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Homework 5Due: 11:59pm on Sunday, April 10, 2011

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Capacitors in Series

Learning Goal: To understand how to calculate capacitance, voltage, and charge for a combination ofcapacitors connected in series.

Consider the combination of capacitors shown in the figure. Three capacitors are connected to each other inseries, and then to the battery. The values of thecapacitances are , , and , and the applied

voltage is . Initially, all of the capacitors are

completely discharged; after the battery is connected,the charge on plate 1 is .

Part A

What are the charges on plates 3 and 6?

Hint A.1 Charges on capacitors connected in series

Hint not displayed

Hint A.2 The charges on a capacitor's plates

Hint not displayed

ANSWER: and

and

and

and

0 and

0 and

MasteringPhysics: Assignment Print View http://session.masteringphysics.com/myct/assignmentPrintView?assignme...

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Correct

Part B

If the voltage across the first capacitor (the one with capacitance ) is , then what are the voltages

across the second and third capacitors?

Hint B.1 Definition of capacitance

Hint not displayed

Hint B.2 Charges on the capacitors

Hint not displayed

ANSWER: and

and

and

0 and

Correct

Part C

Find the voltage across the first capacitor.

Hint C.1 How to analyze voltages

Hint not displayed

Express your answer in terms of .

ANSWER: =

Correct

Part D

Find the charge on the first capacitor.

Express your answer in terms of and .


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ANSWER: =

Correct

Part E

Using the value of just calculated, find the equivalent capacitance for this combination of capacitors

in series.

Hint E.1 Using the definition of capacitance

Hint not displayed


ANSWER: =

Correct

The formula for combining three capacitors in series is

.

How do you think this formula may be generalized to capacitors?

Energy in Capacitors and Electric Fields

Learning Goal: To be able to calculate the energy of a charged capacitor and to understand the concept ofenergy 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 theenergy 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 aparallel-plate capacitor as a vehicle for obtaining the formula for the energy density associated with anelectric field. It will be useful to recall the definition of capacitance, , and the formula for the

capacitance of a parallel-plate capacitor,

, where is the area of each of the plates and is the plate separation. As usual, is the

permittivity of free space.

First, consider a capacitor of capacitance that has a charge and potential difference .

Part A

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.


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ANSWER:

=

Correct

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.


ANSWER:

=

Correct

All three of these formulas are equivalent:

.

Depending on the problem, one or another may be more convenient to use. However, any one of themwould give you the correct answer. Note that these formulas work for any type of capacitor.

Part C

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

Hint not displayed

Hint C.2 Identify which formula to use

Hint not displayed

ANSWER: = 0.5

Correct

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


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new energy of the capacitor is . Find the ratio .

Hint D.1 Determine what remains constant

Hint not displayed

Hint D.2 Identify which formula to use

Hint not displayed

ANSWER: = 2

Correct

In this part of the problem, you will express the energy of various types of capacitors in terms of theirgeometry and voltage.

Part E

A parallel-plate capacitor has area and plate separation , and it is charged to voltage . Use the

formulas from the problem introduction to obtain the formula for the energy of the capacitor.

Express your answer in terms of , , , and appropriate constants.

ANSWER:

=

Correct

Let us now recall that the energy of a capacitor can be thought of as the energy of the electric field insidethe capacitor. The energy of the electric field is usually described in terms of energy density , the energyper unit volume.A parallel-plate capacitor is a convenient device for obtaining the formula for the energy density of an electricfield, 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

A parallel-plate capacitor has area and plate separation , and it is charged so that the electric field

inside is . Use the formulas from the problem introduction to find the energy of the capacitor.

Hint F.1 How to approach the problem

Hint not displayed

Express your answer in terms of , , , and appropriate constants.


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ANSWER:

=

Correct

As mentioned before, we can think of the energy of the capacitor as the energy of the electric fieldinside the capacitor.

Part G

Find the energy density of the electric field in a parallel-plate capacitor. The magnitude of the electric fieldinside the capacitor is .

Hint G.1 How to approach the problem

Hint not displayed

Hint G.2 Volume between the plates

Hint not displayed

Express your answer in terms of and appropriate constants.

ANSWER:

=

Correct

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

describes the energy density in any electrostatic field, whether created by a capacitor or any othersource.

