Ch 23 Hw 4k3b6p
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12/8/2015
Ch 23 HW
Ch 23 HW Due: 11:59pm on Tuesday, November 10, 2015 You will receive no credit for items you complete after the assignment is due. Grading Policy
Electric Potential Energy versus Electric Potential Learning Goal: To understand the relationship and differences between electric potential and electric potential energy. In this problem we will learn about the relationships between electric force F ⃗ , electric field E⃗, potential energy U , and electric potential V . To understand these concepts, we will first study a system with which you are already familiar: the uniform gravitational field.
Gravitational Force and Potential Energy First we review the force and potential energy of an object of mass m in a uniform gravitational field that points ^ downward (in the −k direction), like the gravitational field near the earth's surface.
Part A ⃗ Find the force F (z) on an object of mass m in the uniform gravitational field when it is at height z
= 0
.
⃗ ^ Express F (z) in of m, z , k , and g .
ANSWER: ⃗ F (z)
=
^ −mgk
Correct Because we are in a uniform field, the force does not depend on the object's location. Therefore, the variable z does not appear in the correct answer.
Part B Now find the gravitational potential energy U (z) of the object when it is at an arbitrary height z . Take zero potential to be at position z = 0 . Keep in mind that the potential energy is a scalar, not a vector. Express U (z) in of m, z , and g . ANSWER: U (z)
=
mgz
Correct
Part C https://session.masteringphysics.com/myct/assignmentPrintView?displayMode=studentView&assignmentID=3924419
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In what direction does the object accelerate when released with initial velocity upward? ANSWER: upward downward upward or downward depending on its mass downward only if the ratio of g to initial velocity is large enough
Correct
Electric Force and Potential Energy Now consider the analogous case of a particle with charge q placed in a uniform electric field of strength E , pointing ^ downward (in the −k direction)
Part D ⃗ Find F (z) , the electric force on the charged particle at height z . ⃗ ^ Express F (z) in of q , E , z , and k .
Hint 1. Relationship between force and electric field The force on a particle of charge q in an electric field E⃗
^ = −E k
is given by F ⃗ =
qE
⃗
.
ANSWER: ⃗ F (z)
=
^ −qEk
Correct
Part E Now find the potential energy U (z) of this charged particle when it is at height z . Take zero potential to be at position z = 0 . Express U (z) in of q , E , and z . ANSWER: U (z)
=
qEz
Correct https://session.masteringphysics.com/myct/assignmentPrintView?displayMode=studentView&assignmentID=3924419
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Part F In what direction does the charged particle accelerate when released with upward initial velocity? ANSWER: upward downward upward or downward depending on its charge downward only if the ratio of qE to initial velocity is large enough
Correct
Electric Field and Electric Potential The electric potential V is defined by the relationship U with charge q .
= qV ,
where U is the electric potential energy of a particle
Part G ^ Find the electric potential V of the uniform electric field E⃗ = Ek . Note that this field is not pointing in the same direction as the field in the previous section of this problem. Take zero potential to be at position z = 0 .
Express V in of q , E , and z . ANSWER: V
=
−Ez
Correct The SI unit for electric potential is the volt (V). The volt is a derived unit, which means that it can be written in of other SI units. In of the fundamental units of length, mass, time and charge, the volt can be expressed as follows: 1V =
kg m
2
s2 C
Part H The electric field can be derived from the electric potential, just as the electrostatic force can be determined from the electric potential energy. The relationship between electric field and electric potential is E⃗ is the gradient operator: ⃗ ∇V =
The partial derivative
∂V ∂x
∂V ∂x
^ i +
∂V ∂y
^ j+
∂V ∂z
⃗ = −∇V
, where ∇⃗
^ k.
means the derivative of V with respect to x, holding all other variables constant.
⃗ https://session.masteringphysics.com/myct/assignmentPrintView?displayMode=studentView&assignmentID=3924419
^
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Consider again the electric potential V z coordinate only, so
∂V ∂x
=
∂V ∂y
= −Ez
= 0 and
∂V ∂z
corresponding to the field E⃗ =
dV dz
^ = Ek
. This potential depends on the
.
Find an expression for the electric field E⃗ in of the derivative of V . ^ ^ Express your answer as a vector in of the unit vectors ^ i , j , and/or k . Use dV /dz for the derivative of V with respect to z .
ANSWER:
E
⃗
=
−
dV dz
^ k
Correct
Part I A positive test charge will accelerate toward regions of ________ electric potential and ________ electric potential energy. Choose the appropriate answer combination to fill in the blanks correctly.
Hint 1. Direction of the electric field In what direction do electric fields point? ANSWER: from regions of higher electric potential to lower electric potential from regions of lower electric potential to higher electric potential
Hint 2. Formula for the force on a charge in an electric field The force F ⃗ on a charge q in an electric field E⃗ is given by ⃗ ⃗ F = qE
This is similar to the equation F
=
. mg for the force on a mass in a uniform gravitational field.
Hint 3. Formula for electric potential energy The electric potential energy U of a charge q at electric potential V is given by . , for the gravitational potential energy of a particle with
U = qV
This is similar to the equation Ug mass m.
= mVg = mgh
ANSWER:
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higher; higher higher; lower lower; higher lower; lower
Correct
Part J A negative test charge will accelerate toward regions of ________ electric potential and ________ electric potential energy. Choose the appropriate answer combination to fill in the blanks correctly.
Hint 1. Direction of the electric field In what direction do electric fields point? ANSWER: from regions of higher electric potential to lower electric potential from regions of lower electric potential to higher electric potential
Hint 2. Formula for the force on a charge in an electric field The force F ⃗ on a charge q in an electric field E⃗ is given by ⃗ ⃗ F = qE
This is similar to the equation F
= mg
. for the force on a mass in a uniform gravitational field.
Hint 3. Formula for electric potential energy The electric potential energy U of a charge q at electric potential V is given by . This is similar to the equation Ug = mVg = mgh for the gravitational potential energy of a particle with mass m. It works even for negative charges. Negative masses are not known to exist. U = qV
ANSWER: higher; higher higher; lower lower; higher lower; lower
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Correct A charge in an electric field will experience a force in the direction of decreasing potential energy. Since the electric potential energy of a negative charge is equal to the charge times the electric potential (U = qV ), the direction of decreasing electric potential energy is the direction of increasing electric potential.
Electric Potential, Potential Energy, and Force Learning Goal: To review relationships among electric potential, electric potential energy, and force on a test charge This problem is a review of the relationship between an electric field E⃗, its associated electric potential V , the electric potential energy U , and the direction of force on a test charge.
Part A Electric field lines always begin at _______ charges (or at infinity) and end at _______ charges (or at infinity). One could also say that the lines we use to represent an electric field indicate the direction in which a _______ test charge would initially move when released from rest. Which of the following fills in the three missing words correctly? ANSWER: (positive; negative; negative) (positive; negative; positive) (negative; positive; negative) (negative; positive; positive)
Correct Note that the electric field vector E⃗ is everywhere tangent to the electric field lines. Like electric field lines, the electric field vector generally points away from positive charges and toward negative charges.
Part B Would a positive test charge released from rest move toward a region of higher or lower electric potential (compared to the electric potential at the point where it is released)?
Hint 1. Potential, field, and force The equation relating an electric field to electric potential is E⃗
⃗ = −∇V
. By definition, ∇⃗V points in the
direction in which V is most rapidly increasing. Thus −∇⃗V (or the electric field) points in the direction in which V is most rapidly decreasing. The force on a charge q in an electric field is given by F ⃗ = positive charge will experience a force in the same direction as the electric field.
qE
⃗
, so a
ANSWER: https://session.masteringphysics.com/myct/assignmentPrintView?displayMode=studentView&assignmentID=3924419
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higher electric potential lower electric potential
Correct
Part C Now imagine that the sign of our test particle is changed from positive to negative, but the electric potential remains the same. Which of the following statements is correct?
Hint 1. Direction of field and force The problem states that the electric potential does not change. The electric field is determined from ⃗ ⃗ E = −∇V
, and the force from F ⃗ = qE⃗ , where q is the charge. When the sign of the test charge changes, but the electric potential remains the same, what happens to the direction of the electric field and the direction of the force on the charge? ANSWER: Both the electric field and force are unchanged. The force is unchanged but the electric field is reversed. Both the electric field and force are reversed. The electric field is unchanged but the force is reversed.
