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Ch 23 HW
Ch 23 HW Due: 11:59pm on Wednesday, October 5, 2016 You will receive no credit for items you complete after the assignment is due. Gr Grading ading Policy
Bouncing Electrons Two electrons, each Two each with with mass and charge , are are relea released sed from from positi positions ons very far from each other other.. With respect to a certain reference refere nce frame, frame, electron A has initial nonzer nonzero o speed speed towar toward d electron electron B in the positiv positiv e x direction, and electron B has initial speed toward electron A in the nega negative tive x x direction. direction. The electrons move directly toward each other along the x the x axis (very hard to do with real electrons). As the electrons approach each other, they slow due to their electric repulsion. This repulsion eventually pushes them away from each other.
Part A Which of the following statements about the motion of the electrons in the given reference frame will be true at the instant the two electrons reach their minimum separation? ANSWER: Electron A is moving faster than electron B. Electron B is moving faster than electron A. Both electrons are moving at the same (nonzero) speed in opposite directions. Both electrons are moving at the same (nonzero) speed in the same direction. Both electrons are momentarily stationary.
Correct If at a given moment the electrons are still moving toward each other, then they will be closer in the next instant. If at a given moment the electrons are moving away from each other, then they were closer in the previous instant. The electrons will be traveling in the same direction at the same speed at the moment they reach their minimum separation. Only in a reference frame in which the total momentum is zero (the center of momentum frame) would the electrons be stationary at their minimum separation.
Part B What is the minimum separation Express your answe answerr in term of ,
that the electrons reach? ,
, and
(where
).
Hint 1. How to approach the problem Since no external or nonconservative forces act on the system of the two electrons, both momentum and total energy (kinetic plus potential) are conserved. Find one expression for the energy when the electrons are far apart, and another when they reach reach their minim minimum um separation . This will give you an equation in which the only unknown is the speed of the electrons at the moment of their minimum separation. Apply conservation of momentum, using the same initial and final states, to obtain a second equation involving the speed of the electrons. Solve the simultaneous ener energy gy and momentum equa equations tions to obtain .
Hint 2. Find the initial energy What is the total ener energy gy of the two electrons when they are initially relea released? sed? Ass ume that the electrons are so far apart that their potential energy is zero.
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ANSWER: =
Hint 3. Find the final energy What is the total ener energy gy of the electrons when they reach their minimum separation (identical) (iden tical) speed of the two electrons is . Express your answe answerr in terms of
,
, ,
, and
(where
? Ass ume that the
).
Hint 1. Find the final kinetic energy What is the final kinetic energy (both electrons)? Express your answer in terms of
, and
.
ANSWER: =
Hint 2. Find the final potential energy What is the final potential energy of this 2-electron system? Express your answer in terms of , , and
.
ANSWER:
=
ANSWER:
=
Hint 4. Find the initial momentum What is the total momentum
of the two electrons whe when n they are initially relea released? sed?
Express your answer as a vector in terms of
, , and .
ANSWER: =
Hint 5. Find the final momentum What is the total momentum of the two electrons whe when n they reach their minimum separation that the (identical) velocity of the two electrons is
? Ass ume
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Ch 23 HW
ANSWER: =
Hint 3. Find the final energy What is the total ener energy gy of the electrons when they reach their minimum separation (identical) (iden tical) speed of the two electrons is . Express your answe answerr in terms of
,
, ,
, and
(where
? Ass ume that the
).
Hint 1. Find the final kinetic energy What is the final kinetic energy (both electrons)? Express your answer in terms of
, and
.
ANSWER: =
Hint 2. Find the final potential energy What is the final potential energy of this 2-electron system? Express your answer in terms of , , and
.
ANSWER:
=
ANSWER:
=
Hint 4. Find the initial momentum What is the total momentum
of the two electrons whe when n they are initially relea released? sed?
Express your answer as a vector in terms of
, , and .
ANSWER: =
Hint 5. Find the final momentum What is the total momentum of the two electrons whe when n they reach their minimum separation that the (identical) velocity of the two electrons is
? Ass ume
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Ch 23 HW
Express your answer as a vector in terms of
and
.
ANSWER: =
Hint 6. Some math help From the momentum equa equations, tions, find
; that is,
. Substitut Substitute e for
in the ener energy gy conservation equa equation tion to
.