Capacitor with a DielectricTwo oppositely charged but otherwise identical conducting plates of area 2.50 square centimeters areseparated by a dielectric 1.80 millimeters thick, with a dielectric constant of . The resultant electric

field in the dielectric is volts per meter.

Part A

Compute the magnitude of the charge per unit area on the conducting plate.

Hint A.1 How to approach the problem

Hint not displayed


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Hint A.2 Calculate the electric field without the dielectric

Hint not displayed

Hint A.3 Charge density in a parallel-plate capacitor

Hint not displayed

Express your answer in coulombs per square meter to three significant figures.

ANSWER: = 3.82×10−5

Correct

Part B

Compute the magnitude of the charge per unit area on the surfaces of the dielectric.

Hint B.1 How to approach the problem

Hint not displayed

Hint B.2 Find the net surface charge density

Hint not displayed

Express your answer using three significant figures.

ANSWER: = 2.76×10−5

Correct

Note that the charges on the dielectric will be polarized to counteract the charges (and electric field)created by the capacitor. For example, near the positive surface of the capacitor the dielectric will havea negative charge. However, this does not mean that the charge on the capacitor plates changes, onlythat the dielectric has an induced charge on each of its surfaces that will oppose the effects of thecharges on the plates.

Part C

Find the total electric-field energy stored in the capacitor.

Hint C.1 How to approach the problem

Hint not displayed

Hint C.2 Calculate the electric-field energy density

Hint not displayed

Hint C.3 Volume where the electric field exists

Hint not displayed


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Express your answer in joules to three significant figures.

ANSWER: = 1.03×10−5

Correct

Because of the dielectric, there is less energy stored than if there were no dielectric. Since we alreadyknow that , it follows that the electric energy stored in the capacitor will be reduced by a

factor of from the value if the dielectric were not present.

A Capacitor with Thick PlatesConsider a capacitor that consists of two metal plates of nonzero thickness separated by a positive distance

. When such a capacitor is connected across the terminals of a battery with emf , two things happen

simultaneously: Charge flows from one plate to another, and a potential difference appears between the twoplates.In this problem, you will determine the amount and distribution of charge on each metallic plate for an ideal(infinite) parallel-plate capacitor. There is competition between the repulsive forces from like charges on aplate and the attractive forces from the opposite charge on the other plate. You will determine which force"wins."Let , , , and be the respective surfacecharge densities on surfaces A, B, C, and D. Take allthe charge densities to be positive for now. Some ofthem are in fact negative, and this will be revealed atthe end of the calculation.

Part A

If we assume that the metallic plates are perfect conductors, the electric field in their interiors must vanish.Given that the electric field due to a charged sheet with surface charge is given by

,

and that it points away from the plane of the sheet, how can the condition that the electric field in plate Ivanishes be written?


Find the electric field due to each surface in this region and then add them up.

ANSWER:


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Correct

Part B

Similarly, how can the condition that the electric field in plate II vanishes be written?


Hint not displayed

ANSWER:

Correct

Part C

There is an electric field in the region between the two plates. The magnitude of this electric field is . This

imposes another condition on the charge densities on the surfaces of the plates. How can this condition beexpressed?

Hint C.1 How to approach the problem

Hint not displayed

ANSWER:


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Correct

Part D

Finally, how can the condition that the total charge of the system is conserved be expressed?

Hint D.1 How to approach the problem

Hint not displayed

ANSWER:

Correct

You now have four equations in four variables. If you solve the four equations simultaneously you obtain,

,

which is the familiar expression for the surface charge density on a parallel-plate capacitor. However,you have now also established that all the charge resides on the insides of the capacitor plates.

Capacitor with Partial DielectricConsider a parallel-plate capacitor that is partially filledwith a dielectric of dielectric constant . The dielectric

has the same same height as the separation of theplates of the capacitor but fills a fraction of the area

of the capacitor. The capacitance of the capacitorwhen the dielectric is completely removed is .


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Part A

What is the capacitance of this capacitor as a function of ?

Hint A.1 Modeling the partly filled capacitor

Hint not displayed

Hint A.2 Find the capacitance of the air-filled portion

Hint not displayed

Hint A.3 Find the capacitance of the dielectric-filled portion

Hint not displayed

Hint A.4 Special cases

Hint not displayed

Express in terms of , , and .

ANSWER: =

Correct

Capacitors with Partial DielectricsConsider two parallel-plate capacitors identical in shape, one aligned so that the plates are horizontal , andthe other with the plates vertical .