Hint 2. Direction of force and potential energy gradient Like other forms of potential energy, the electric potential energy U is always intimately related to the electric force by the equation F ⃗ =
⃗ −∇U
. Note that this relationship does not depend on the sign of the
⃗ ∇U
test charge. By definition, points in the direction in which U is most rapidly increasing. Thus −∇⃗U (or the force) points in the direction in which U is most rapidly decreasing. ANSWER: The direction of the force will change and it will point to regions of higher potential energy. The direction of the force will not change and it will point to regions of higher potential energy. The direction of the force will not change and it will point to regions of lower potential energy. The direction of the force will change and it will point to regions of lower potential energy.
Correct
PhET Tutorial: Charges and Electric Potential https://session.masteringphysics.com/myct/assignmentPrintView?displayMode=studentView&assignmentID=3924419
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Learning Goal: To understand the spatial distribution of the electric potential for a variety of simple charge configurations, and to understand how the electric field and electric potential (voltage) are related. For this problem, use the PhET simulation Charges and Fields. This simulation allows you to place multiple positive and negative pointcharges in any configuration and look at the resulting electric field and corresponding electric potential.
Start the simulation. You can click and drag positive charges (red) or negative charges (blue) into the main screen. If you select Show Efield in the green menu, red arrows will appear, showing the direction of the electric field. Faint red arrows indicate that the electric field is weaker than at locations where the arrows are brighter (this simulation does not use arrow length as a measure of field magnitude). You can drag a positive charge into the main screen and place the crosshairs of the electric potential (voltage) measurement tool on top of any point for which you would like to see the voltage (for this tutorial, voltage will be taken to be the same as electric potential). Clicking plot on this device shows an equipotential line. All points along an equipotential line have the same potential. Click Clear All before beginning Part A.
Part A The electric potential (voltage) at a specific location is equal to the potential energy per unit charge a charged object would have if it were at that location. If the zero point of the voltage is at infinity, the numerical value of the voltage is equal to the numerical value of work done to bring in a unit charge from infinity to that location. Select Show numbers and grid in the green menu, and drag one positive charge to the middle of the screen, right on top of two intersecting bold grid lines. Using the voltage meter, you should find that 1 m away from the charge, the voltage is 9 V. What is the voltage 2 m away from the charge? Express your answer numerically in volts to two significant figures. ANSWER: 4.5 V
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Correct Unlike the magnitude of the electric field, the electric potential (voltage) is not proportional to the inverse of the distance squared.
Part B What is the voltage 3 m away from the charge? ANSWER: 1 V 3 V 9 V
Correct Based on this result, and the previous question, the electric potential (voltage) is inversely proportional to the distance r from the charge: V ∝ 1/r. Recall that the magnitude of the electric field E ∝ 1/r 2 .
Part C Another way to study voltage and its relationship to electric field is by producing equipotential lines. Just like every point on a contour line has the same elevation in a topographical map, every point on an equipotential line has the same voltage. Click plot on the voltage tool to produce an equipotential line. Produce many equipotential lines by clicking plot as you move the tool around. You should produce a graph that looks similar to the one shown below.
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Place several EField Sensors at a few points on different equipotential lines, and look at the relationship between the electric field and the equipotential lines. Which statement is true? ANSWER: At any point, the electric field is parallel to the equipotential line at that point. At any point, the electric field is perpendicular to the equipotential line at that point, and it is directed toward lines of lower voltages. At any point, the electric field is perpendicular to the equipotential line at that point, and it is directed toward lines of higher voltages.
Correct All points on an equipotential line have the same voltage; thus, no work would be done in moving a test charge along an equipotential line. No work is done because the electric field, and thus the force on the test charge, is perpendicular to the displacement of the test charge being moved along the equipotential line.
Part D Equipotential lines are usually shown in a manner similar to topographical contour lines, in which the difference in the value of consecutive lines is constant. Clear the equipotential lines using the Clear button on the voltage tool. Place the first equipotential line 1 m away from the charge. It should have a value of roughly 9 V. Now, produce several additional equipotential lines, increasing and decreasing by an interval of 3 V (e.g., one with 12 V, one with 15 V, and one with 6 V). Don’t worry about getting these exact values. You can be off by a few tenths of a volt. Which statement best describes the distribution of the equipotential lines? ANSWER: The equipotential lines are equally spaced. The distance between each line is the same for all adjacent lines. The equipotential lines are closer together in regions where the electric field is weaker. The equipotential lines are closer together in regions where the electric field is stronger.
Correct Near the positive charge, where the electric field is strong, the voltage lines are close to each other. Farther from the charge, the electric field is weaker and the lines are farther apart.
Part E Now, remove the positive charge by dragging it back to the basket, and drag one negative charge toward the middle of the screen. Determine how the voltage is different from that of the positive charge. How does the voltage differ from that of the positive charge? ANSWER:
https://session.masteringphysics.com/myct/assignmentPrintView?displayMode=studentView&assignmentID=3924419
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The voltages become negative instead of positive and keep the same magnitudes. The voltage distribution does not change. The voltages are positive, but the magnitude increases with increasing distance.
Correct The voltage is still inversely proportional to the distance from the charge, but the voltage is negative everywhere rather than positive.
Part F Now, remove the negative charge, and drag two positive charges, placing them 1 m apart, as shown below.
What is the voltage at the midpoint of the two charges? ANSWER: Exactly twice the voltage produced by only one of the charges at the same point Zero Greater than zero, but less than twice the voltage produced by only one of the charges at the same point
https://session.masteringphysics.com/myct/assignmentPrintView?displayMode=studentView&assignmentID=3924419
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Correct Because voltage is a scalar quantity, there are no vector components with opposite directions canceling out, as for electric fields. The voltage is simply the sum of the voltages due to each of the individual charges. Since both charges are positive, the voltage due to each charge (at all locations) is positive.
Part G Now, make an electric dipole by replacing one of the positive charges with a negative charge, so the final configuration looks like the figure shown below.
What is the voltage at the midpoint of the dipole?
Hint 1. How to approach the problem The total voltage is equal to the sum of the voltages due to each of the charges. The voltage due to each charge simply depends on the distance from the charge and the charge’s value. Voltage is a scalar quantity, so it does not add like a vector. ANSWER:
The voltage at the midpoint of the dipole is
positive. zero. negative.
https://session.masteringphysics.com/myct/assignmentPrintView?displayMode=studentView&assignmentID=3924419
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Correct Because the voltage due to the negative charge has the opposite sign of the voltage due to the positive charge at the midpoint, the net voltage is zero. The electric field, however, is not zero here!
Part H Make several equipotential lines similar to the figure below.
Try to have the equipotential lines equally spaced in voltage. Then, use an EField Sensor to measure the electric field at a few points while looking at the relationship between the electric field and the equipotential lines. Which of the following statements is true? ANSWER: The electric field strength is greatest where the voltage is the greatest. The electric field strength is greatest where the voltage is the smallest. The electric field strength is greatest where the equipotential lines are very close to each other.
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Correct Locations where the voltage is changing steeply are locations with a strong electric field. The magnitude of the electric field is equal to the rate the voltage is changing with distance. Mathematically, this idea is conveyed by |Es | = dV /ds, where Es is the component of the electric field in the direction of a small displacement ds . (As you learned earlier, the electric field is directed in the direction where the voltage decreases.)
PhET Interactive Simulations University of Colorado http://phet.colorado.edu
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 conducting spheres are each given a charge Q . 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 E0 , 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:
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1/3 E0 3 E0 9 E0 1/9 E0 E0
Correct The larger sphere has nine times the surface area of the smaller one, and this reduces the surface charge density by a factor of nine.
Bouncing Electrons Two electrons, each with mass m and charge q , are released from positions very far from each other. With respect to a certain reference frame, electron A has initial nonzero speed v toward electron B in the positive x direction, and electron B has initial speed 3v toward electron A in the negative x direction. The electrons move directly toward each other along 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 r min that the electrons reach? Express your answer in term of q , m, v , and k (where k =
1 4πϵ 0
).