ANSWER:
=
Correct An experienced experienced physicist physici st might approach approach this problem problem by considering the sys tem of electrons in a reference reference frame in which the initial momentum is zero. In this frame the initial speed of each each electron is . Try solving the proble problem m this way. way. Make sure that you obtain the same result for , and and decide for yourself which which approach approach is easier. easier.
Energy Stored in a Charge Configuration Four p point oint charges, charges, A, A , B, C, and D, D, are placed placed at the corners corners of a square with with side length , and D has charge . Throughou Throu ghoutt this prob problem, lem, use
in place of
. Charges Charges A, B , and C have have charge
.
Part A If you calculate
, the amount amount of of work work it took to assemble assemble this charge configuration if the point point charges charges were were initially
infinitely far apar apart, t, you will find that the contribution for each charge is propor proportional tional to numeric numer ic value that multiplies the above factor, in
.
. In the space provided, provided, enter the
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The Coulomb force is conservative. If we define the potential energy of the system to be zero when the charges are infinitely far apart, the amount of work needed to place any one charge in a configuration is equal to its electric potential energy. Imagine moving charge A, then B, then C, and finally D into place. Find the work required to add each charge to the configuration by calculating the potential energy of each just after it is added. Add the work required for each charge to find the total work required.
Hint 2. Electric potential and potential energy Recall that the electric potential at a point at distance
from a charge
is
, where
. Note that
this equation implicitly defines the electric potential to be zero at . The electric potential energy of a charge is equal to , where is the electric potential at the position of the charge before the charge is placed there. To find the potential at a point due to multiple charges, sum the potentials at that point due to each charge.
Hint 3. Work required to place charge A What is
, the work required to assemble the charge distribution shown in the figure?
Express your answer in terms of some or all of the variables , , and .
Hint 1. Find the potential at the location of charge A What is
, the electric potential at the upper left corner of the square before charge A is placed there?
Express your answer in terms of some or all of the variables , , and
.
ANSWER: =
0
ANSWER: = 0
Hint 4. Work required to place charge B What is
, the amount of work required to add charge B to the configuration, as shown in the figure?
Express your answer in terms of some or all of the variables , , and
.
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Hint 1. Find the potential at the location of charge B What is
, the potential at the upper right corner due to charge A, before charge B is placed there?
Express your answer in terms of some or all of the variables , , and
.
ANSWER: =
ANSWER:
=
Hint 5. Work required to place charge C What is
, the amount of work required to add charge C to the configuration, as shown in the figure?
Express your answer in terms of some or all of the variables , , and .
Hint 1. Find the potential at the location of charge C
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What is
, the potential at the lower right corner of the square before charge C is placed there?
Hint 1. How to approach this part The potential at C is the sum of the individual potentials due to the charges at B and A.
Hint 2. Find the potential at C due to the charge at B What is the potential at C due to the charge at B? Express your answer in terms of some or all of the variables , , and
.
ANSWER: =
Hint 3. Find the potential at C due to the charge at A What is the potential at C due to the charge at A? Express your answer in terms of some or all of the variables , , and ANSWER:
=
ANSWER:
=
ANSWER:
=
Hint 6. Find the work required to place charge D What is
, the amount of work required to add charge D to the configuration?
Express your answer in terms of some or all of the variables , , and
.
.
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Hint 1. Find the potential at the position of charge D What is
, the potential at the lower left corner of the square before charge D is placed there?
Express your answer in terms of some or all of the variables , , and
.
ANSWER:
=
ANSWER:
=
ANSWER:
= 0
Correct The hints led you through the problem by adding one charge at a time. A little thought shows that this is equivalent to simply adding the energies of all possible pairs: . Note that this is not equivalent to adding the potential energies of each charge. Adding the potential energies will give you double the correct answer because you will be counting each charge twice.
Part B
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Which of the following figures depicts a charge configuration that requires less work to assemble than the configuration in the problem introduction? Assume that all charges have the same magnitude .
ANSWER: figure a figure b figure c
Correct
Potential of a Charged Ring A ring with radius
and a uniformly distributed total charge
lies in the xy plane, centered at the origin.
Part A What is the potential
due to the ring on the z axis as a function of ?
Express your answer in terms of
, ,
, and
or
.
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Hint 1. How to approach the problem The formula for the electric potential produced by a static charge distribution involves the amount of charge and the distance from the charge to the position where the potential is measured. All points on the ring are equidistant from a given point on the z axis. This enables you to calculate the electric potential simply, without doing an integral.