Part A


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The horizontal capacitor is filled halfway with a material that has dielectric constant . What fraction of

the area of the vertical capacitor should be filled (as shown in the figure) with the same dielectric so thatthe two capacitors have equal capacitance?

Hint A.1 Capacitance of a parallel-plate capacitor filled with air

The capacitance of a capacitor depends solely on its geometry. What is the capacitance of a

capacitor with area and separation ?

Express your answer in terms of , , and .

ANSWER: =

Correct

Hint A.2 Effect of a Dielectric

Inserting a dielectric material between the plates of a capacitor decreases the electric field between theplates, because the molecules in the dielectric align themselves like the electrons in a conductor. The netresult of inserting the dielectric is an increase in the capacitor's capacitance by a factor of the dielectricconstant: .

Hint A.3 Modeling the horizontal capacitor

The horizontal capacitor can be modeled as two smaller capacitors: one with half the separationcompletely filled with air, and one with half the separation completely filled with the dielectric. These twosmaller capacitors are connected in series. What is the capacitance of the horizontal capacitor, with

original area and separation ?

Hint A.3.1 Connecting capacitors in series

When capacitors are connected in series, the charge on each capacitor is the same, and the potentialsadd. Therefore,

and so

.

This can be re-written as

.

Note: Recall that a capacitor with half the plate separation as another has twice the capacitance.

Express your answer in terms of , , , and .

ANSWER: =

Answer Requested


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As a check that this all makes sense, note that if , then (the capacitance with

no dielectric), and if , then (half of the capacitor is a conductor, and the net

capacitance is that of a capacitor without a dielectric but a separation ).

Hint A.4 Modeling the vertical capacitor

The vertical capacitor can be treated as two smaller capacitors: one with a fraction of the area

filled with air ( ), and one with a fraction of the area filled with the dielectric (

). These two smaller capacitors are connected in parallel. What is the net capacitance

of the vertical capacitor, which has original area and separation .

Hint A.4.1 Connecting capacitors in parallel

Hint not displayed


ANSWER: = Answer not displayed


ANSWER: =

Correct

Dielectrics and Capacitance Ranking TaskSix parallel-plate capacitors of identical plate separation have different plate areas , different capacitances

, and different dielectrics filling the space between the plates. Below is a generic diagram of what each one

of these capacitors might look like.


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Part A

Rank the following capacitors on the basis of the dielectric constant of the material between the plates.

Hint A.1 Capacitance of a parallel-plate capacitor

Hint not displayed

Rank from largest to smallest. To rank items as equivalent, overlap them.

ANSWER:

View Correct

All of the capacitors from Part A are now attached to batteries with the same potential difference.

Part B

Rank the capacitors on the basis of the charge stored on the positive plate.

Hint B.1 Definition of capacitance

Hint not displayed

Rank from largest to smallest. To rank items as equivalent, overlap them.

ANSWER:

View Correct

Energy of a CapacitorA parallel-plate vacuum capacitor is connected to abattery and charged until the stored electric energy is

. The battery is removed, and then a dielectric


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material with dielectric constant is inserted into the

capacitor, filling the space between the plates. Finally,the capacitor is fully discharged through a resistor(which is connected across the capacitor terminals).

Part A

Find , the the energy dissipated in the resistor.

Hint A.1 How to approach the question

The energy dissipated in the resistor equals the energy stored in the dielectric-filled capacitor.

Hint A.2 Energy of a capacitor

The electric potential energy of a capacitor with capacitance and charge is given by

.

Hint A.3 Effects of the dielectric

Inserting a dielectric into the capacitor in the manner described (with the charging battery removed) doesnot change the charge on the capacitor since its plate is isolated. However, the presence of the dielectricdoes increase the capacitance by a factor of .

Express your answer in terms of and other given quantities.

ANSWER: =

Correct

The energy of the capacitor in this case drops from to as the dielectric plate is inserted. This

energy loss is associated with mutual attraction of the plate and the capacitor. As the plate goes intothe capacitor, the potential energy of the "capacitor and plate system" decreases, much like thepotential energy of a stretched spring as it contracts back to its relaxed state.

Part B

Consider the same situation as in the previous part, except that the charging battery remains connectedwhile the dielectric is inserted. The battery is thendisconnected and the capacitor is discharged. Forthis situation, what is , the energy dissipated in the

resistor?


