the paper. The current in the coil must be
.
(b)At position 2 the magnetic field is directed into the paper and is decreasing as the coil moves away from the
wire. The induced magnetic field, therefore, must be directed into the paper, so the current in the coil must be
.
74.In some places, insect “zappers,” with their blue lights, are a familiar sight on a summer's night. These devices use a
high voltage to electrocute insects. One such device uses an ac voltage of 4320 V, which is obtained from a standard
120.0-V outlet by means of a transformer. If the primary coil has 21 turns, how many turns are in the secondary coil?
75.
A 3.0capacitor has a voltage of 35 V between its plates. What must be the current in a 5.0-mH inductor so
that the energy stored in the inductor equals the energy stored in the capacitor?
Answer:
0.86 A
REASONING AND SOLUTION The energy stored in a capacitor is given by Equation 19.11a as
The energy stored in an inductor is given by Equation 22.10 as
and solving for the current , we get
.
. Setting these two equations equal to each other
*76.
*77.
At its normal operating speed, an electric fan motor draws only 15.0% of the current it draws when it just begins
to turn the fan blade. The fan is plugged into a 120.0-V socket. What back emf does the motor generate at its normal
operating speed?
Parts a and b of the drawing show the same uniform and constant (in time) magnetic field
directed
perpendicularly into the paper over a rectangular region. Outside this region, there is no field. Also shown is a
rectangular coil (one turn), which lies in the plane of the paper. In part a the long side of the coil
is just
at the edge of the field region, while in part b the short side
is just at the edge. It is known that
. In both parts of the drawing the coil is pushed into the field with the same velocity
until it is
completely within the field region. The magnitude of the average emf induced in the coil in part a is 0.15 V. What is
its magnitude in part b?
Answer:
0.050 V
*78.
Indicate the direction of the electric field between the plates of the parallel plate capacitor shown in the
drawing if the magnetic field is decreasing in time. Give your reasoning.
Problem 78
*79.
A piece of copper wire is formed into a single circular loop of radius 12 cm. A magnetic field is oriented parallel
to the normal to the loop, and it increases from 0 to 0.60 T in a time of 0.45 s. The wire has a resistance per unit length
of
Answer:
. What is the average electrical energy dissipated in the resistance of the wire?
REASONING The energy dissipated in the resistance is given by Equation 6.10b as the power dissipated
multiplied by the time ,
. The power, according to Equation 20.6c, is the square of the induced emf
divided by the resistance ,
induction, Equation 22.3.
. The induced emf can be determined from Faraday's law of electromagnetic
SOLUTION Expressing the energy consumed as
, and substituting in
, we find
The induced emf is given by Faraday's law as
unit length
, and the resistance is equal to the resistance per
times the length of the circumference of the loop,
. Thus, the energy dissipated is
*80.The purpose of this problem is to show that the work W needed to establish a final current
(Equation 22.10). In Section 22.8 we saw that the amount of work
through an inductor by an amount
versus I. Notice that
is
needed to change the current
, where L is the inductance. The drawing shows a graph of
is the area of the shaded vertical rectangle whose height is
this fact to show that the total work W needed to establish a current
**81.
in an inductor is
is
and whose width is
. Use
.
A solenoid has a cross-sectional area of
, consists of 400 turns per meter, and carries a current of 0.40
A. A 10-turn coil is wrapped tightly around the circumference of the solenoid. The ends of the coil are connected to a
1.5-Ω resistor. Suddenly, a switch is opened, and the current in the solenoid dies to zero in a time of 0.050 s. Find the
average current induced in the coil.
Answer:
**82.A 60.0-Hz generator delivers an average power of 75 W to a single light bulb. When an induced current exists in the
rotating coil of a generator, a torque—called a countertorque—is exerted on the coil. Determine the maximum
countertorque in the generator coil.
(Hint: The peak current, peak emf, and maximum countertorque all occur at the same instant.)
Copyright © 2012 John Wiley & Sons, Inc. All rights reserved.