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Physics 220
Assignment 6
Assignment is due at 11:59pm on Tuesday, October 17, 2006
Credit for problems submitted late will decrease to 0% over the course of 240 hour(s) after the deadline has passed. The wrong answer penalty is 1% per part. Multiple choice questions are penalized as described in the online help. There is no opened hint penalty or unopened hint bonus. You are allowed unlimited attempts per answer.
Class 16
Capacitance: A Review
Description: Multiple-choice questions on concept of capacitance. Use after Introduction to Capacitance. Some questions are repetitions of those in other problems. This fact may get some students frustrated. Learning Goal: To review the meaning of capacitance and ways of changing the capacitance of a parallel-plate capacitor. Capacitance is one of the central concepts in electrostatics. Understanding its meaning and the difference between its definition and the ways of calculating capacitance can be challenging at first. This tutorial is meant to help you become more comfortable with capacitance. Recall the fundamental formula for capacitance: , where is the capacitance in farads, is the charge stored on the plates in coulombs, and is the potential difference (or voltage) between the plates. In the following problems it may help to keep in mind that the voltage is related to the strength of the electric field and the distance between the plates, , by . Part A What property of objects is best measured by their capacitance? ANSWER: the ability to conduct electric current ability to store charge the ability to distort an external electrostatic field the
Capacitance is a measure of the ability of a system of two conductors to store electric charge and energy. It is defined as . This ratio remains constant as long as the system retains its geometry and the amount of dielectric does not change. Capacitors are special devices designed to combine a large capacitance with a small size. However, any pair of conductors separated by a dielectric (or vacuum) has some capacitance. Even an isolated electrode has a small capacitance. That is, if a charge is placed on it, its potential with respect to ground would change, and the ratio is its capacitance . Part B Consider an air-filled charged capacitor. How can its capacitance be increased? Hint B.1 What does capacitance depend on?
Capacitance depends on the inherent properties of the system of conductors, such as its geometry and the presence of dielectric, not on the charge placed on the conductors. Specifically, capacitance depends on the area of the conducting plates and the distance between the plates and is given by
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, where is a constant called the permittivity of free space. Increase the charge on the capacitor. Decrease the charge on the capacitor. Increase the spacing between the plates of the capacitor. Decrease the spacing between the plates of the capacitor. Increase the length of the wires leading to the capacitor plates.
ANSWER:
Part C Consider a charged parallel-plate capacitor. How can its capacitance be halved? A. B. C. D. E. F. Double the charge. Double the plate area. Double the plate separation. Halve the charge. Halve the plate area. Halve the plate separation.
Enter the letters of the correct actions in alphabetical order. For example, if actions A and B would achieve the desired result enter AB. ANSWER:
Part D Consider a charged parallel-plate capacitor. Which combination of changes would quadruple its capacitance? ANSWER: Double the charge and double the plate area. Double the charge and double the plate separation. Halve the charge and double the plate separation. Halve the charge and double the plate area. Halve the plate separation and double the plate area. Double the plate separation and halve the plate area.
The Capacitor as an Energy-Storing Device
Description: A tutorial containing quantitative questions related to the "work-energy" relationships as applied to capacitors (including the insertion of a dielectric slab). The problem "Energy of a capacitor in the presence of dielectric" can be used as a follow-up to this. Learning Goal: To understand that the charge stored by capacitors represents energy; to be able to calculate the stored energy and its changes under different circumstances. An air-filled parallel-plate capacitor has plate area creates a constant voltage . Part A Find the energy Hint A.1 stored in the capacitor. and plate separation . The capacitor is connected to a battery that
Formula for the energy of a capacitor . as epsilon_0.
Recall that the equation for the energy of a capacitor is Express your answer in terms of , ,
, and . Remember to enter
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ANSWER:
=
Part B The capacitor is now disconnected from the battery, and the plates of the capacitor are then slowly pulled apart until the separation reaches . Find the new energy of the capacitor after this process. Part B.1 What quantity remains constant?
What characteristic of the capacitor does not change in this process? ANSWER: charge voltage between the plates capacitance energy
In particular, look carefully at the quantities that do change. Part B.2 Find the charge on the capacitor that resides on the plates of the capacitor? as epsilon_0.
What is the charge
Express your answer in terms of some or all of the variables , , and . Remember to enter ANSWER: =
Part B.3
How does the capacitance change?
How does the capacitance change in this process? ANSWER: remains constant increases by a factor of 3 decreases by a factor of 3 increases by a factor of 9 decreases by a factor of 9
Part B.4
What is the formula for the energy? for this situation?
Which of the following formulas is most useful in finding ANSWER:
Express your answer in terms of , , ANSWER: =
, and .
The increase in energy of the capacitor comes from the external work that must be done to pull the plates apart. Keep in mind that the plates have opposite charges and attract each other; some work must be done by an external
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agent to pull them apart. Part C The capacitor is now reconnected to the battery, and the plate separation is restored to . A dielectric plate is slowly moved into the capacitor until the entire space between the plates is filled. Find the energy of the dielectric-filled capacitor. The capacitor remains connected to the battery. The dielectric constant is . Express your answer in terms of , , ANSWER: = , , and .