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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 their minimum separation r min . 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 energy and momentum equations to obtain r min .
Hint 2. Find the initial energy What is the total energy Einitial of the two electrons when they are initially released? Assume that the electrons are so far apart that their potential energy is zero. Express your answer in of m and v . ANSWER: Einitial
=
5mv
2
Hint 3. Find the final energy What is the total energy Ef inal of the electrons when they reach their minimum separation r min ? Assume that the (identical) speed of the two electrons is u. Express your answer in of m, u, q , r min , and k (where k =
1 4πϵ 0
).
Hint 1. Find the final kinetic energy What is the final kinetic energy (both electrons)? Express your answer in of u, and m. ANSWER: Kf inal
=
mu
2
Hint 2. Find the final potential energy What is the final potential energy of this 2electron system? Express your answer in of k , q , and r min . ANSWER:
Uf inal
=
kq
2
rmin
ANSWER:
https://session.masteringphysics.com/myct/assignmentPrintView?displayMode=studentView&assignmentID=3924419
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Ef inal
=
mu
2
+
kq
2
rmin
Hint 4. Find the initial momentum ⃗ What is the total momentum p initial of the two electrons when they are initially released?
Express your answer as a vector in of m, v , and ^ i. ANSWER: p⃗
initial
=
^ −2mv i
Hint 5. Find the final momentum What is the total momentum p f⃗ inal of the two electrons when they reach their minimum separation r min ? Assume that the (identical) velocity of the two electrons is u⃗. Express your answer as a vector in of m and u⃗. ANSWER: p⃗
f inal
=
2mu⃗
Hint 6. Some math help From the momentum equations, ∣∣u⃗∣∣ equation to find r min .
= v
; that is, u
= v
. Substitute for u in the energy conservation
ANSWER:
r min
kq
=
2
4mv 2
Correct An experienced physicist might approach this problem by considering the system of electrons in a reference frame in which the initial momentum is zero. In this frame the initial speed of each electron is 2v . Try solving the problem this way. Make sure that you obtain the same result for r min , and decide for yourself which approach is easier.
Energy Stored in a Charge Configuration Four point charges, A, B, C, and D, are placed at the corners of a square with side length L. Charges A, B, and C have charge + q , and D has charge − q . Throughout this problem, use k in place of
1 4πϵ 0
.
https://session.masteringphysics.com/myct/assignmentPrintView?displayMode=studentView&assignmentID=3924419
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Part A If you calculate W , the amount of work it took to assemble this charge configuration if the point charges were initially infinitely far apart, you will find that the contribution for each charge is proportional to provided, enter the numeric value that multiplies the above factor, in W .
kq L
2
. In the space
Hint 1. How to approach the problem 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 r from a charge q is V
=k
q r
, where k =
1 4πϵ 0
.
Note that this equation implicitly defines the electric potential to be zero at r = ∞. The electric potential energy of a charge q is equal to qV , where V 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 W A , the work required to assemble the charge distribution shown in the figure? Express your answer in of some or all of the variables k , q , and L.
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Hint 1. Find the potential at the location of charge A What is VA , the electric potential at the upper left corner of the square before charge A is placed there? Express your answer in of some or all of the variables k , q , and L. ANSWER: VA
= 0
ANSWER: WA
= 0
Hint 4. Work required to place charge B What is W B , the amount of work required to add charge B to the configuration, as shown in the figure? Express your answer in of some or all of the variables k , q , and L.
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Hint 1. Find the potential at the location of charge B What is VB , the potential at the upper right corner due to charge A, before charge B is placed there? Express your answer in of some or all of the variables k , q , and L. ANSWER:
VB
kq
=
L
ANSWER:
WB
=
kq
2
L
Hint 5. Work required to place charge C What is W C , the amount of work required to add charge C to the configuration, as shown in the figure? Express your answer in of some or all of the variables k , q , and L.
Hint 1. Find the potential at the location of charge C What is VC , 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 of some or all of the variables k , q , and L. ANSWER: https://session.masteringphysics.com/myct/assignmentPrintView?displayMode=studentView&assignmentID=3924419
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=
VCB
kq L
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 of some or all of the variables k , q , and L. ANSWER: =
VCA
kq √2 L
ANSWER:
VC
kq
=
L
(1 +
1 √2
)
ANSWER:
WC
=
kq L
2
(1 +
1 √2
)
Hint 6. Find the work required to place charge D What is W D , the amount of work required to add charge D to the configuration? Express your answer in of some or all of the variables k , q , and L.
Hint 1. Find the potential at the position of charge D What is VD , the potential at the lower left corner of the square before charge D is placed there? https://session.masteringphysics.com/myct/assignmentPrintView?displayMode=studentView&assignmentID=3924419
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Express your answer in of some or all of the variables k , q , and L. ANSWER:
VD
=
kq L
(2 +
1 √2
)
ANSWER:
WD
=
−
kq
2
(2 +
L
1 √2
)
ANSWER:
W
= 0 ×(
kq L
2
)
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: W =
q
2
4πϵ 0
(
1 R BA
+
1 R BC
+
1 R CA
−
1 R DA
−
1 R DB
−
1 R DC
).
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 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 q .
ANSWER: https://session.masteringphysics.com/myct/assignmentPrintView?displayMode=studentView&assignmentID=3924419
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figure a figure b figure c
Correct
Electric Force and Potential: Spherical Symmetry Learning Goal: To understand the electric potential and electric field of a point charge in three dimensions Consider a positive point charge q , located at the origin of threedimensional space. Throughout this problem, use k in place of
1 4πϵ 0
.
Part A Due to symmetry, the electric field of a point charge at the origin must point _____ from the origin. Answer in one word. ANSWER: radial
Correct
Part B Find E(r) , the magnitude of the electric field at distance r from the point charge q . Express your answer in of r , k , and q . ANSWER:
E(r)
=
kq r2
Correct
Part C Find V (r) , the electric potential at distance r from the point charge q . Express your answer in of r , k , and q . ANSWER: https://session.masteringphysics.com/myct/assignmentPrintView?displayMode=studentView&assignmentID=3924419
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V (r)
=
kq r
Correct
Part D Which of the following is the correct relationship between the magnitude of a radial electric field E(r) and its associated electric potential V (r) ? More than one answer may be correct for the particular case of a point charge at the origin, but you should choose the correct general relationship. ANSWER:
E(r) = E(r) =
dV (r) dr V (r) r
E(r) = − E(r) = −
dV (r) dr
V (r) r
Correct Now consider the figure, which shows several functions of the variable r .
Part E Which curve could indicate the magnitude of the electric field due to a charge q located at the origin (r
= 0
)?
Hint 1. How to approach the problem Look at the general expression for the electric field determined earlier. What sort of rdependence does it https://session.masteringphysics.com/myct/assignmentPrintView?displayMode=studentView&assignmentID=3924419
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have? ANSWER: A B C D E F
Correct
Part F Which curve could indicate the electric potential due to a positive charge q located at the origin (r
= 0
)?
Hint 1. How to approach the problem Look at the general expression for the electric potential determined earlier. What sort of rdependence does it have? How does its sign depend on q ? ANSWER: A B C D E F
Correct
Part G Which curve could indicate the electric potential due to a negative charge q located at the origin (r
= 0
)?
ANSWER:
https://session.masteringphysics.com/myct/assignmentPrintView?displayMode=studentView&assignmentID=3924419
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A B C D E F
Correct
Part H For either a positive or a negative charge, the electric field points from regions of ______ electric potential. ANSWER: higher to lower lower to higher
Correct
Potential of a Charged Ring A ring with radius R and a uniformly distributed total charge Q lies in the xy plane, centered at the origin.
Part A What is the potential V (z) due to the ring on the z axis as a function of z ? 1 https://session.masteringphysics.com/myct/assignmentPrintView?displayMode=studentView&assignmentID=3924419 =
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Express your answer in of Q , z , R, and ϵ 0 or k =
1 4πϵ 0
.