Hint 2. The potential due to a point charge If you incorporate the symmetry of the problem, you will need only to know the formula for potential of a point charge:
, where
charge, and
is the potential at distance
from the point charge,
is the magnitude of the
is the permittivity of free space.
ANSWER:
=
Correct
Part B What is the magnitude of the electric field
on the z axis as a function of , for
Express your answer in terms of some or all of the quantities
, ,
, and
? or
.
Hint 1. Determine the direction of the field By symmetry, the electric field has only one Cartesian component. In what direction does the electric field point? ANSWER:
Hint 2. The relationship between electric field and potential You can obtain the electric field from a potential by the following expression: ,
where
is the electric field vector,
derivative of
ANSWER:
with respect to .
is the electric potential,
is the gradient operator, and
means the partial
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Ch 23 HW
|
|=
Correct Notice that while the potential is a strictly decreasing function of , the electric field first increases till
and
then starts to decrease. Why does the electric field exhibit such a behavior? Though the contribution to the electric field from each point on the ring strictly decreases as a function of , the vector cancellation from points on opposite sides of the ring becomes very strong for small . these vector cancellations. On the other hand
, even though all the individual
on account of 's point in (almost) the
same direction there, because the contribution to the electric field, per unit length of the ring
as
.
Exercise 23.6 The two sides of the DNA double helix are connected by pairs of bases (adenine, thymine, cytosine, and guanine). Because of the geometric shape of these molecules, adenine bonds with thymine and cytosine bonds with guanine. The figure shows the thymine–adenine bond. Each charge shown is and the distance is 0.110 .
Part A Calculate the electric potential energy of the adenine–thymine bond, using combinations of molecules . Express your answer with the appropriate units. ANSWER: = −9.71×10−19
Correct
and
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Part B Compare this energy with the potential energy of the proton–electron pair in the hydrogen atom. The electron in the hydrogen atom is 0.0529 from the proton. ANSWER:
= 0.223
Correct
Exercise 23.24 At a certain distance from a point charge, the potential and electric field magnitude due to that charge are 4.98 , respectively. (Take the potential to be zero at infinity.)
Part A What is the distance to the point charge? ANSWER: = 0.415
Correct
Part B What is the magnitude of the charge? ANSWER: = 2.30×10−10
Correct
Part C Is the electric field directed toward or away from the point charge? ANSWER: Toward Away
Correct
and 12.0
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Exercise 23.23
Part A An electron is to be accelerated from a velocity of 4.50×106 difference must the electron pass to accomplish this?
to a velocity of 8.00×106
. Through what potential
ANSWER: = -125
Correct
Part B Through what potential difference must the electron pass if it is to be slowed from 8.00×10 6
to a halt?
ANSWER: = 182
Correct
Exercise 23.48 A metal sphere with radius = 1.30 is supported on an insulating stand at the center of a hollow, metal, spherical shell with radius = 9.60 . Charge is put on the inner sphere and charge on the outer spherical shell. The magnitude of is chosen to make the potential difference between the spheres 460 , with the inner sphere at higher potential.
Part A Calculate . ANSWER: = 7.70×10−10
Correct
Part B Are the electric field lines and equipotential s urfaces mutually perpendicular? ANSWER:
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Yes No
Correct
Part C Are the equipotential surfaces closer together when the magnitude of
is largest?
ANSWER: Equipotential surfaces closer together when the magnitude of
is largest.
Equipotential surfaces closer together when the magnitude of
is smallest.
Correct
Problem 23.89 An alpha particle with kinetic energy 15.0 makes a collision with lead nucleus, but it is not "aimed" at the center of the lead nucleus, and has an initial nonzero angular momentum (with respect to the stationary lead nucleus) of magnitude −12
where is the magnitude of the initial momentum of the alpha particle and =1.20×10 . (Assume that the lead nucleus remains stationary and that it may be treated as a point charge. The atomic number of lead is 82. The alpha particle is a helium nucleus, with atomic number 2.)
Part A What is the distance of closest approach? ANSWER: = 1.21×10−12
Correct
Part B Repeat for =1.50×10−13
.
ANSWER: =
Incorrect; Try Again; 27 attempts remaining
,
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Part C Repeat for =1.20×10−14
.