Equivalent Capacitance
Description: Find the equivalent capacitance of a network of capacitors with series and parallel connections. Consider the combination of capacitors shown in the diagram, where = 5.00 . = 3.00 , = 11.0 , = 3.00 , and
Part A Find the equivalent capacitance Hint A.1 of the network of capacitors.
How to reduce the network of capacitors
To find the equivalent capacitance of the given network of capacitors, it is most convenient to reduce the network in successive stages. First, replace the capacitors , , and , which are in parallel, with a single capacitor with an equivalent capacitance. By doing so, you will reduce the network to a series connection of two capacitors. At this point, you only need to find their equivalent capacitance. Part A.2 Find the capacitance equivalent to , , and , , and .
Find the capacitance
equivalent to the parallel connection of the capacitors and
Part A.2.a Find the capacitance equivalent to Find the capacitance
equivalent to the parallel connection of the capacitors
and
.
Hint A.2.a.i Two capacitors in parallel Consider two capacitors of capacitance capacitance given by and connected in parallel. They are equivalent to a capacitor with
. Express your answer in microfarads.
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ANSWER: If you replace the capacitors parallel connection between and and with a capacitor of capacitance . , the resulting network would be a
Express your answer in microfarads. ANSWER: =
If you replace the capacitors , , and with a capacitor of capacitance , the resulting network would be a . Its equivalent capacitance is also the equivalent capacitance of the original series connection between and network. Hint A.3 Two capacitors in series
Consider two capacitors of capacitance and connected in series. They are equivalent to a capacitor of capacitance that satisfies the following relation: . Express your answer in microfarads. ANSWER: =
Part B Two capacitors of capacitance = 6.00 and = 3.00 are added to the network, as shown in the diagram. Find of the new network of capacitors. the equivalent capacitance
Hint B.1
How to reduce the extended network of capacitors
To determine the equivalent capacitance of the extended network of capacitors, it is again convenient to reduce the network in successive stages. First, determine the equivalent capacitance of the series connection of the capacitors and . Then, combine it with the equivalent capacitance of the parallel connection of , , and , and replace the five capacitors with their equivalent capacitor. The resulting network will consist of two capacitors in series. At this point, you only need to find their equivalent capacitance. Part B.2 Find the equivalent capacitance of , , , , and , , , , and .
Find the equivalent capacitance
of the combination of capacitors and
Part B.2.a Find the equivalent capacitance of
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Find the equivalent capacitance
of the series connection of
and
.
Hint B.2.a.i Two capacitors in Consider series two capacitors of capacitance and connected in series. They are equivalent to a capacitor of capacitance that satisfies the following relation: . Express your answer in microfarads. ANSWER: =
Part B.2.b Find the equivalent capacitance of Find the equivalent capacitance
,
, and , , and
of the parallel connection of
Hint B.2.b.i Three capacitors in parallel Consider three capacitors of capacitance given by with capacitance , , and connected in parallel. They are equivalent to a capacitor
. Express your answer in microfarads. ANSWER: =
Express your answer in microfarads. ANSWER: = with a capacitor of capacitance , the resulting network . Its equivalent capacitance is also the equivalent capacitance of
If you replace the capacitors , , , , and would be a series connection between and the original network. Hint B.3 Two capacitors in series
Consider two capacitors of capacitance and connected in series. They are equivalent to a capacitor of capacitance that satisfies the following relation: . Express your answer in microfarads. ANSWER: =
Problem 30.22
A switch that connects a battery to a . charged to Part A capacitor is closed. Several seconds later you find that the capacitor plates are
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What is the emf of the battery? ANSWER: V
Problem 30.28
Initially, the switch in the figure is open and the capacitor is uncharged.
Part A How much charge flows through the switch after the switch is closed? ANSWER:
Problem 30.32
40.0 Part A What is the electric field strength? ANSWER: V/m of energy is stored in a 2.50 2.50 2.50 region of uniform electric field.
Problem 30.58
You need a capacitance of Part A What additional capacitor do you need to produce a total capacitance of ANSWER: , but you don't happen to have a capacitor. You do have a capacitor.
Part B Should you join the two capacitors in parallel or in series? ANSWER: in parallel in series
Problem 30.62
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Part A What is the equivalent capacitance of the three capacitors in the figure ?
ANSWER:
Class 17
RC Circuit and Current Conceptual Question
Description: Short conceptual problem about current through a charging and discharging RC circuit. In the diagram below,the two resistors, open. and , are identical and the capacitor is initially uncharged with the switch
Part A How does the current through Hint A.1 compare with the current through immediately after the switch is first closed?
Using Kirchhoff's junction rule for currents
At all times, the current through (entering the junction) must be equal to the sum of the currents through and the capacitors branch (exiting the junction). The relative sizes of the currents through and the capacitors branch are determined by the effective resistances of these branches. ANSWER: The current through <=> the current through .
Part B
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How does the current through Hint B.1
compare with the current through
a very long time after the switch has been closed?
