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MasteringPhysics: Assignment Print View

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http://session.masteringphysics.com/myct/assignmentPrintView?assignme...

Homework 5 Due: 11:59pm on Sunday, April 10, 2011 Note: You will receive no credit for late submissions. To learn more, read your instructor's Grading Policy [Switch to Standard Assignment View]

Capacitors in Series Learning Goal: To understand how to calculate capacitance, voltage, and charge for a combination of capacitors connected in series. Consider the combination of capacitors shown in the figure. Three capacitors are connected to each other in series, and then to the battery. The values of the capacitances 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

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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 Hint C.1

across the first capacitor.

How to analyze voltages Hint not displayed

Express your answer in of

.

ANSWER: = Correct

Part D Find the charge

on the first capacitor.


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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. Using the definition of capacitance

Hint E.1

Hint not displayed Express your answer in of

.

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 of energy 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 parallel-plate 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

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 of

and

by using the definition of capacitance and the

formula for the energy in a capacitor.

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and

.

ANSWER: = Correct

Part B Find the energy

of the capacitor in of

and

by using the definition of capacitance and the

formula for the energy in a capacitor. Express your answer in of

and

.

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 them would 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

.

Determine what remains constant

Hint D.1

Hint not displayed Identify which formula to use

Hint D.2

Hint not displayed

ANSWER:

= 2 Correct

In this part of the problem, you will express the energy of various types of capacitors in of their geometry and voltage. Part E A parallel-plate capacitor has area

and plate separation , and it is charged to voltage

formulas from the problem introduction to obtain the formula for the energy Express your answer in of

,

,

. Use the

of the capacitor.

, 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 inside the capacitor. The energy of the electric field is usually described in 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 A parallel-plate capacitor has area inside is Hint F.1

and plate separation , and it is charged so that the electric field

. Use the formulas from the problem introduction to find the energy

of the capacitor.

How to approach the problem Hint not displayed


,

,

, 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 field inside the capacitor.

Part G Find the energy density inside the capacitor is Hint G.1

of the electric field in a parallel-plate capacitor. The magnitude of the electric field .


Hint G.2

Volume between the plates Hint not displayed


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 other source.

Capacitor with a Dielectric Two oppositely charged but otherwise identical conducting plates of area 2.50 square centimeters are separated 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 Hint A.1

on the conducting plate.


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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:

−5 = 3.82×10 Correct

Part B Compute the magnitude of the charge per unit area Hint B.1

on the surfaces of the dielectric.


Hint B.2

Find the net surface charge density Hint not displayed

Express your answer using three significant figures. ANSWER:

−5 = 2.76×10 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 have a negative charge. However, this does not mean that the charge on the capacitor plates changes, only that the dielectric has an induced charge on each of its surfaces that will oppose the effects of the charges on the plates. Part C Find the total electric-field energy Hint C.1

stored in the capacitor.


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:

−5 = 1.03×10 Correct

Because of the dielectric, there is less energy stored than if there were no dielectric. Since we already know 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 Plates Consider 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 two plates. 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 a plate and the attractive forces from the opposite charge on the other plate. You will determine which force "wins." Let , , , and be the respective surface charge densities on surfaces A, B, C, and D. Take all the charge densities to be positive for now. Some of them are in fact negative, and this will be revealed at the 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 I vanishes be written? Hint A.1

How to approach the problem

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 B.1


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 be expressed? Hint C.1


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


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 Dielectric Consider a parallel-plate capacitor that is partially filled with a dielectric of dielectric constant . The dielectric has the same same height as the separation of the plates of the capacitor but fills a fraction of the area of the capacitor. The capacitance of the capacitor when the dielectric is completely removed is .

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Part A What is the capacitance Hint A.1

of this capacitor as a function of

?

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 ANSWER:

in of

,

, and

.

= Correct

Capacitors with Partial Dielectrics Consider two parallel-plate capacitors identical in shape, one aligned so that the plates are horizontal , and the 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 that the 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 capacitor with area

of a

and separation ?


,

, 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 the plates, because the molecules in the dielectric align themselves like the electrons in a conductor. The net result of inserting the dielectric is an increase in the capacitor's capacitance by a factor of the dielectric constant: .

Hint A.3

Modeling the horizontal capacitor

The horizontal capacitor can be modeled as two smaller capacitors: one with half the separation completely filled with air, and one with half the separation completely filled with the dielectric. These two smaller 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 potentials add. 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 of

,

,

, and

.