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Hint B.1 Energy of a capacitor

Hint not displayed

Hint B.2 Effects of the dielectric

Hint not displayed

Express your answer in terms of and other given quantities.

ANSWER: =

Correct

In this case, the energy increase comes from the battery. The battery does positive work on thecapacitor by moving more electrons from one plate to another to maintain the constant potentialdifference as the capacitance increases.

Equivalent CapacitanceConsider the combination of capacitors shown in the diagram, where = 3.00 , = 11.0 ,

= 3.00 , and = 5.00 .

Part A

Find the equivalent capacitance of the network of capacitors.

Hint A.1 How to reduce the network of capacitors


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Hint not displayed

Hint A.2 Find the capacitance equivalent to , , and

Hint not displayed

Hint A.3 Two capacitors in series

Hint not displayed

Express your answer in microfarads.

ANSWER: = 2.59Correct

Part B

Two capacitors of capacitance = 6.00 and = 3.00 are added to the network, as shown in the

diagram. Find the equivalent capacitance of the

new network of capacitors.

Hint B.1 How to reduce the extended network of capacitors

Hint not displayed

Hint B.2 Find the equivalent capacitance of , , , , and

Hint not displayed

Hint B.3 Two capacitors in series

Hint not displayed

Express your answer in microfarads.



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Force between Capacitor PlatesConsider a parallel-plate capacitor with plates of area and with separation .

Part A

Find , the magnitude of the force each plate experiences due to the other plate as a function of ,

the potential drop across the capacitor.


Hint not displayed

Hint A.2 Method 1: Use the stored energy to find the force

Hint not displayed

Hint A.3 Method 2: Use the product of the charge and the field to find the force

Hint not displayed

Express your answer in terms of given quantities and .

ANSWER: =

Correct

The Capacitor as an Energy-Storing Device

Learning Goal: To understand that the charge stored by capacitors represents energy; to be able tocalculate the stored energy and its changes under different circumstances.

An air-filled parallel-plate capacitor has plate area and plate separation . The capacitor is connected to a

battery that creates a constant voltage .

Part A

Find the energy stored in the capacitor.

Hint A.1 Formula for the energy of a capacitor

Hint not displayed

Express your answer in terms of , , , and . Remember to enter as epsilon_0.

ANSWER: =

Correct


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Part B

The capacitor is now disconnected from the battery, and the plates of the capacitor are then slowly pulledapart until the separation reaches . Find the new energy of the capacitor after this process.

Hint B.1 What quantity remains constant?

Hint not displayed

Hint B.2 Find the charge on the capacitor

Hint not displayed

Hint B.3 How does the capacitance change?

Hint not displayed

Hint B.4 What is the formula for the energy?

Hint not displayed


ANSWER: =

Correct

The increase in energy of the capacitor comes from the external work that must be done to pull theplates apart. Keep in mind that the plates have opposite charges and attract each other; some workmust be done by an external 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 , , , , and .

ANSWER: =

Correct

± Potential Difference and Electric-Field Energy of a Spherical CapacitorA spherical capacitor is formed from two concentric spherical conducting shells separated by vacuum. Theinner sphere has radius 10.0 centimeters, and the separation between the spheres is 1.50 centimeters. The


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magnitude of the charge on each sphere is 3.30 nanocoulombs.

Part A

What is the magnitude of the potential difference between the two spheres?


Hint not displayed

Hint A.2 Choosing the Gaussian surface

Hint not displayed

Hint A.3 Find the electric field

Hint not displayed

Hint A.4 How to use the electric field to calculate the potential difference

Hint not displayed


Part B

What is the electric-field energy stored in the capacitor?

Hint B.1 How to calculate the electric-field energy

Hint not displayed

ANSWER: 6.38×10−8

Correct

± Energy of a Capacitor in the Presence of a DielectricA dielectric-filled parallel-plate capacitor has plate area = 10.0 , plate separation = 7.00 and

dielectric constant = 5.00. The capacitor is connected to a battery that creates a constant voltage

= 7.50 . Throughout the problem, use = 8.85×10−12 .

Part A

Find the energy of the dielectric-filled capacitor.

Hint A.1 Energy of a charged capacitor in terms of its capacitance and voltage


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Hint not displayed

Hint A.2 Capacitance of a dielectric-filled capacitor

Hint not displayed

Hint A.3 Find the capacitance

Hint not displayed

Express your answer numerically in joules.