Using Kirchhoff's junction rule for currents
At all times, the current through (entering the junction) must be equal to the sum of the currents through and the capacitors branch (exiting the junction). The relative sizes of the currents through and the capacitors branch are determined by the effective resistances of these branches. Hint B.2 Current associated with a fully charged capacitor
When a capacitor is fully charged, the current in the branch of the circuit containing the capacitor is zero. ANSWER: The current through <=> the current through
Part C How does the current through compare with the current through being closed a very long time)? Hint C.1 Effect of a discharging capacitor immediately after the switch is opened (after
The charge is "held" on the plates of the capacitor by the potential difference of the battery. When the battery is removed from the system, the negative charge on one plate of the capacitor will flow toward the positive charge on the other plate if a conducting path (circuit loop) exists. During this discharging phase, the capacitor behaves in a way very analogous to a battery. ANSWER: The current through <=> the current through .
Capacitor Supplies Current to Bulb
Description: A charged capacitor is connected in series to a flashlight bulb. Multiple choice questions about what happens, then a numerical question about how long it takes the capacitor to discharge. A large capacitor is charged with on one electrode and on the other. At time , the capacitor is connected in series to two ammeters and a bulb. The ammeter connected to the positive side of the capacitor reads , and the ammeter connected to the negative side of the capacitor reads . Both ammeters will read positive if current flows in a clockwise sense through the circuit (from the + to the - terminal of the meter).
Part A Immediately after time , what happens to the charge on the capacitor plates?
a. Electrons flow through the circuit from the positive to the negative side of the capacitor. b. Electrons flow through the circuit from the negative to the positive side of the capacitor.
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c. The positive and negative charges attract each other, so they stay in the capacitor. d. Current flows clockwise through the circuit. e. Current flows counterclockwise through the circuit. Hint A.1 What is meant by current flow?
Conventionally, current flow refers to the motion of positive charges in a circuit. This convention was adopted before it was discovered that electrons are the actual free charge carriers in metals, but we use it even today. In most respects, a negative chatge moving to the right can be thought of instead as a positive charge moving to the left. So the direction of the current is actually opposite to the direction in which electrons in the circuit move. The collective motion of electrons (negative charges) in a circuit is usually referred to as the "electron current". List the letters corresponding to the correct statements in alphabetical order. Do not use commas. For instance, if you think that only statements a and c are correct, write ac. ANSWER:
Part B At any given instant after , what is the relationship between the current flowing through the two ammeters, , and the current through the bulb, ? ANSWER: and
This is a fundamental result that reflects conservation of charge. In a circuit where elements are arranged in series, the voltage changes as current flows through the circuit, but the current is constant. Otherwise, charge would accumulate in the circuit. In a circuit where elements are arranged in parallel, the opposite is true; all parallel branches have the same voltage, although the current may be different in different branches. This result is formalized in Kirchoff's junction law -- the algebraic sum of currents entering any junction must be zero. (In this law, a current leaving a junction is considered negative).
Part C What is the sign of the quantity ANSWER: Positive ?
Negative
Part D
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Light bulbs are often assumed to obey Ohm's law, but this is not strictly true, because their resistance increases as the filament heats up at higher voltages. A typical flashlight bulb at full brilliance draws a current of approximately 0.5 while attached to a 3 source. For this problem, assume that the changing resistance causes the current to be 0.5 for any voltage between 2 and 3 . Suppose this flashlight bulb is attached to a capacitor as shown in the circuit from the problem introduction. If the capacitor has a capacitance of 3 (an unusually large but not unrealistic value) and is initially charged to 3 , how long will it take for the voltage across the flashlight bulb to drop to 2 (where the bulb will be orange and dim)? Call this . time Hint D.1 How to approach this problem
Determine the initial charge on the capacitor, and the charge on the capacitor when the voltage drops to 2 . Then determine how long current must flow to move this amount of charge. Part D.2 What is Initial charge on capacitor , the initial charge on the capacitor (when it has a voltage of 3 )?
Express your answer numerically, to the nearest integer. ANSWER: = coulombs
Part D.3 What is
Charge on capacitor at 2 V , the charge on the capacitor when the voltage has dropped to 2 ?
Express your answer numerically, to the nearest integer. ANSWER: = coulombs
Hint D.4
Relationship between charge and current (0.5 ), 1 of charge will move through the bulb every 2 .
With the bulb drawing a current of 0.5 Express ANSWER:
numerically, in seconds, to the nearest integer. = seconds
Problem 31.41
Part A What value resistor will discharge a 2.30 ANSWER: capacitor to 20.0 % of its initial charge in 2.40 ?
Problem 31.73
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A 10.0 capacitor that had been charged to 30 V is discharged through a resistor. The figure shows the capacitor voltage as a function of time.
Part A What is the value of the resistance? ANSWER:
Problem 31.76
The capacitor in the figure begins to charge after the switch closes at .
Part A What is ANSWER: a very long time after the switch has closed?
0
Part B What is ANSWER: in terms of , , and ?
Part C In this circuit, does or ?
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ANSWER:
Part D Find an expression for the current ANSWER: at time .
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