ANSWER: = Answer Requested

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As a check that this all makes sense, note that if no dielectric), and if

, then

, 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

(the capacitance with

).

Modeling the vertical capacitor

The vertical capacitor can be treated as two smaller capacitors: one with a fraction filled with air (

), and one with a fraction

of the area

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 Express your answer in of ANSWER:

,

,

, and

.

= Answer not displayed


.

ANSWER: = Correct

Dielectrics and Capacitance Ranking Task Six 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 Capacitor A parallel-plate vacuum capacitor is connected to a battery 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) does not change the charge on the capacitor since its plate is isolated. However, the presence of the dielectric does increase the capacitance by a factor of . Express your answer in 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 into the capacitor, the potential energy of the "capacitor and plate system" decreases, much like the potential 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 connected while the dielectric is inserted. The battery is then disconnected and the capacitor is discharged. For this situation, what is , the energy dissipated in the resistor?

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Energy of a capacitor

Hint B.1

Hint not displayed Effects of the dielectric

Hint B.2

Hint not displayed Express your answer in of ANSWER:

and other given quantities.

= Correct

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

Equivalent Capacitance Consider the combination of capacitors shown in the diagram, where = 3.00

, and

= 5.00

= 3.00

,

= 11.0

,

.

Part A Find the equivalent capacitance Hint A.1

of the network of capacitors.

How to reduce the network of capacitors

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Hint not displayed Hint A.2

Find the capacitance equivalent to

,

, and


Two capacitors in series Hint not displayed

Express your answer in microfarads. ANSWER:

= 2.59 Correct

Part B Two capacitors of capacitance

= 6.00

diagram. Find the equivalent capacitance

and

= 3.00

are added to the network, as shown in the

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. ANSWER:

= 2.54 Correct

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Force between Capacitor Plates Consider 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 A.1


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 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 to calculate the stored energy and its changes under different circumstances. An air-filled parallel-plate capacitor has plate area battery that creates a constant voltage

and plate separation . The capacitor is connected to a

.

Part A Find the energy Hint A.1

stored in the capacitor.

Formula for the energy of a capacitor Hint not displayed


,

,

, and

. 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 pulled apart 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


,

,

, and

.

ANSWER: = Correct

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 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

.


,

,

,

, and

.

ANSWER: = Correct

± Potential Difference and Electric-Field Energy of a Spherical Capacitor A spherical capacitor is formed from two concentric spherical conducting shells separated by vacuum. The inner 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 Hint A.1

between the two spheres?


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

ANSWER:

= 38.7 Correct

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 Dielectric A dielectric-filled parallel-plate capacitor has plate area dielectric constant = 7.50

= 10.0

, plate separation

= 7.00

and

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

. Throughout the problem, use

= 8.85×10−12

.

Part A Find the energy Hint A.1

of the dielectric-filled capacitor.

Energy of a charged capacitor in of its capacitance and voltage

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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:

−10 = 1.78×10 Correct

Part B The dielectric plate is now slowly pulled out of the capacitor, which remains connected to the battery. Find the energy of the capacitor at the moment when the capacitor is half-filled with the dielectric. Hint B.1


Hint B.2

Modeling the capacitor Hint not displayed

Hint B.3

Finding the energy Hint not displayed


−10 = 1.07×10 Correct

Part C The capacitor is now disconnected from the battery, and the dielectric plate is slowly removed the rest of the way out of the capacitor. Find the new energy of the capacitor, . Hint C.1


Hint C.2

Energy of a charged capacitor in of its capacitance and charge Hint not displayed

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Hint C.3


Calculate the charge in the capacitor Hint not displayed


−10 = 3.20×10 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, how much work is done by the external agent acting on the dielectric? Hint D.1

Conservation of energy Hint not displayed


−10 = 2.13×10 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 circuit A 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 current

flowing through the circuit.

Express your answer in of ANSWER:

in of the

and

.

= 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 of the voltage. This will produce a differential equation relating the voltage to its derivative. Rewrite the right-hand side of this relation, replacing Hint B.1

with an expression involving the time derivative of the voltage. How to approach the problem Hint not displayed

Hint B.2

Find the relation between

and Hint not displayed


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

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Method 2: Separation of variables

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Hint not displayed Express your answer in of

,

,

, and .