ANSWER: =
Incorrect; Try Again; 28 attempts remaining
Exercise 23.17 Point charges and are placed at adjacent corners of a square for which the length of each side is 4.50 . Point is at the center of the square, and point is at the empty corner closest to . Take the electric potential to be zero at a distance far from both charges.
Part A What is the electric potential at point
due to
and
?
Express your answer with the appropriate units. ANSWER: = 0
Correct
Part B What is the electric potential at point ? Express your answer with the appropriate units. ANSWER: = −1.17×105
Correct
Part C A point charge by and ?
= -3.00
moves from point
to point . How much work is done on
Express your answer with the appropriate units. ANSWER:
by the electric forces exerted
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= -0.351
Correct
Exercise 23.14 A particle with a charge of the left; after it has moved 6.00
is in a uniform elect ric field directed t o the left. It is released from rest and moves to , its kinetic energy is found to be .
Part A What work was done by the electric force? ANSWER: = 1.50×10−6
Correct
Part B What is the potential of the starting point with respect to the endpoint? ANSWER: = 357
Correct
Part C What is the magnitude of
?
ANSWER: = 5950
Correct
Exercise 23.45 In a certain region of space, the electric potential is constants.
, where
,
, and
are positive
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Part A Calculate the
-component of the electric field.
Express your answer in terms of the given quantities. ANSWER: =
Correct
Part B Calculate the -component of the electric field. Express your answer in terms of the given quantities. ANSWER: =
Correct
Part C Calculate the -component of the electric field. Express your answer in terms of the given quantities. ANSWER: = 0
Correct
Part D At which points is the electric field equal to zero? ANSWER:
any value of any value of There is no point at which the electric field equal to zero.
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Correct
Problem 23.64 A Geiger counter detects radiation such as alpha particles by using the fact that the radiation ionizes the air along its path. A thin wire lies on the axis of a hollow metal cylinder and is insulated from it . A large potential difference is established between the wire and the outer cylinder, with the wire at higher potential; this sets up a strong electric field directed radially outward. When ionizing radiation enters the device, it ionizes a few air molecules. The free electrons produced are accelerated by the electric field toward the wire and, on the way there, ionize many more air molecules. Thus a current pulse is produced that can be detected by appropriate electronic circuitry and converted to an audible "click." Suppose the radius of the central wire is 145 and the radius of the hollow cylinder is 1.80 .
Part A What potential difference between the wire and the cylinder produces an electric field of at a distance of 1.20 from the axis of the wire? (Assume that the wire and cylinder are both very long in comparison to their radii.) ANSWER: = 1160
Correct
Exercise 23.5 A small metal sphere, carrying a net charge of = -2.60 , is held in a stationary position by insulating supports. A second small metal sphere, with a net charge of = -8.00 and mass 1.90 , is projected toward . When the two spheres are 0.800 apart, is moving toward with speed 22.0 . Assume that the two spheres can be treated as point charges. You can ignore the force of gravity.
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Part A What is the speed of
when the spheres are 0.420
apart?
ANSWER: = 16.2
Correct
Part B How close does
get to
?
ANSWER: = 0.270
Correct
Exercise 23.8
Part A Three equal 1.60point charges are placed at the corners of an equilateral triangle with sides 0.700 long. What is the potential energy of the system? (Take as zero the potential energy of the three charges when they are infinitely far apart.) Express your answer with the appropriate units. ANSWER: = 9.86×10−2
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Correct
Change in Electric Potential Ranking Task In the diagram below, there are two charges of and and six points (a through f) at various distances from the two charges. You will be asked to rank changes in the electric potential along paths between pairs of points.
Part A Using the diagram to the left, rank each of the given paths on the basis of the change in electric potential. Rank the largestmagnitude positive change (increase in electric potential) as largest and the largest-magnitude negative change (decrease in electric potential) as smallest. Rank from largest to smallest. To rank items as equivalent, overlap them.
Hint 1. Change in electric potential Determining the change in electric potential along some path involves determining the electric potential at the two end points of the path, and subtracting:
Hint 2. Determine the algebraic sign of the change in potential The path from point d to point a results in a positive change in electric potential. Which of the other paths also involves a positive change in electric potential (i.e., electric potential that increases along the path)? ANSWER: from b to a from f to e from c to d from c to e from c to b
Hint 3. Conceptualizing changes in electric potential
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Since positive charges create large positive electric potentials in their vicinity and negative charges create negative potentials in their vicinity, electric potential is sometimes visualized as a sort of “elevation.” Positive charges represent mountain peaks and negative charges deep valleys. In this picture, when you are close to a positive charge, you are high “up” and have a large positive potential. Conversely, near a negative charge you are deep in a “valley” and have a negative potential. Thus, changes in electric potential can be thought of as changes in elevation. The change is positive if you are moving “uphill” and the change is negative if you move “downhill.” The farther you travel either uphill or downhill, the larger the magnitude of the change in electric potential.