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 q point charge: V (r) = , where V (r) is the potential at distance r from the point charge, q is the 4πϵ 0 r
magnitude of the charge, and ϵ 0 is the permittivity of free space. ANSWER:
V (z)
Q
=
4πϵ 0 √R
2
+z
2
Correct
Part B What is the magnitude of the electric field E⃗ on the z axis as a function of z , for z
> 0
?
Express your answer in of some or all of the quantities Q , z , R, and ϵ 0 or k =
1 4πϵ 0
.
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: ^ i
^ j
^ k
Hint 2. The relationship between electric field and potential You can obtain the electric field from a potential by the following expression: ⃗ ⃗ E = −∇ V = −
∂V ∂x
^ i −
∂V ∂y
^ j−
∂V ∂z
^ k,
https://session.masteringphysics.com/myct/assignmentPrintView?displayMode=studentView&assignmentID=3924419
∂f
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where E⃗ is the electric field vector, V is the electric potential, ∇⃗ is the gradient operator, and the partial derivative of f with respect to t.
∂f ∂t
means
ANSWER: kQz 3
| E⃗(z) | = (R
2
+z
2
)
2
Correct Notice that while the potential is a strictly decreasing function of z , the electric field first increases till z=
R
and then starts to decrease.
√2
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 z , the vector cancellation from points on opposite sides of the ring becomes very strong for small z . ⃗ E (z = 0) = 0
on of these vector cancellations. On the other hand E⃗(z→∞)→0, even though all
the individual dE⃗ 's point in (almost) the same direction there, because the contribution to the electric field, per unit length of the ring
dE dr
⃗
→0 as z→∞ .
± Equipotential Surfaces in a Capacitor
Part A Is the electric potential energy of a particle with charge q the same at all points on an equipotential surface?
Hint 1. Formula for electric potential energy For a particle with charge q at potential V , the electric potential energy is qV . ANSWER: Yes No
Correct For a particle with charge q on an equipotential surface at potential V , the electric potential energy has a constant value qV .
Part B https://session.masteringphysics.com/myct/assignmentPrintView?displayMode=studentView&assignmentID=3924419
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What is the work required to move a charge around on an equipotential surface at potential V with constant speed?
Hint 1. A formula for work The total work done on an object is equal to the change in its energy (potential + kinetic). ANSWER: Work = 0
Correct Since the speed of the charge is constant as it moves along the equipotential surface, and the electric potential energy is constant on that surface, there is no change in the total energy of the charge. This also means that no work is done by the charge as it moves along the equipotential surface.
Part C What is the work done by the electric field on a charge as it moves along an equipotential surface at potential V ?
Hint 1. Work done by an electric field on a charged particle The force due to an electric field is a conservative force. As such, the work done by such a force is equal to the change in the potential energy of the particle it is acting on. ANSWER: Work done by the electric field = 0
Correct Just as in Part B, since there is no change in the electric potential energy, no work is done by the electric field as the charge moves along the equipotential surface.
Part D ⃗ The work W E done by the uniform electric field E⃗ in displacing a particle with charge q along the path d is given by ⃗ ⃗ ⃗ ⃗ W E = qE ⋅ d = q|E ||d | cos θ
,
⃗ ⃗ where θ is the angle between E⃗ and d . Since in general, E is not equal to zero, for points on an equipotential surface, what must θ be for W E to equal 0?
Express your answer in radians. ANSWER:
https://session.masteringphysics.com/myct/assignmentPrintView?displayMode=studentView&assignmentID=3924419
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= 1.57 rad
Correct You have shown that equipotential surfaces are always perpendicular to the electric field at their surface.
Now assume that a parallelplate capacitor is attached across the terminals of a battery as shown in . The electric field ⃗
in the region between the plates points in the negative z direction, from higher to lower voltage. E
Part E Find the electric potential V (x, y, z) at a point X system O = (0, 0, 0) is at potential 0.
= (x, y, z)
inside the capacitor if the origin of the coordinate
Express your answer in of some or all of the variables E , x, y, and z .
Hint 1. The relation between electric potential and the electric field ield E⃗ in a region of space is ⃗ ⃗ V = − ∫ E ⋅ dℓ ,
where the line integral may be taken along any path ℓ .
Hint 2. Expressing an infinitesimal length element In general, a small length vector along the path of choice can be written as ⃗ ^ dℓ = dx ^ ı + dy ^ ȷ + dz k
.
Substitute this expression into the integral for V .
Hint 3. Analysis of the equation ^ are perpendicular (where w ^ is one of the Cartesian coordinate axes), then Recall that if E⃗ and w ⃗ ^ = 0 E ⋅w
^ , since θ = π/2. According to the setup of this part, only one of the directions (^ ı , ^ ȷ , k ) will not be perpendicular to the electric field as defined.
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ANSWER: V (x, y, z)
=
Ez
Correct Therefore, the equation of an equipotential surface at a potenial V0 is given by z=
V0 E
.
This is the equation of a plane that is parallel to the plates of the capacitor and perpendicular to the electric field. In particular, the lower plate, which is at zero potential, corresponds to the surface z = 0 .
Part F What is the distance Δz between two surfaces separated by a potential difference ΔV ? Express your answer in of E and ΔV .
Hint 1. How to approach the problem Use the equation you found in Part E to find equations that represent the potentials V1 and V2 of the planes located at z 1 and z 2 . Use these expressions to find an equation for Δz . ANSWER: Δz
=
ΔV E
Correct
Exercise 23.6 https://session.masteringphysics.com/myct/assignmentPrintView?displayMode=studentView&assignmentID=3924419
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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 ±e and the H − N distance is 0.110 nm.
Part A Calculate the electric potential energy of the adenine–thymine bond, using combinations of molecules O − H − N and N − H − N . Express your answer with the appropriate units. ANSWER: Uadenine−thymine
= −9.71×10−19 J
Correct
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 nm from the proton. ANSWER: U adenine−thymine UH
= 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 V and V/m https://session.masteringphysics.com/myct/assignmentPrintView?displayMode=studentView&assignmentID=3924419
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12.0 V/m , respectively. (Take the potential to be zero at infinity.)
Part A What is the distance to the point charge? ANSWER: d
= 0.415 m
Correct
Part B What is the magnitude of the charge? ANSWER: q
= 2.30×10−10 C
Correct
Part C Is the electric field directed toward or away from the point charge? ANSWER: Toward Away
Correct
Exercise 23.23
Part A An electron is to be accelerated from a velocity of 2.50×106 m/s to a velocity of 9.00×106 m/s . Through what potential difference must the electron to accomplish this? ANSWER: V1 − V2
= 213 V
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Correct
Part B Through what potential difference must the electron if it is to be slowed from 9.00×106 m/s to a halt? ANSWER: V1 − V2
= 231 V
Correct
Exercise 23.32 A very long insulating cylinder of charge of radius 3.00 cm carries a uniform linear density of 13.0 nC/m .
Part A If you put one probe of a voltmeter at the surface, how far from the surface must the other probe be placed so that the voltmeter reads 155 V ? ANSWER: d
= 2.82 cm
Correct
Exercise 23.48 A metal sphere with radius r a = 1.30 cm is ed on an insulating stand at the center of a hollow, metal, spherical shell with radius r b = 9.60 cm . Charge +q is put on the inner sphere and charge −q on the outer spherical shell. The magnitude of q is chosen to make the potential difference between the spheres 440 V , with the inner sphere at higher potential.
Part A Calculate q . ANSWER: q
= 7.36×10−10 C
Correct https://session.masteringphysics.com/myct/assignmentPrintView?displayMode=studentView&assignmentID=3924419
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Part B Are the electric field lines and equipotential surfaces mutually perpendicular? ANSWER: Yes No
Correct
Part C Are the equipotential surfaces closer together when the magnitude of E⃗ is largest? ANSWER: Equipotential surfaces closer together when the magnitude of E⃗ is largest. Equipotential surfaces closer together when the magnitude of E⃗ is smallest.
Correct
Problem 23.89 An alpha particle with kinetic energy 14.5 MeV 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 L = p0 b, where p0 is the magnitude of the initial momentum of the alpha particle and b=1.40×10−12 m . (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: r
= 1.41×10−12 m
Answer Requested
Part B https://session.masteringphysics.com/myct/assignmentPrintView?displayMode=studentView&assignmentID=3924419
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Repeat for b=1.00×10−13 m . ANSWER: r
= 1.08×10−13 m
Answer Requested
Part C Repeat for b=1.30×10−14 m . ANSWER: r
= 2.35×10−14 m
Answer Requested Score Summary: Your score on this assignment is 92.9%. You received 13.93 out of a possible total of 15 points.