ANSWER: = 1.78×10−10

Correct

Part B

The dielectric plate is now slowly pulled out of the capacitor, which remains connected to the battery. Findthe energy of the capacitor at the moment when the capacitor is half-filled with the dielectric.

Hint B.1 What quantity remains constant?

Hint not displayed

Hint B.2 Modeling the capacitor

Hint not displayed

Hint B.3 Finding the energy

Hint not displayed


ANSWER: = 1.07×10−10

Correct

Part C

The capacitor is now disconnected from the battery, and the dielectric plate is slowly removed the rest ofthe way out of the capacitor. Find the new energy of the capacitor, .

Hint C.1 What quantity remains constant?

Hint not displayed

Hint C.2 Energy of a charged capacitor in terms of its capacitance and charge

Hint not displayed


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Hint C.3 Calculate the charge in the capacitor

Hint not displayed


ANSWER: = 3.20×10−10

Correct

Comparing the expressions for and , one can see that ; in other words, the energy of

the capacitor increases as the plate is being pulled out.

Part D

In the process of removing the remaining portion of the dielectric from the disconnected capacitor, howmuch work is done by the external agent acting on the dielectric?

Hint D.1 Conservation of energy

Hint not displayed


ANSWER: = 2.13×10−10

Correct

Since for every dielectric , the work done in the last process is positive; in other words, an

external agent must apply a force to pull the plate out; the capacitor would exert a net force that would"resist" the pullout.

An R-C Circuit

Learning Goal: To understand the behavior of the current and voltage in a simple R-C circuitA capacitor with capacitance is initially charged with charge . At time a resistor with resistance

is connected across the capacitor.


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Part A

Use the Kirchhoff loop rule and Ohm's law to express the voltage across the capacitor in terms of the

current flowing through the circuit.


ANSWER: =

Correct

Part B

We would like to use the relation to find the voltage and current in the circuit as functions of

time. To do so, we use the fact that current can be expressed in terms of the voltage. This will produce adifferential equation relating the voltage to its derivative. Rewrite the right-hand side of this relation,

replacing with an expression involving the time derivative of the voltage.


Hint not displayed

Hint B.2 Find the relation between and

Hint not displayed

Express your answer in terms of and quantities given in the problem introduction.

ANSWER: =

Correct

Part C

Now solve the differential equation for the initial conditions given in the problem

introduction to find the voltage as a function of time for any time .

Hint C.1 Find the voltage at time

Hint not displayed

Hint C.2 Method 1: Guessing the form of the solution

Hint not displayed

Hint C.3 Method 2: Separation of variables


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Hint not displayed


ANSWER: =

Correct

If there were a battery in the circuit with EMF , the equation for would be

. This differential equation is no longer homogeneous in (homogeneous

means that if you multiply any solution by a constant it is still a solution). However, it can be solvedsimply by the substitution . The effect of this substitution is to eliminate the term

and yield an equation for that is identical to the equation you solved for . If a battery is

added, the initial condition is usually that the capacitor has zero charge at time . The solution

under these conditions will look like . This solution implies that the voltage

across the capacitor is zero at time (since the capacitor was uncharged then) and rises

asymptotically to (with the result that current essentially stops flowing through the circuit).

Part D

Given that the voltage across the capacitor as a function of time is , what is the current

flowing through the resistor as a function of time (for )? It might be helpful to look again at Part A

of this problem.

Hint D.1 Apply Ohm's law

Hint not displayed

Express your answer in terms of and any quantities given in the problem introduction.

ANSWER: =

Correct

± Charging and Discharging a Capacitor in an R-C Circuit

Learning Goal: To understand the dynamics of a series R-C circuit.Consider a series circuit containing a resistor of resistance and a capacitor of capacitance connected to

a source of EMF with negligible internal resistance. The wires are also assumed to have zero resistance.

Initially, the switch is open and the capacitor discharged.Let us try to understand the processes that take placeafter the switch is closed. The charge of the capacitor,the current in the circuit, and, correspondingly, the


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voltages across the resistor and the capacitor, will bechanging. Note that at any moment in time during thelife of our circuit, Kirchhoff's loop rule holds and indeed,it is helpful: , where is the voltage

across the resistor, and is the voltage across the

capacitor.

Part A

Immediately after the switch is closed, what is the voltage across the capacitor?