ANSWER:
Reset
Help
from b to a fr om c to b
fr om d to a
fr om c to d
from c to e from f to e
The correct ranking cannot be determined.
Correct
Exercise 23.12 An object with charge
= −4.00×10−9
the charge has moved to point
is placed in a region of uniform electric field and is released from rest at point
, 0.500
−7
.
, what is the electric potential at point
?
to the right, it has kinetic energy 4.00×10
Part A If the electric potential at point
is +30.0
Express your answer with the appropriate units. ANSWER: = 130
. After
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Correct
Part B What is the magnitude of the electric field? Express your answer with the appropriate units. ANSWER: = 200
Correct
Part C What is the direction of the electric field? ANSWER: from point
to point
from point
to point
perpendicular to the line
Correct
Exercise 23.19 Two point charges = 2.30 and = -6.00 are 0.100 apart. Point from and 0.060 from (the figure ). Take the electric potential to be zero at infinity.
Part A
is midway between them; point
is 0.080
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Find the potential at point
.
ANSWER: = -665
Correct
Part B Find the potential at point
.
ANSWER: = -640
Correct
Part C Find the work done by the electric field on a charge of 2.90
that travels from point
to point
.
Express your answer using two significant figures. ANSWER: = 7.2×10−8
Correct
A Millikan-type Experiment Consider an oil droplet of mass and charge . We want to determine the charge on the droplet in a Millikan-type experiment. We will do this in several steps. Assume, for simplicity, that the charge is positive and that the electric field between the plates points upward.
Part A An electric field is established by applying a potential difference to the plates. It is found that a field of strength will cause the droplet to be suspended motionless. Write an expression for the droplet's charge . Let be the acceleration due to gravity. Express your answer in terms of the suspending field
and the droplet's weight
.
Hint 1. A body in static equilibrium Recall Newton's 1st law, which states that a body is in equilibrium when the net force acting on it is zero. Then, for the droplet to be suspended motionless (i.e., to be in static equilibrium), the net force acting on it has to be zero. Note that the only forces acting on the droplet are the gravitational force (downward) and the electric force (upward).
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Hint 2. Find the electric force Recall the definition of electric field as electric force per unit charge. Write an expression for the electric force exerted on a point charge by an electric field . Express your answer in terms of
and .
ANSWER: =
ANSWER: =
Correct
Part B The field is easily determined by knowing the spacing between the plates and measuring the potential difference applied to them. The larger problem is to determine the mass of a microscopic droplet. Consider a mass falling through a viscous medium in which there is a retarding or drag force. For very small particles, the retarding force is given by , where is a constant and is the droplet's speed. The negative sign tells us that the drag force vector points upward when the droplet is falling. A falling droplet quickly reaches a constant speed, called the terminal speed . Write an expression for the terminal speed . Express your answer in terms of
, , and .
Hint 1. A body in dynamic equilibrium From Newton's 1st law, if the net force acting on a body initially in motion is zero, the body continues to move with constant velocity, and the body is said to be in dynamic equilibrium. It follows that the net force acting on a falling droplet that moves at constant velocity is zero. Note that the only forces acting on the falling oil droplet are the gravitational force and the drag force.
ANSWER: =
Correct
Part C A spherical object of radius moving slowly through the air is known to experience a retarding force , where is the viscosity of the air. Use this and your answer to Part B to find the radius of a spherical droplet of density falling with a terminal speed . Express your answer in terms of ,
, , and .
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Hint 1. How to approach the problem In this part of the problem, the situation is simply a specific case of the more general situation analyzed in Part B, where a body of mass that moves at constant speed through a viscous medium experiences a retarding force proportional to the speed of the body. Here, a droplet of given density falls at constant speed through air and experiences a retarding force of magnitude , which is proportional to the droplet speed. The constant of proportionality depends on the viscosity of air and the radius of the droplet. To solve this part of the problem, simply apply Newton's 1st law to the droplet and solve for the droplet's radius.
