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2a. Electric Potential
Due: 11:00pm on Sunday, January 17, 2010
Note: To understand how points are awarded, read your instructor's Grading Policy.
Electric Fields and Equipotential Surfaces
Description: Find the work done to move a unit charge from and to given points on a diagram showing equipotential surfaces, and compare the magnitude of the electric field at these points. The dashed lines in the diagram represent cross sections of equipotential surfaces drawn in 1increments.
Part A What is the work done by the electric force to move a 1charge from A to B?
Hint A.1 Find the potential difference between A and B What is the potential difference Hint A.1.1 Equipotential surfaces Recall that an equipotential surface is a surface on which the electric potential is the same at every point. Express your answer in volts. ANSWER: = between point A and point B?
Hint A.2 Potential difference and work Recall that the potential difference (in volts) between a point the electric force to move a 1Express your answer in joules. charge from to . and a point equals the work (in joules) done by
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ANSWER:
=
Part B What is the work Hint B.1 done by the electric force to move a 1charge from A to D?
Find the potential difference between A and D between point A and point D?
What is the potential difference Express your answer in volts. ANSWER: =
Hint B.2
Potential difference and work and a point equals the work (in joules) done by charge from to .
Recall that the potential difference (in volts) between a point the electric force to move a 1Express your answer in joules. ANSWER: =
Part C The magnitude of the electric field at point C is Hint C.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.
Energy Stored in a Charge Configuration
Description: Find the work required to assemble four charges of the same magnitude--three positive, one negative-at the corners of a square. Four point charges, A, B, C, and D, are placed at the corners of a square with side length have charge , and D has charge . . Charges A, B, and C
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Throughout this problem, use
in place of
.
Part A If you calculate , the amount of work it took to assemble this charge configuration if the point charges were . In the space
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 Hint A.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 A.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 A.3 Work required to place charge A What is , the work required to assemble the charge distribution shown in the figure?
Hint A.3.1 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
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ANSWER:
=
Express your answer in terms of some or all of the variables , , and ANSWER:
.
=
Hint A.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?
Hint A.4.1 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: =
Express your answer in terms of some or all of the variables , , and ANSWER: =
.
Hint A.5 Work required to place charge C
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What is
, the amount of work required to add charge C to the configuration, as shown in the figure?
Hint A.5.1 Find the potential at the location of charge C What is , the potential at the lower right corner of the square before charge C is placed there?
Hint A.5.1.1 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 A.5.1.2 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 A.5.1.3 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: =
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Express your answer in terms of some or all of the variables , , and ANSWER: =
.
Hint A.6 Find the work required to place charge D What is , the amount of work required to add charge D to the configuration?
Hint A.6.1 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: =
Express your answer in terms of some or all of the variables , , and ANSWER: =
.
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Because D has a negative charge, it is attracted to charges A, B, and C. This accounts for the negative sign on the work. We would have to do positive work to remove charge D from the configuration. ANSWER: =
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 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
Energy in Capacitors and Electric Fields
Description: Several questions about the energy of charged capacitors, energy density, and energy of an electrostatic field. Students are asked to calculate the energy of a charged capacitor by two different methods. Learning Goal: To be able to calculate the energy of a charged capacitor and to understand the concept of energy
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associated with an electric field. The energy of a charged capacitor is given by , where is the charge of the capacitor and is the
potential difference across the capacitor. The energy of a charged capacitor can be described as the energy associated with the electric field created inside the capacitor. In this problem, you will derive two more formulas for the energy of a charged capacitor; you will then use a parallelplate capacitor as a vehicle for obtaining the formula for the energy density associated with an electric field. It will be useful to recall the definition of capacitance, , and the formula for the capacitance of a parallel-plate capacitor, , where free space. First, consider a capacitor of capacitance Part A Find the energy of the capacitor in terms of and by using the definition of capacitance and the formula for that has a charge and potential difference . is the area of each of the plates and is the plate separation. As usual, is the permittivity of
the energy in a capacitor. Express your answer in terms of ANSWER: = and .
Part B Find the energy of the capacitor in terms of and by using the definition of capacitance and the formula for the energy in a capacitor. Express your answer in terms of ANSWER: = and .
All three of these formulas are equivalent: .