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Ch 23 HW
Ch 23 HW Due: 11:59pm on Tuesday, November 10, 2015 You will receive no credit for items you complete after the assignment is due. Grading Policy
Electric Potential Energy versus Electric Potential Learning Goal: To understand the relationship and differences between electric potential and electric potential energy. In this problem we will learn about the relationships between electric force F ⃗ , electric field E⃗, potential energy U , and electric potential V . To understand these concepts, we will first study a system with which you are already familiar: the uniform gravitational field.
Gravitational Force and Potential Energy First we review the force and potential energy of an object of mass m in a uniform gravitational field that points ^ downward (in the −k direction), like the gravitational field near the earth's surface.
Part A ⃗ Find the force F (z) on an object of mass m in the uniform gravitational field when it is at height z
= 0
.
⃗ ^ Express F (z) in of m, z , k , and g .
ANSWER: ⃗ F (z)
=
^ −mgk
Correct Because we are in a uniform field, the force does not depend on the object's location. Therefore, the variable z does not appear in the correct answer.
Part B Now find the gravitational potential energy U (z) of the object when it is at an arbitrary height z . Take zero potential to be at position z = 0 . Keep in mind that the potential energy is a scalar, not a vector. Express U (z) in of m, z , and g . ANSWER: U (z)
=
mgz
Correct
Part C https://session.masteringphysics.com/myct/assignmentPrintView?displayMode=studentView&assignmentID=3924419
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In what direction does the object accelerate when released with initial velocity upward? ANSWER: upward downward upward or downward depending on its mass downward only if the ratio of g to initial velocity is large enough
Correct
Electric Force and Potential Energy Now consider the analogous case of a particle with charge q placed in a uniform electric field of strength E , pointing ^ downward (in the −k direction)
Part D ⃗ Find F (z) , the electric force on the charged particle at height z . ⃗ ^ Express F (z) in of q , E , z , and k .
Hint 1. Relationship between force and electric field The force on a particle of charge q in an electric field E⃗
^ = −E k
is given by F ⃗ =
qE
⃗
.
ANSWER: ⃗ F (z)
=
^ −qEk
Correct
Part E Now find the potential energy U (z) of this charged particle when it is at height z . Take zero potential to be at position z = 0 . Express U (z) in of q , E , and z . ANSWER: U (z)
=
qEz
Correct https://session.masteringphysics.com/myct/assignmentPrintView?displayMode=studentView&assignmentID=3924419
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Part F In what direction does the charged particle accelerate when released with upward initial velocity? ANSWER: upward downward upward or downward depending on its charge downward only if the ratio of qE to initial velocity is large enough
Correct
Electric Field and Electric Potential The electric potential V is defined by the relationship U with charge q .
= qV ,
where U is the electric potential energy of a particle
Part G ^ Find the electric potential V of the uniform electric field E⃗ = Ek . Note that this field is not pointing in the same direction as the field in the previous section of this problem. Take zero potential to be at position z = 0 .
Express V in of q , E , and z . ANSWER: V
=
−Ez
Correct The SI unit for electric potential is the volt (V). The volt is a derived unit, which means that it can be written in of other SI units. In of the fundamental units of length, mass, time and charge, the volt can be expressed as follows: 1V =
kg m
2
s2 C
Part H The electric field can be derived from the electric potential, just as the electrostatic force can be determined from the electric potential energy. The relationship between electric field and electric potential is E⃗ is the gradient operator: ⃗ ∇V =
The partial derivative
∂V ∂x
∂V ∂x
^ i +
∂V ∂y
^ j+
∂V ∂z
⃗ = −∇V
, where ∇⃗
^ k.
means the derivative of V with respect to x, holding all other variables constant.
⃗ https://session.masteringphysics.com/myct/assignmentPrintView?displayMode=studentView&assignmentID=3924419
^
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Consider again the electric potential V z coordinate only, so
∂V ∂x
=
∂V ∂y
= −Ez
= 0 and
∂V ∂z
corresponding to the field E⃗ =
dV dz
^ = Ek
. This potential depends on the
.
Find an expression for the electric field E⃗ in of the derivative of V . ^ ^ Express your answer as a vector in of the unit vectors ^ i , j , and/or k . Use dV /dz for the derivative of V with respect to z .
ANSWER:
E
⃗
=
−
dV dz
^ k
Correct
Part I A positive test charge will accelerate toward regions of ________ electric potential and ________ electric potential energy. Choose the appropriate answer combination to fill in the blanks correctly.
Hint 1. Direction of the electric field In what direction do electric fields point? ANSWER: from regions of higher electric potential to lower electric potential from regions of lower electric potential to higher electric potential
Hint 2. Formula for the force on a charge in an electric field The force F ⃗ on a charge q in an electric field E⃗ is given by ⃗ ⃗ F = qE
This is similar to the equation F
=
. mg for the force on a mass in a uniform gravitational field.
Hint 3. Formula for electric potential energy The electric potential energy U of a charge q at electric potential V is given by . , for the gravitational potential energy of a particle with
U = qV
This is similar to the equation Ug mass m.
= mVg = mgh
ANSWER:
https://session.masteringphysics.com/myct/assignmentPrintView?displayMode=studentView&assignmentID=3924419
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higher; higher higher; lower lower; higher lower; lower
Correct
Part J A negative test charge will accelerate toward regions of ________ electric potential and ________ electric potential energy. Choose the appropriate answer combination to fill in the blanks correctly.
Hint 1. Direction of the electric field In what direction do electric fields point? ANSWER: from regions of higher electric potential to lower electric potential from regions of lower electric potential to higher electric potential
Hint 2. Formula for the force on a charge in an electric field The force F ⃗ on a charge q in an electric field E⃗ is given by ⃗ ⃗ F = qE
This is similar to the equation F
= mg
. for the force on a mass in a uniform gravitational field.
Hint 3. Formula for electric potential energy The electric potential energy U of a charge q at electric potential V is given by . This is similar to the equation Ug = mVg = mgh for the gravitational potential energy of a particle with mass m. It works even for negative charges. Negative masses are not known to exist. U = qV
ANSWER: higher; higher higher; lower lower; higher lower; lower
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Correct A charge in an electric field will experience a force in the direction of decreasing potential energy. Since the electric potential energy of a negative charge is equal to the charge times the electric potential (U = qV ), the direction of decreasing electric potential energy is the direction of increasing electric potential.
Electric Potential, Potential Energy, and Force Learning Goal: To review relationships among electric potential, electric potential energy, and force on a test charge This problem is a review of the relationship between an electric field E⃗, its associated electric potential V , the electric potential energy U , and the direction of force on a test charge.
Part A Electric field lines always begin at _______ charges (or at infinity) and end at _______ charges (or at infinity). One could also say that the lines we use to represent an electric field indicate the direction in which a _______ test charge would initially move when released from rest. Which of the following fills in the three missing words correctly? ANSWER: (positive; negative; negative) (positive; negative; positive) (negative; positive; negative) (negative; positive; positive)
Correct Note that the electric field vector E⃗ is everywhere tangent to the electric field lines. Like electric field lines, the electric field vector generally points away from positive charges and toward negative charges.
Part B Would a positive test charge released from rest move toward a region of higher or lower electric potential (compared to the electric potential at the point where it is released)?
Hint 1. Potential, field, and force The equation relating an electric field to electric potential is E⃗
⃗ = −∇V
. By definition, ∇⃗V points in the
direction in which V is most rapidly increasing. Thus −∇⃗V (or the electric field) points in the direction in which V is most rapidly decreasing. The force on a charge q in an electric field is given by F ⃗ = positive charge will experience a force in the same direction as the electric field.
qE
⃗
, so a
ANSWER: https://session.masteringphysics.com/myct/assignmentPrintView?displayMode=studentView&assignmentID=3924419
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higher electric potential lower electric potential
Correct
Part C Now imagine that the sign of our test particle is changed from positive to negative, but the electric potential remains the same. Which of the following statements is correct?