ANSWER:

zero

Correct

Part B

Immediately after the switch is closed, what is the voltage across the resistor?

ANSWER:

zero

Correct

Part C

Immediately after the switch is closed, what is the direction of the current in the circuit?

ANSWER: clockwise

counterclockwise

There is no current because the capacitor does not allow the current to passthrough.

Correct


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While no charge can physically pass through the gap between the capacitor plates, it can flow in therest of the circuit. The current in the capacitor can be thought of as a different sort of current, notinvolved with the flow of charge, but with an electric field that is increasing with time. This current iscalled the displacement current. You will learn more about this later. Of course, when the charge ofthe capacitor is not changing, then there is no current.

Part D

After the switch is closed, which plate of the capacitor eventually becomes positively charged?

ANSWER: the top plate

the bottom plate

both plates

neither plate because electrons are negatively charged

Correct

Part E

Eventually, the process approaches a steady state. In that steady state, the charge of the capacitor is notchanging. What is the current in the circuit in the steady state?

Hint E.1 Charge and current

Hint not displayed

ANSWER:

zero

Correct

Part F

In the steady state, what is the charge of the capacitor?

Hint F.1 Voltage in the steady state


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Hint not displayed

Express your answer in terms of any or all of , , and .

ANSWER: =

Correct

Part G

How much work is done by the voltage source by the time the steady state is reached?

Hint G.1 Charge and EMF

Hint not displayed

Express your answer in terms any or all of , , and .

ANSWER: =

Correct

In order to charge the capacitor, a total amount of charge had to move across the potential

difference of the EMF source. The source did work to move this charge equal to .

Recall that a charged capacitor stores an amount of energy . This is only half the work done

by the EMF source. The remaining was dissipated in the resistor. So such a simple charging

circuit has a high loss percentage, independent of the value of the resistance of the circuit.Even though energy is dissipated across the resistor as the capacitor charges, note that the work donedepends on , but not on ! This is because it is the capacitor that determines the amount of charge

flow through the circuit. Charge flow stops when . The resistance does however affect the rate

of charge flow i.e. the current. You will calculate this effect in the parts that follow.

Now that we have a feel for the state of the circuit in its steady state, let us obtain expressions for thecharge of the capacitor and the current in the resistor as functions of time. We start with the loop rule:

. Note that , , and . Using these equations, we

obtain , and then, .

Part H

Integrate both sides of the equation to obtain an expression for .

Hint H.1 Constant of integration

Hint not displayed


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Express your answer in terms of any or all of , , , and . Enter exp(x) for .

ANSWER: =

Correct

Part I

Now differentiate to obtain an expression for the current .

Express your answer in terms of any or all of , , , and . Enter exp(x) for .

ANSWER: =

Correct

Theoretically, the steady state is never reached: The exponential functions approach their limits as asymptotically. However, it does not take very long for the values of and to get very

close to their limiting values. The next few questions illustrate this point. Note that the quantity has

dimensions of time and is called the time constant, or the relaxation time. It is often denoted by .Using , one can rewrite the expressions for charge and current as follows:

and

.

Graphs of these functions are shown in the figure.

Part J


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Find the time that it would take the charge of the capacitor to reach 99.99% of its maximum value giventhat and .

Hint J.1 Find an expression for the time

Find the time that it takes the charge of the capacitor to reach 99.99% of its maximum value.

Hint J.1.1 How to approach this part

Hint not displayed

Express your answer in terms of . Use three significant figures for any numerical terms.

ANSWER: =

Correct

It would take the same amount of time for the current to drop to 0.01% of its initial (maximum) value;compare the expressions for and to see this for yourself.

Express your answer numerically in seconds. Use three significant figures in your answer.

ANSWER: = 5.53×10−2

Correct

Notice how quickly the circuit approaches steady state for these typical values of resistance andcapacitance!

Let us now consider a different R-C circuit. This time, the capacitor is initially charged ( ), and there

is no source of EMF in the circuit. We will assumethat the top plate of the capacitor initially holdspositive charge. For this circuit, Kirchhoff's loop rulegives , or equivalently, .

Part K

Find the current as a function of time for this circuit.

Hint K.1 Find the charge on the capacitor


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Find the charge on the capacitor as a function of time for this circuit.

Hint K.1.1 The relationship between charge and current

Hint not displayed

Express your answer in terms of , , , and . Enter exp(x) for .