Depending on the problem, one or another may be more convenient to use. However, any one of them would give you the correct answer. Note that these formulas work for any type of capacitor.
Part C
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A parallel-plate capacitor is connected to a battery. The energy of the capacitor is
. The capacitor remains
connected to the battery while the plates are slowly pulled apart until the plate separation doubles. The new energy of the capacitor is . Find the ratio . Hint C.1 Determine what remains constant As the plates are being pulled apart slowly, what quantity or quantities remain constant? ANSWER:
capacitance only voltage only charge only both voltage and capacitance both voltage and charge
Since the geometry of the capacitor is changing, its capacitance changes, too. However, the voltage remains constant, since it must equal the provided voltage by the battery. Hint C.2 Identify which formula to use Which formula for energy is most convenient to use in this case? ANSWER:
ANSWER:
=
Part D A parallel-plate capacitor is connected to a battery. The energy of the capacitor is . The capacitor is then
disconnected from the battery and the plates are slowly pulled apart until the plate separation doubles. The new energy of the capacitor is . Find the ratio . Hint D.1 Determine what remains constant As the plates are being pulled apart, what quantity or quantities remain constant? ANSWER:
capacitance only voltage only charge only both voltage and charge both voltage and capacitance
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The charge remains constant, since the capacitor is disconnected and the charge therefore literally has nowhere to go. Hint D.2 Identify which formula to use Which formula for energy is most convenient to use in this case? ANSWER:
ANSWER:
=
In this part of the problem, you will express the energy of various types of capacitors in terms of their geometry and voltage. Part E A parallel-plate capacitor has area and plate separation , and it is charged to voltage of the capacitor. . Use the formulas from the problem introduction to obtain the formula for the energy Express your answer in terms of ANSWER: = ,,
, and appropriate constants.
Let us now recall that the energy of a capacitor can be thought of as the energy of the electric field inside the capacitor. The energy of the electric field is usually described in terms of energy density , the energy per unit volume. A parallel-plate capacitor is a convenient device for obtaining the formula for the energy density of an electric field, since the electric field inside it is nearly uniform. The formula for energy density can then be written as ,
where
is the energy of the capacitor and
is the volume of the capacitor (not its voltage).
Part F
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A parallel-plate capacitor has area
and plate separation , and it is charged so that the electric field inside is of the capacitor.
.
Use the formulas from the problem introduction to find the energy Hint F.1 How to approach the problem
Recall that for the uniform electric field
between the plates of a parallel-plate capacitor,
, where
is
the potential difference between the plates and to rewrite the equation for energy
is the distance between the two plates. You can use this relation in terms of the electric field and the geometry of the capacitor (i.e.,
the area of the plates and the distance between them). Express your answer in terms of ANSWER: = ,, , and appropriate constants.
As mentioned before, we can think of the energy of the capacitor as the energy of the electric field inside the capacitor. Part G Find the energy density the capacitor is . of the electric field in a parallel-plate capacitor. The magnitude of the electric field inside
Hint G.1 How to approach the problem Since the electric field outside a parallel-plate capacitor is essentially zero, the volume that you are looking for is the volume of the space between the two plates. Hint G.2 Volume between the plates Recall that the volume is given by of a solid with two parallel bases of the same shape and sides perpendicular to the bases is the area of each of the bases and is the distance between the bases. Note that , where
the space between the plates of a parallel-plate capacitor is such a solid. Express your answer in terms of ANSWER: = and appropriate constants.
Note that the answer for
does not contain any reference to the geometry of the capacitor:
and
do not
appear in the formula. In fact, the formula
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describes the energy density in any electrostatic field, whether created by a capacitor or any other source.
Ionic Potentials across Cell Membranes Conceptual Question
Description: Short conceptual problem dealing with ionic potentials across cell membranes. In its resting state, the membrane surrounding a neuron is permeable to potassium ions but not permeable to sodium ions. Thus, positive K ions can flow through the membrane in an attempt to equalize K concentration, but Na ions cannot. This leads to an excess of Na ions outside of the cell. If the space outside the cell is defined as zero electric potential, then the electric potential of the interior of the cell is negative. This resting potential is typically about 80 . A schematic of this situation is shown in the figure.