Hint 1. Direction of field and force The problem states that the electric potential does not change. The electric field is determined from ⃗ ⃗ E = −∇V
, and the force from F ⃗ = qE⃗ , where q is the charge. When the sign of the test charge changes, but the electric potential remains the same, what happens to the direction of the electric field and the direction of the force on the charge? ANSWER: Both the electric field and force are unchanged. The force is unchanged but the electric field is reversed. Both the electric field and force are reversed. The electric field is unchanged but the force is reversed.
Hint 2. Direction of force and potential energy gradient Like other forms of potential energy, the electric potential energy U is always intimately related to the electric force by the equation F ⃗ =
⃗ −∇U
. Note that this relationship does not depend on the sign of the
⃗ ∇U
test charge. By definition, points in the direction in which U is most rapidly increasing. Thus −∇⃗U (or the force) points in the direction in which U is most rapidly decreasing. ANSWER: The direction of the force will change and it will point to regions of higher potential energy. The direction of the force will not change and it will point to regions of higher potential energy. The direction of the force will not change and it will point to regions of lower potential energy. The direction of the force will change and it will point to regions of lower potential energy.
Correct
PhET Tutorial: Charges and Electric Potential https://session.masteringphysics.com/myct/assignmentPrintView?displayMode=studentView&assignmentID=3924419
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Learning Goal: To understand the spatial distribution of the electric potential for a variety of simple charge configurations, and to understand how the electric field and electric potential (voltage) are related. For this problem, use the PhET simulation Charges and Fields. This simulation allows you to place multiple positive and negative pointcharges in any configuration and look at the resulting electric field and corresponding electric potential.
Start the simulation. You can click and drag positive charges (red) or negative charges (blue) into the main screen. If you select Show Efield in the green menu, red arrows will appear, showing the direction of the electric field. Faint red arrows indicate that the electric field is weaker than at locations where the arrows are brighter (this simulation does not use arrow length as a measure of field magnitude). You can drag a positive charge into the main screen and place the crosshairs of the electric potential (voltage) measurement tool on top of any point for which you would like to see the voltage (for this tutorial, voltage will be taken to be the same as electric potential). Clicking plot on this device shows an equipotential line. All points along an equipotential line have the same potential. Click Clear All before beginning Part A.
Part A The electric potential (voltage) at a specific location is equal to the potential energy per unit charge a charged object would have if it were at that location. If the zero point of the voltage is at infinity, the numerical value of the voltage is equal to the numerical value of work done to bring in a unit charge from infinity to that location. Select Show numbers and grid in the green menu, and drag one positive charge to the middle of the screen, right on top of two intersecting bold grid lines. Using the voltage meter, you should find that 1 m away from the charge, the voltage is 9 V. What is the voltage 2 m away from the charge? Express your answer numerically in volts to two significant figures. ANSWER: 4.5 V
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Correct Unlike the magnitude of the electric field, the electric potential (voltage) is not proportional to the inverse of the distance squared.
Part B What is the voltage 3 m away from the charge? ANSWER: 1 V 3 V 9 V
Correct Based on this result, and the previous question, the electric potential (voltage) is inversely proportional to the distance r from the charge: V ∝ 1/r. Recall that the magnitude of the electric field E ∝ 1/r 2 .
Part C Another way to study voltage and its relationship to electric field is by producing equipotential lines. Just like every point on a contour line has the same elevation in a topographical map, every point on an equipotential line has the same voltage. Click plot on the voltage tool to produce an equipotential line. Produce many equipotential lines by clicking plot as you move the tool around. You should produce a graph that looks similar to the one shown below.
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Place several EField Sensors at a few points on different equipotential lines, and look at the relationship between the electric field and the equipotential lines. Which statement is true? ANSWER: At any point, the electric field is parallel to the equipotential line at that point. At any point, the electric field is perpendicular to the equipotential line at that point, and it is directed toward lines of lower voltages. At any point, the electric field is perpendicular to the equipotential line at that point, and it is directed toward lines of higher voltages.
Correct All points on an equipotential line have the same voltage; thus, no work would be done in moving a test charge along an equipotential line. No work is done because the electric field, and thus the force on the test charge, is perpendicular to the displacement of the test charge being moved along the equipotential line.
Part D Equipotential lines are usually shown in a manner similar to topographical contour lines, in which the difference in the value of consecutive lines is constant. Clear the equipotential lines using the Clear button on the voltage tool. Place the first equipotential line 1 m away from the charge. It should have a value of roughly 9 V. Now, produce several additional equipotential lines, increasing and decreasing by an interval of 3 V (e.g., one with 12 V, one with 15 V, and one with 6 V). Don’t worry about getting these exact values. You can be off by a few tenths of a volt. Which statement best describes the distribution of the equipotential lines? ANSWER: The equipotential lines are equally spaced. The distance between each line is the same for all adjacent lines. The equipotential lines are closer together in regions where the electric field is weaker. The equipotential lines are closer together in regions where the electric field is stronger.
Correct Near the positive charge, where the electric field is strong, the voltage lines are close to each other. Farther from the charge, the electric field is weaker and the lines are farther apart.
Part E Now, remove the positive charge by dragging it back to the basket, and drag one negative charge toward the middle of the screen. Determine how the voltage is different from that of the positive charge. How does the voltage differ from that of the positive charge? ANSWER:
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The voltages become negative instead of positive and keep the same magnitudes. The voltage distribution does not change. The voltages are positive, but the magnitude increases with increasing distance.
Correct The voltage is still inversely proportional to the distance from the charge, but the voltage is negative everywhere rather than positive.
Part F Now, remove the negative charge, and drag two positive charges, placing them 1 m apart, as shown below.
What is the voltage at the midpoint of the two charges? ANSWER: Exactly twice the voltage produced by only one of the charges at the same point Zero Greater than zero, but less than twice the voltage produced by only one of the charges at the same point
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Correct Because voltage is a scalar quantity, there are no vector components with opposite directions canceling out, as for electric fields. The voltage is simply the sum of the voltages due to each of the individual charges. Since both charges are positive, the voltage due to each charge (at all locations) is positive.
Part G Now, make an electric dipole by replacing one of the positive charges with a negative charge, so the final configuration looks like the figure shown below.
What is the voltage at the midpoint of the dipole?
Hint 1. How to approach the problem The total voltage is equal to the sum of the voltages due to each of the charges. The voltage due to each charge simply depends on the distance from the charge and the charge’s value. Voltage is a scalar quantity, so it does not add like a vector. ANSWER:
The voltage at the midpoint of the dipole is
positive. zero. negative.
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Correct Because the voltage due to the negative charge has the opposite sign of the voltage due to the positive charge at the midpoint, the net voltage is zero. The electric field, however, is not zero here!
Part H Make several equipotential lines similar to the figure below.
Try to have the equipotential lines equally spaced in voltage. Then, use an EField Sensor to measure the electric field at a few points while looking at the relationship between the electric field and the equipotential lines. Which of the following statements is true? ANSWER: The electric field strength is greatest where the voltage is the greatest. The electric field strength is greatest where the voltage is the smallest. The electric field strength is greatest where the equipotential lines are very close to each other.
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Correct Locations where the voltage is changing steeply are locations with a strong electric field. The magnitude of the electric field is equal to the rate the voltage is changing with distance. Mathematically, this idea is conveyed by |Es | = dV /ds, where Es is the component of the electric field in the direction of a small displacement ds . (As you learned earlier, the electric field is directed in the direction where the voltage decreases.)
PhET Interactive Simulations University of Colorado http://phet.colorado.edu
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 conducting spheres are each given a charge Q . 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 E0 , 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:
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1/3 E0 3 E0 9 E0 1/9 E0 E0
Correct The larger sphere has nine times the surface area of the smaller one, and this reduces the surface charge density by a factor of nine.
Bouncing Electrons Two electrons, each with mass m and charge q , are released from positions very far from each other. With respect to a certain reference frame, electron A has initial nonzero speed v toward electron B in the positive x direction, and electron B has initial speed 3v toward electron A in the negative x direction. The electrons move directly toward each other along 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 r min that the electrons reach? Express your answer in term of q , m, v , and k (where k =
1 4πϵ 0
).