ANSWER: = Answer not displayed

Express your answer in terms of , , , and . Enter exp(x) for .

ANSWER: =

All attempts used; correct answer displayed

The negative value of the current can be explained by the fact that the positive charge on thecapacitor's top plate decreases. Graphs ofthese functions are shown in the figure.

Changing Capacitance Yields a CurrentEach plate of a parallel-plate capacator is a square with side length , and the plates are separated by adistance . The capacitor is connected to a source of voltage . A plastic slab of thickness and dielectric

constant is inserted slowly between the plates over the time period until the slab is squarely between

the plates. While the slab is being inserted, a current runs through the battery/capacitor circuit.


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Part A

Assuming that the dielectric is inserted at a constant rate, find the current as the slab is inserted.

Hint A.1 What is the effect of the dielectric on capacitance?

Hint not displayed

Hint A.2 What is the current in the circuit?

Hint not displayed

Hint A.3 What is the initial capacitance?

Hint not displayed

Hint A.4 What is the change in capacitance?

Hint not displayed

Express your answer in terms of any or all of the given variables , , , , , and , the

permittivity of free space.

ANSWER:

=

Correct

± Capacitor Supplies Current to BulbA large capacitor has a charge + on one plate and on the other. At time , the capacitor is

connected in series to two ammeters and a light bulb. Immediately after the circuit is closed, the ammeterconnected to the positive plate of the capacitor reads and the ammeter connected to the negative plate of

the capacitor reads .

Each ammeter reads positive if current flows throughthe circuit in a clockwise direction (from the to the

terminal of the meter).


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Part A

Immediately after time , what happens to the charge on the capacitor plates?

Hint A.1 What particles carry current in wires?

Hint not displayed

Hint A.2 Direction of current

Hint not displayed

Check all that apply.

ANSWER: Individual charges flow through the circuit from the positive to the negativeplate of the capacitor.

Individual charges flow through the circuit from the negative to the positiveplate of the capacitor.

The positive and negative charges attract each other, so they stay in thecapacitor.

Current flows clockwise through the circuit.

Current flows counterclockwise through the circuit.

Correct

Part B

At any given instant after , what is the relationship between the current flowing through the two

ammeters, and , and the current through the bulb, ?

ANSWER:

Correct

This is a fundamental result that reflects the law of conservation of charge. In a circuit where allelements are arranged in series, the current is the same through all the elements. Otherwise, theelectric charge would, in effect, "accumulate" or "disappear" somewhere in the circuit.In a circuit where all elements are arranged in parallel, the current may be different in differentbranches. This result is formalized in Kirchhoff's junction law: The algebraic sum of currents enteringany junction must be zero.


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Part C

What is the relationship between the current and charge? As the charge on the positive plate of the

capacitor decreases, what happens to the value of the current?

ANSWER:

The current

increases.

decreases.

does not change.

Correct

Part D

Light bulbs are often assumed to obey Ohm's law. However, this is not really true because their resistanceincreases substantially as the filament heats up in its "working" state.A typical flashlight bulb at full brilliance draws a current of approximately 0.5 when connected to a 3-

voltage 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

Hint not displayed

Hint D.2 Find the initial charge on the capacitor

Hint not displayed

Hint D.3 Find the final charge on the capacitor

Hint not displayed

Hint D.4 Relationship between charge and current

Hint not displayed

Express numerically in seconds to the nearest integer.

ANSWER: = 6

Correct

A Resistor and a Capacitor in a Series AC Circuit


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A resistor with resistance and a capacitor with capacitance are connected in series to an AC voltage

source. The time-dependent voltage across the capacitor is given by .

Part A

What is the amplitude of the total current in the circuit?


For a single-loop circuit, the current flowing through the capacitor is equal to that flowing through theresistor, and is equal to the total current in the circuit. Hence you can work out the current in the circuit byusing Ohm's law to find the current through the capacitor.

Hint A.2 Applying Ohm's law to a capacitor

To apply Ohm's law to a capacitor, you must use the capacitor's reactance in place of resistance. Ohm'slaw for a capacitor is , where is the amplitude of the current through the capacitor, is the

amplitude of the voltage, and is the reactance of the capacitor.

Hint A.3 The reactance of a capacitor

The reactance of a capacitor is given by , where is the angular frequency of the voltage

across the capacitor. Substituting this expression for into Ohm's law for the capacitor will enable you

to calculate the amplitude of the current in the capacitor.