In response to a stimulus, the membrane can become permeable to Na ions. As Na ions rush into the cell, the interior of the cell reaches an electric potential of about 40 . This process is termed depolarization. In response to depolarization, the membrane again becomes impermeable to Na ions, and the K ions flow out of the interior of the cell (driven by the positive electric potential inside of the cell), reestablishing the resting potential. This is termed repolarization. Only a small percentage of the available Na and K ions participate in each depolarization/repolarization cycle, so the cell can respond to many stimuli in succession without depleting its "stock" of available Na and K ions. A graph of an electric potential inside a cell vs. time is shown in the next figure for a single depolarization/repolarization cycle.
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Part A During the resting phase, what is the electric potential energy of a typical Na ion outside of the cell? Hint A.1 The electron volt Electric potential energy is defined as . The electric charge on individual particles is always a multiple of the fundamental charge (the charge on a single proton). Rather than substituting a numerical value for , it is often more convenient to use the constant as a unit. Thus, a proton located at a potential of 100 has energy , which can be written as
or . Thus, the proton has 100 electron volts of energy. (Electron volts can be converted to the more traditional unit of energy, the joule, by multiplying by the conversion factor and recalling that . Thus, ANSWER: .)
40
+40 80 +80 0
Part B During the resting phase, what is the electrical potential energy of a typical K ion inside of the cell? Hint B.1 The electron volt
Electric potential energy is defined as . The electric charge on individual particles is always a multiple of the fundamental charge (the charge on a single proton). Rather than substituting a numerical value for , it is often more convenient to use the constant as a unit. Thus, a proton located at a potential of 100 has energy
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, which can be written as
or . Thus, the proton has 100 electron volts of energy. (Electron volts can be converted to the more traditional unit of energy, the joule, by multiplying by the conversion factor and recalling that . Thus, .)
ANSWER:
40
+40 80 +80 0
Part C During depolarization, what is the work done (by the electric field) on the first few Na ions that enter the cell? Hint C.1 The electron volt Electric potential energy is defined as . The electric charge on individual particles is always a multiple of the fundamental charge (the charge on a single proton). Rather than substituting a numerical value for , it is often more convenient to use the constant as a unit. Thus, a proton located at a potential of 100 has energy , which can be written as
or .
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Thus, the proton has 100 electron volts of energy. (Electron volts can be converted to the more traditional unit of energy, the joule, by multiplying by the conversion factor and recalling that . Thus, .)
Hint C.2 Algebraic sign of the work In general, work is defined as the product of the force applied parallel (or antiparallel) to the displacement of an object. Thus, . The work done by a force is positive if the force and the displacement are parallel; it is negative if the force and displacement are opposite in direction. Hint C.3 Magnitude of the work Work transfers energy into or out of a system. Therefore, in the absence of other energy transfers, the magnitude of the work done on an object is equal to the magnitude of the objects change in energy. Since the primary form of energy present in this example is electric potential energy, the magnitude of the work done is equal to the change in the ions electric potential energy. ANSWER:
40
+40 80 +80 120 +120 0
Part D During repolarization, what is the work done (by the electric field) on the first few K ions that exit the cell? Hint D.1 The electron volt
Electric potential energy is defined as . The electric charge on individual particles is always a multiple of the fundamental charge (the charge on a single proton). Rather than substituting a numerical value for , it is often more convenient to use the constant as a unit. Thus, a proton located at a potential of 100 has energy , which can be written as
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or . Thus, the proton has 100 electron volts of energy. (Electron volts can be converted to the more traditional unit of energy, the joule, by multiplying by the conversion factor and recalling that . Thus, .)
Hint D.2 Algebraic sign of the work In general, work is defined as the product of the force applied parallel (or antiparallel) to the displacement of an object. Thus, . The work done by a force is positive if the force and the displacement are parallel; the work done is negative if the force and displacement are opposite in direction. Hint D.3 Magnitude of the work Work transfers energy into or out of a system. Therefore, in the absence of other energy transfers, the magnitude of the work done on an object is equal to the magnitude of the objects change in energy. Since the primary form of energy present in this example is electric potential energy, the magnitude of the work done is equal to the change in the ions electric potential energy. ANSWER:
40
+40 80 +80 120 +120 0
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