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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 their minimum separation r min . 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 energy and momentum equations to obtain r min .
Hint 2. Find the initial energy What is the total energy Einitial of the two electrons when they are initially released? Assume that the electrons are so far apart that their potential energy is zero. Express your answer in of m and v . ANSWER: Einitial
=
5mv
2
Hint 3. Find the final energy What is the total energy Ef inal of the electrons when they reach their minimum separation r min ? Assume that the (identical) speed of the two electrons is u. Express your answer in of m, u, q , r min , and k (where k =
1 4πϵ 0
).
Hint 1. Find the final kinetic energy What is the final kinetic energy (both electrons)? Express your answer in of u, and m. ANSWER: Kf inal
=
mu
2
Hint 2. Find the final potential energy What is the final potential energy of this 2electron system? Express your answer in of k , q , and r min . ANSWER:
Uf inal
=
kq
2
rmin
ANSWER:
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Ef inal
=
mu
2
+
kq
2
rmin
Hint 4. Find the initial momentum ⃗ What is the total momentum p initial of the two electrons when they are initially released?
Express your answer as a vector in of m, v , and ^ i. ANSWER: p⃗
initial
=
^ −2mv i
Hint 5. Find the final momentum What is the total momentum p f⃗ inal of the two electrons when they reach their minimum separation r min ? Assume that the (identical) velocity of the two electrons is u⃗. Express your answer as a vector in of m and u⃗. ANSWER: p⃗
f inal
=
2mu⃗
Hint 6. Some math help From the momentum equations, ∣∣u⃗∣∣ equation to find r min .
= v
; that is, u
= v
. Substitute for u in the energy conservation
ANSWER:
r min
kq
=
2
4mv 2
Correct An experienced physicist might approach this problem by considering the system of electrons in a reference frame in which the initial momentum is zero. In this frame the initial speed of each electron is 2v . Try solving the problem this way. Make sure that you obtain the same result for r min , and decide for yourself which approach is easier.
Energy Stored in a Charge Configuration Four point charges, A, B, C, and D, are placed at the corners of a square with side length L. Charges A, B, and C have charge + q , and D has charge − q . Throughout this problem, use k in place of
1 4πϵ 0
.
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Part A If you calculate W , the amount of work it took to assemble this charge configuration if the point charges were initially infinitely far apart, you will find that the contribution for each charge is proportional to provided, enter the numeric value that multiplies the above factor, in W .
kq L
2
. In the space
Hint 1. How to approach the problem 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 r from a charge q is V
=k
q r
, where k =
1 4πϵ 0
.
Note that this equation implicitly defines the electric potential to be zero at r = ∞. The electric potential energy of a charge q is equal to qV , where V 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 W A , the work required to assemble the charge distribution shown in the figure? Express your answer in of some or all of the variables k , q , and L.
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Hint 1. Find the potential at the location of charge A What is VA , the electric potential at the upper left corner of the square before charge A is placed there? Express your answer in of some or all of the variables k , q , and L. ANSWER: VA
= 0
ANSWER: WA
= 0
Hint 4. Work required to place charge B What is W B , the amount of work required to add charge B to the configuration, as shown in the figure? Express your answer in of some or all of the variables k , q , and L.
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Hint 1. Find the potential at the location of charge B What is VB , the potential at the upper right corner due to charge A, before charge B is placed there? Express your answer in of some or all of the variables k , q , and L. ANSWER:
VB
kq
=
L
ANSWER:
WB
=
kq
2
L
Hint 5. Work required to place charge C What is W C , the amount of work required to add charge C to the configuration, as shown in the figure? Express your answer in of some or all of the variables k , q , and L.
Hint 1. Find the potential at the location of charge C What is VC , 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 of some or all of the variables k , q , and L. ANSWER: https://session.masteringphysics.com/myct/assignmentPrintView?displayMode=studentView&assignmentID=3924419
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=
VCB
kq L
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 of some or all of the variables k , q , and L. ANSWER: =
VCA
kq √2 L
ANSWER:
VC
kq
=
L
(1 +
1 √2
)
ANSWER:
WC
=
kq L
2
(1 +
1 √2
)
Hint 6. Find the work required to place charge D What is W D , the amount of work required to add charge D to the configuration? Express your answer in of some or all of the variables k , q , and L.
Hint 1. Find the potential at the position of charge D What is VD , the potential at the lower left corner of the square before charge D is placed there? https://session.masteringphysics.com/myct/assignmentPrintView?displayMode=studentView&assignmentID=3924419
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Express your answer in of some or all of the variables k , q , and L. ANSWER:
VD
=
kq L
(2 +
1 √2
)
ANSWER:
WD
=
−
kq
2
(2 +
L
1 √2
)
ANSWER:
W
= 0 ×(
kq L
2
)
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: W =
q
2
4πϵ 0
(
1 R BA
+
1 R BC
+
1 R CA
−
1 R DA
−
1 R DB
−
1 R DC
).
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 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 q .
ANSWER: https://session.masteringphysics.com/myct/assignmentPrintView?displayMode=studentView&assignmentID=3924419
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figure a figure b figure c
Correct
Electric Force and Potential: Spherical Symmetry Learning Goal: To understand the electric potential and electric field of a point charge in three dimensions Consider a positive point charge q , located at the origin of threedimensional space. Throughout this problem, use k in place of
1 4πϵ 0
.
Part A Due to symmetry, the electric field of a point charge at the origin must point _____ from the origin. Answer in one word. ANSWER: radial
Correct
Part B Find E(r) , the magnitude of the electric field at distance r from the point charge q . Express your answer in of r , k , and q . ANSWER:
E(r)
=
kq r2
Correct
Part C Find V (r) , the electric potential at distance r from the point charge q . Express your answer in of r , k , and q . ANSWER: https://session.masteringphysics.com/myct/assignmentPrintView?displayMode=studentView&assignmentID=3924419
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V (r)
=
kq r
Correct
Part D Which of the following is the correct relationship between the magnitude of a radial electric field E(r) and its associated electric potential V (r) ? More than one answer may be correct for the particular case of a point charge at the origin, but you should choose the correct general relationship. ANSWER:
E(r) = E(r) =
dV (r) dr V (r) r
E(r) = − E(r) = −
dV (r) dr
V (r) r
Correct Now consider the figure, which shows several functions of the variable r .
Part E Which curve could indicate the magnitude of the electric field due to a charge q located at the origin (r
= 0
)?
Hint 1. How to approach the problem Look at the general expression for the electric field determined earlier. What sort of rdependence does it https://session.masteringphysics.com/myct/assignmentPrintView?displayMode=studentView&assignmentID=3924419
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have? ANSWER: A B C D E F
Correct
Part F Which curve could indicate the electric potential due to a positive charge q located at the origin (r
= 0
)?
Hint 1. How to approach the problem Look at the general expression for the electric potential determined earlier. What sort of rdependence does it have? How does its sign depend on q ? ANSWER: A B C D E F
Correct
Part G Which curve could indicate the electric potential due to a negative charge q located at the origin (r
= 0
)?
ANSWER:
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A B C D E F
Correct
Part H For either a positive or a negative charge, the electric field points from regions of ______ electric potential. ANSWER: higher to lower lower to higher
Correct
Potential of a Charged Ring A ring with radius R and a uniformly distributed total charge Q lies in the xy plane, centered at the origin.
Part A What is the potential V (z) due to the ring on the z axis as a function of z ? 1 https://session.masteringphysics.com/myct/assignmentPrintView?displayMode=studentView&assignmentID=3924419 =
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Express your answer in of Q , z , R, and ϵ 0 or k =
1 4πϵ 0
.
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 q point charge: V (r) = , where V (r) is the potential at distance r from the point charge, q is the 4πϵ 0 r
magnitude of the charge, and ϵ 0 is the permittivity of free space. ANSWER:
V (z)
Q
=
4πϵ 0 √R
2
+z
2
Correct
Part B What is the magnitude of the electric field E⃗ on the z axis as a function of z , for z
> 0
?