ANSWER:
=
Correct
Part D Oil has a density of . An oil droplet is suspended between two plates apart by adjusting the potential difference between them to . When the voltage is removed, the droplet falls and quickly reaches constant speed. It is timed with a stopwatch and falls in . The viscosity of air is . What is the droplet's charge ? Express the charge in coulombs to two signi ficant figures. Take the free-fall acceleration to be
.
Hint 1. How to approach the problem In Part A, you found an expression for the charge of a droplet that is in equilibrium between two plates separated by a potential difference. In particular, you found that the charge of the droplet is given by the weight of the droplet divided by the electric field established between the plates. Therefore, to find the charge of the droplet, you need to know its mass and the value of the electric field between the plates.
Hint 2. Find the mass of the droplet Recall the definition of density and write an expression for the mass . Express your answer in terms of
, , and
of a spherical droplet of radius
and density
.
Hint 1. Definition of density Consider a mass by .
of a homogeneous material, whose volume is
Hint 2. Volume of a sphere The volume of a sphere of radius
ANSWER:
is given by
.
. The density
of that material is given
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=
Hint 3. Find the droplet's radius In Part C you found an expression for the radius of a droplet of given density that falls through air at constant speed. In particular, you found that the droplet's radius depends on the viscosity of air, the speed of the droplet, the density of the droplet, and the acceleration due to gravity. Find the droplet's radius is the droplet's speed.
for the situation described in this part of the problem. Note that the only unknown variable
Express your answer in meters to three significant figures. Take the free-fall acceleration to be
.
Hint 1. Find the speed of the droplet The oil droplet falls
in
. Recall the definition of speed and find the droplet's speed .
Express your answer in meters per second to three significant figures. ANSWER: = 4.093×10−4
ANSWER: = 2.000×10−6
Hint 4. Find the electric field between two parallel plates The potential difference between two parallel charged plates is a function of the electric field established between the plates and the distance between them. Write an expression for the electric field difference between them of . Express your answer in terms of ANSWER: =
ANSWER: = 2.40×10−18
Correct
between two parallel plates a distance and .
apart and with a potential
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How many units of the fundamental electric charge does this droplet possess? Express your answer as an integer.
Hint 1. The fundamental electric charge The fundamental electric charge is the magnitude of the charge of an electron, which is .
ANSWER: 15
Correct
Exercise 23.35 A very small sphere with positive charge uniform linear charge density 1.00
7.00 .
is released from rest at a point 1.30
from a very long line of
Part A What is the kinetic energy of the sphere when it is 4.40 by the line of charge?
from the line of charge if the only force on it is the force exerted
Express your answer with the appropriate units. ANSWER: = 0.153
Correct
Electric Fields and Equipotential Surfaces The dashed lines in the diagram represent cross sections of equipotential surfaces drawn in 1
increments.
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Part A What is the work done by the electric force to move a 1
charge from A to B?
Express your answer in joules.
Hint 1. Find the potential difference between A and B What is the potential difference
between point A and point B?
Express your answer in volts.
Hint 1. Equipotential surfaces Recall that an equipotential surface is a surface on which the electric potential is the same at every point.
ANSWER: = 0
Hint 2. Potential difference and work Recall that the electric potential energy difference between any two points is equal to the negative of the work done by the electric force as a charged object moves between those two points. If we combine this with the relationship between electric potential energy and electric potential we have: .
ANSWER: 0
Correct
Part B
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What is the work done by the electric force to move a 1
charge from A to D?
Express your answer in joules.
Hint 1. Find the potential difference between A and D What is the potential difference
between point A and point D?
Express your answer in volts. ANSWER: = -1
Hint 2. Potential difference and work Recall that the electric potential energy difference between any two points is equal to the negative of the work done by the electric force as a charged object moves between those two points. If we combine this with the relationship between electric potential energy and electric potential we have:
ANSWER: 1
Correct
Part C The magnitude of the electric field at point C is
Hint 1. Electric field and equipotential surfaces Since the diagram shows equal potential differences between adjacent surfaces, equal amounts of work are done to move a particular charge from one surface to the next adjacent one. It follows then that if the equipotentials are closer together, the electric force does the same amount of work in a smaller displacement than if the equipotentials were farther apart. Therefore, the electric force, as well as the corresponding electric field, has a larger magnitude.
ANSWER: greater than the magnitude of the electric field at point B. less than the magnitude of the electric field at point B. equal to the magnitude of the electric field at point B. unknown because the value of the electric potential at point C is unknown.