Express your answer in terms of any or all of , , , and .

ANSWER: =

Correct

Part B

What is the amplitude of the voltage across the resistor?

Hint B.1 Relating to

Hint not displayed


ANSWER: =

Correct

Part C

If , , , and , what is ?


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Hint C.1 Calculating the answer

Use the equation obtained in Part B to work out the answer. Be careful of powers of ten in yourcalculation.

Express your answer numerically, in millivolts, to the nearest integer.

ANSWER: = 3Correct mV

Energy Flow and AC Phasor Diagrams Conceptual QuestionIn each of the phasor diagrams shown in the figure, the phasors rotate counterclockwise with angularfrequency .

Use this phasor diagram to answer Parts A and B.

Part A

Chose the best completion for the following sentence: This phasor diagram represents an AC circuitconsisting of _________

Hint A.1 Any easy way to remember AC phase relations

Hint not displayed

ANSWER:

Correct

Part B

At the instant shown, which of the following statements is correct?

Hint B.1 Energy flow

Hint not displayed


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Hint B.2 Find the algebraic signs of and

Hint not displayed

ANSWER: Energy is flowing from the AC source to the capacitor.

Energy is flowing from the capacitor to the AC source.

Correct

Use the phasor diagram in the figure to answer parts C, D, and E.

Part C

Chose the best completion for the following sentence: This phasor diagram represents an AC circuitconsisting of _________

ANSWER:

Correct

Part D


Hint D.1 Energy flow to a resistor

Hint not displayed

ANSWER: Energy is flowing from the AC source to the resistor.

Energy is flowing from the resistor to the AC source.

Correct


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Part E


Hint E.1 Find the algebraic signs of and

Hint not displayed

ANSWER: Energy is flowing from the AC source to the capacitor.

Energy is flowing from the capacitor to the AC source.

Correct

Use the phasor diagram in the figure to answer Parts F, G, and H.

Part F

Chose the best completion for the following sentence: This phasor diagram represents an AC circuitconsisting of ________

ANSWER:

Correct

Part G


ANSWER: Energy is flowing from the AC source to the resistor.

Energy is flowing from the resistor to the AC source.

Correct


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Part H


ANSWER: Energy is flowing from the AC source to the inductor.

Energy is flowing from the inductor to the AC source.

Correct

Reactance and CurrentConsider the two circuits shown in the figure. Thecurrent in circuit 1, containing an inductor ofself-inductance , has an angular frequency , while

the current in circuit 2, containing a capacitor ofcapacitance , has an angular frequency . If we

increase and decrease , both bulbs growdimmer.

Part A

If we keep and constant, we can achieve the exact same effect of decreasing the brightness of eachbulb by performing which of the following sets of actions?


Hint not displayed

Hint A.2 Inductive reactance

Hint not displayed

Hint A.3 Capacitive reactance

Hint not displayed

Hint A.4 Determine how can be changed

Hint not displayed

Hint A.5 Determine how can be changed


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Hint not displayed

ANSWER: increasing and decreasing

increasing both and

decreasing and increasing

decreasing both and

Correct

As you found out, the reactance of these circuits can be changed not only by varying the frequency ofthe current, but also by changing the characteristics of the elements in them, i.e., by changing theinductance of the inductor and the capacitance of the capacitor.

Part B

Now combine the capacitor, the inductor, and the bulbs in a single circuit, as shown in the figure. Whathappens to the brightness of each bulb if youincrease the frequency of the current in the newcircuit while keeping and constant?


Hint not displayed

Hint B.2 Find which element experiences a decrease in current at higher frequencies

Hint not displayed

ANSWER: Both bulbs become brighter.

The brightness of each bulb remains constant.

Bulb 1 becomes brighter than bulb 2.

Bulb 2 becomes brighter than bulb 1.

Both bulbs grow dimmer.


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Correct

Since inductive reactance is proportional to frequency, for a given voltage, high-frequency currents willhave a much smaller amplitude through the inductor than through the capacitor. That is, the inductortends to block high-frequency currents. The opposite situation occurs if the frequency is decreased.The capacitor will block low-frequency currents and bulb 2 will grow dimmer.

Score Summary:Your score on this assignment is 97.7%.You received 97.7 out of a possible total of 100 points.


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