Express your answer in of some or all of the quantities Q , z , R, and ϵ 0 or k =
1 4πϵ 0
.
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: ^ i
^ j
^ k
Hint 2. The relationship between electric field and potential You can obtain the electric field from a potential by the following expression: ⃗ ⃗ E = −∇ V = −
∂V ∂x
^ i −
∂V ∂y
^ j−
∂V ∂z
^ k,
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∂f
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where E⃗ is the electric field vector, V is the electric potential, ∇⃗ is the gradient operator, and the partial derivative of f with respect to t.
∂f ∂t
means
ANSWER: kQz 3
| E⃗(z) | = (R
2
+z
2
)
2
Correct Notice that while the potential is a strictly decreasing function of z , the electric field first increases till z=
R
and then starts to decrease.
√2
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 z , the vector cancellation from points on opposite sides of the ring becomes very strong for small z . ⃗ E (z = 0) = 0
on of these vector cancellations. On the other hand E⃗(z→∞)→0, even though all
the individual dE⃗ 's point in (almost) the same direction there, because the contribution to the electric field, per unit length of the ring
dE dr
⃗
→0 as z→∞ .
± Equipotential Surfaces in a Capacitor
Part A Is the electric potential energy of a particle with charge q the same at all points on an equipotential surface?
Hint 1. Formula for electric potential energy For a particle with charge q at potential V , the electric potential energy is qV . ANSWER: Yes No
Correct For a particle with charge q on an equipotential surface at potential V , the electric potential energy has a constant value qV .
Part B https://session.masteringphysics.com/myct/assignmentPrintView?displayMode=studentView&assignmentID=3924419
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What is the work required to move a charge around on an equipotential surface at potential V with constant speed?
Hint 1. A formula for work The total work done on an object is equal to the change in its energy (potential + kinetic). ANSWER: Work = 0
Correct Since the speed of the charge is constant as it moves along the equipotential surface, and the electric potential energy is constant on that surface, there is no change in the total energy of the charge. This also means that no work is done by the charge as it moves along the equipotential surface.
Part C What is the work done by the electric field on a charge as it moves along an equipotential surface at potential V ?
Hint 1. Work done by an electric field on a charged particle The force due to an electric field is a conservative force. As such, the work done by such a force is equal to the change in the potential energy of the particle it is acting on. ANSWER: Work done by the electric field = 0
Correct Just as in Part B, since there is no change in the electric potential energy, no work is done by the electric field as the charge moves along the equipotential surface.
Part D ⃗ The work W E done by the uniform electric field E⃗ in displacing a particle with charge q along the path d is given by ⃗ ⃗ ⃗ ⃗ W E = qE ⋅ d = q|E ||d | cos θ
,
⃗ ⃗ where θ is the angle between E⃗ and d . Since in general, E is not equal to zero, for points on an equipotential surface, what must θ be for W E to equal 0?
Express your answer in radians. ANSWER:
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= 1.57 rad
Correct You have shown that equipotential surfaces are always perpendicular to the electric field at their surface.
Now assume that a parallelplate capacitor is attached across the terminals of a battery as shown in . The electric field ⃗
in the region between the plates points in the negative z direction, from higher to lower voltage. E
Part E Find the electric potential V (x, y, z) at a point X system O = (0, 0, 0) is at potential 0.
= (x, y, z)
inside the capacitor if the origin of the coordinate
Express your answer in of some or all of the variables E , x, y, and z .
Hint 1. The relation between electric potential and the electric field ield E⃗ in a region of space is ⃗ ⃗ V = − ∫ E ⋅ dℓ ,
where the line integral may be taken along any path ℓ .
Hint 2. Expressing an infinitesimal length element In general, a small length vector along the path of choice can be written as ⃗ ^ dℓ = dx ^ ı + dy ^ ȷ + dz k
.
Substitute this expression into the integral for V .
Hint 3. Analysis of the equation ^ are perpendicular (where w ^ is one of the Cartesian coordinate axes), then Recall that if E⃗ and w ⃗ ^ = 0 E ⋅w
^ , since θ = π/2. According to the setup of this part, only one of the directions (^ ı , ^ ȷ , k ) will not be perpendicular to the electric field as defined.
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ANSWER: V (x, y, z)
=
Ez
Correct Therefore, the equation of an equipotential surface at a potenial V0 is given by z=
V0 E
.
This is the equation of a plane that is parallel to the plates of the capacitor and perpendicular to the electric field. In particular, the lower plate, which is at zero potential, corresponds to the surface z = 0 .
Part F What is the distance Δz between two surfaces separated by a potential difference ΔV ? Express your answer in of E and ΔV .
Hint 1. How to approach the problem Use the equation you found in Part E to find equations that represent the potentials V1 and V2 of the planes located at z 1 and z 2 . Use these expressions to find an equation for Δz . ANSWER: Δz
=
ΔV E
Correct
Exercise 23.6 https://session.masteringphysics.com/myct/assignmentPrintView?displayMode=studentView&assignmentID=3924419
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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 ±e and the H − N distance is 0.110 nm.
Part A Calculate the electric potential energy of the adenine–thymine bond, using combinations of molecules O − H − N and N − H − N . Express your answer with the appropriate units. ANSWER: Uadenine−thymine
= −9.71×10−19 J
Correct
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 nm from the proton. ANSWER: U adenine−thymine UH
= 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 V and V/m https://session.masteringphysics.com/myct/assignmentPrintView?displayMode=studentView&assignmentID=3924419
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12.0 V/m , respectively. (Take the potential to be zero at infinity.)
Part A What is the distance to the point charge? ANSWER: d
= 0.415 m
Correct
Part B What is the magnitude of the charge? ANSWER: q
= 2.30×10−10 C
Correct
Part C Is the electric field directed toward or away from the point charge? ANSWER: Toward Away
Correct
Exercise 23.23
Part A An electron is to be accelerated from a velocity of 2.50×106 m/s to a velocity of 9.00×106 m/s . Through what potential difference must the electron to accomplish this? ANSWER: V1 − V2
= 213 V
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Correct
Part B Through what potential difference must the electron if it is to be slowed from 9.00×106 m/s to a halt? ANSWER: V1 − V2
= 231 V
Correct
Exercise 23.32 A very long insulating cylinder of charge of radius 3.00 cm carries a uniform linear density of 13.0 nC/m .
Part A If you put one probe of a voltmeter at the surface, how far from the surface must the other probe be placed so that the voltmeter reads 155 V ? ANSWER: d
= 2.82 cm
Correct
Exercise 23.48 A metal sphere with radius r a = 1.30 cm is ed on an insulating stand at the center of a hollow, metal, spherical shell with radius r b = 9.60 cm . Charge +q is put on the inner sphere and charge −q on the outer spherical shell. The magnitude of q is chosen to make the potential difference between the spheres 440 V , with the inner sphere at higher potential.
Part A Calculate q . ANSWER: q
= 7.36×10−10 C
Correct https://session.masteringphysics.com/myct/assignmentPrintView?displayMode=studentView&assignmentID=3924419
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Ch 23 HW
Part B Are the electric field lines and equipotential surfaces mutually perpendicular? ANSWER: Yes No
Correct
Part C Are the equipotential surfaces closer together when the magnitude of E⃗ is largest? ANSWER: Equipotential surfaces closer together when the magnitude of E⃗ is largest. Equipotential surfaces closer together when the magnitude of E⃗ is smallest.
Correct
Problem 23.89 An alpha particle with kinetic energy 14.5 MeV 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 L = p0 b, where p0 is the magnitude of the initial momentum of the alpha particle and b=1.40×10−12 m . (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: r
= 1.41×10−12 m
Answer Requested
Part B https://session.masteringphysics.com/myct/assignmentPrintView?displayMode=studentView&assignmentID=3924419
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Ch 23 HW
Repeat for b=1.00×10−13 m . ANSWER: r
= 1.08×10−13 m
Answer Requested
Part C Repeat for b=1.30×10−14 m . ANSWER: r
= 2.35×10−14 m
Answer Requested Score Summary: Your score on this assignment is 92.9%. You received 13.93 out of a possible total of 15 points.
https://session.masteringphysics.com/myct/assignmentPrintView?displayMode=studentView&assignmentID=3924419
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