Correct
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Ch 23 HW
In a certain region of space, the electric potential is constants.
Part A Calculate the
-component of the electric field.
Express your answer in terms of the given quantities. ANSWER: =
Correct
Part B Calculate the -component of the electric field. Express your answer in terms of the given quantities. ANSWER: =
Correct
Part C Calculate the -component of the electric field. Express your answer in terms of the given quantities. ANSWER: = 0
Correct
Part D At which point is the electric field equal to zero? ANSWER:
, where
,
, and
are positive
10/4/2016
Ch 23 HW
Correct
Problem 23.47 A point charge = 4.05 is placed at the origin, and a second point charge = -3.10 20.5 . A third point charge = 2.00 is to be placed on the -axis between and energy of the three charges when they are infinitely far apart.)
is placed on t he -axis at . (Take as zero the potential
Part A What is the potential energy of the sys tem of the three charges if
is placed at
10.5
?
ANSWER: = −4.15×10−7
Correct
Part B Where should
be placed between
and
to make the potential energy of the sys tem equal to zero?
ANSWER: = 7.45
Correct
Problem 23.69 Charge
4.00
is distributed uniformly over the volume of an insulating sphere that has radius
Part A What is the potential difference between the center of the sphere and the surface of the sphere? Express your answer with the appropriate units. ANSWER: = 3.00×105
Correct
Problem 23.79
= 6.00
.
10/4/2016
Ch 23 HW
The electric potential in a region that is within 2.00 of the origin of a rectangular coordinate sys tem is given by , where , , , , , , and are constants. The units of , , , and are such that if , , and are in meters, then is in volts. You measure and each component of the electric field at four points and obtain these results: Point 1
(0, 0, 0)
10.0
0
0
0
2
(1.00, 0, 0)
4.0
14.0
0
0
3
(0, 1.00, 0)
6.0
0
14.0
0
4
(0, 0, 1.00)
8.0
0
0
14.0
Part A Use the data in the table to calculate
.
ANSWER: = -6.0
Correct
Part B Use the data in the table to calculate
.
ANSWER: = -4.0
Correct
Part C Use the data in the table to calculate
.
ANSWER: = -2.0
Correct
Part D Use the data in the table to calculate ANSWER:
.
10/4/2016
Ch 23 HW
= 10.0
Correct
Part E Use the data in the table to calculate . ANSWER: = 2.3
Correct
Part F Use the data in the table to calculate
.
ANSWER: = 3.5
Correct
Part G Use the data in the table to calculate
.
ANSWER: = 7.0
Correct
Part H What is
at the point (0, 0, 0)?
Express your answer with the appropriate units. ANSWER: = 10.0
Correct
10/4/2016
Ch 23 HW
Part I What is the magnitude of
at the point (0, 0, 0)?
Express your answer with the appropriate units. ANSWER: = 0
Correct
Part J What is
at the point (0.50
, 0.50
, 0.50
)?
Express your answer with the appropriate units. ANSWER: =
Part K What is the magnitude of
at the point (0.50
, 0.50
, 0.50
)?
Express your answer with the appropriate units. ANSWER: =
Incorrect; Try Again; 29 attempts remaining
Part L What is
at the point (1.00
, 1.00
, 1.00
)?
Express your answer with the appropriate units. ANSWER: =
Incorrect; Try Again; 29 attempts remaining
Part M What is the magnitude of
at the point (1.00
, 1.00
, 1.00
)?
10/4/2016
Ch 23 HW
Express your answer with the appropriate units. ANSWER: =
Incorrect; Try Again; 29 attempts remaining
Video Tutor: Charged Conductor with Teardrop Shape First, launch the video below. You will be asked to use your knowledge of physics to predict the outcome of an experiment. Then, close the video window and answer the question at right. You can watch the video again at any point.
Part A Two conduct ing spheres are each given a charge . The radius of the larger sphere is three times greater than that of the smaller sphere. If the electric field just outside of the smaller sphere is , then the electric field just outside of the larger sphere is
Hint 1. How to approach the problem. The electric field just outside of a conductor is proportional to the conductor's surface charge density. The surface charge density of a sphere is calculated by dividing the total charge on the sphere by the sphere's surface area. Think about how changing the radius of a sphere changes the sphere's surface area.
ANSWER:
1/3 1/9 3 9