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 solved simply 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

added, the initial condition is usually that the capacitor has zero charge at time under these conditions will look like across the capacitor is zero at time asymptotically to

. If a battery is . The solution

. This solution implies that the voltage (since the capacitor was uncharged then) and rises

(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 flowing through the resistor as a function of time (for

, what is the current

)? It might be helpful to look again at Part A

of this problem. Hint D.1

Apply Ohm's law Hint not displayed


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 a source of EMF

and a capacitor of capacitance

connected to

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 place after 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 be changing. Note that at any moment in time during the life 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 through. Correct

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While no charge can physically through the gap between the capacitor plates, it can flow in the rest of the circuit. The current in the capacitor can be thought of as a different sort of current, not involved with the flow of charge, but with an electric field that is increasing with time. This current is called the displacement current. You will learn more about this later. Of course, when the charge of the 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 not changing. 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 Hint F.1

of the capacitor?

Voltage in the steady state

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Hint not displayed Express your answer in of any or all of ANSWER:

=

,

, and

.

Correct

Part G How much work Hint G.1

is done by the voltage source by the time the steady state is reached?

Charge and EMF Hint not displayed

Express your answer in any or all of ANSWER:

,

, and

.

= Correct

In order to charge the capacitor, a total amount of charge difference

had to move across the potential

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

Recall that a charged capacitor stores an amount of energy by the EMF source. The remaining

.

. This is only half the work done

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 done depends 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 the charge of the capacitor and the current in the resistor as functions of time. We start with the loop rule: . Note that obtain

, and then,

,

, and

. Using these equations, we

.

Part H Integrate both sides of the equation

Hint H.1

to obtain an expression for

.

Constant of integration Hint not displayed

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Express your answer in 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 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 given

that

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 of . Use three significant figures for any numerical . 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:

−2 = 5.53×10 Correct

Notice how quickly the circuit approaches steady state for these typical values of resistance and capacitance! 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 assume that the top plate of the capacitor initially holds positive charge. For this circuit, Kirchhoff's loop rule gives , or equivalently, .

Part K Find the current Hint K.1

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as a function of time for this circuit.

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 of ANSWER:

,

, , and

. Enter exp(x) for

.

= Answer not displayed


,

, , 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 the capacitor's top plate decreases. Graphs of these functions are shown in the figure.

Changing Capacitance Yields a Current Each plate of a parallel-plate capacator is a square with side length , and the plates are separated by a distance constant

. The capacitor is connected to a source of voltage

. A plastic slab of thickness

is inserted slowly between the plates over the time period

and dielectric

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 Hint A.1

as the slab is inserted.

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 of any or all of the given variables

,

, ,

,

, and

, the

permittivity of free space. ANSWER: = Correct

± Capacitor Supplies Current to Bulb A 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 ammeter connected 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 through the circuit in a clockwise direction (from the to the terminal of the meter).

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Part A Immediately after time Hint A.1

, what happens to the charge on the capacitor plates?

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 negative plate of the capacitor. Individual charges flow through the circuit from the negative to the positive plate of the capacitor. The positive and negative charges attract each other, so they stay in the capacitor. Current flows clockwise through the circuit. Current flows counterclockwise through the circuit. Correct

Part B At any given instant after ammeters,

and

, what is the relationship between the current flowing through the two

, 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 all elements are arranged in series, the current is the same through all the elements. Otherwise, the electric 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 different branches. This result is formalized in Kirchhoff's junction law: The algebraic sum of currents entering any 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:

increases. decreases. The current

does not change. Correct

Part D Light bulbs are often assumed to obey Ohm's law. However, this is not really true because their resistance increases 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 3voltage source. For this problem, assume that the changing resistance causes the current to be 0.5 any voltage between 2 and 3

for

.

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

orange and dim)? Call this time Hint D.1

(where the bulb will be

.

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 ANSWER:

numerically in seconds to the nearest integer. = 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 Hint A.1

of the total current

in the circuit?

How to approach the problem

For a single-loop circuit, the current flowing through the capacitor is equal to that flowing through the resistor, and is equal to the total current in the circuit. Hence you can work out the current in the circuit by using 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's law for a capacitor is , where is the amplitude of the current through the capacitor, is the amplitude of the voltage, and

Hint A.3

is the reactance of the capacitor.

The reactance of a capacitor

The reactance of a capacitor is given by

, where

across the capacitor. Substituting this expression for

is the angular frequency of the voltage

into Ohm's law for the capacitor will enable you

to calculate the amplitude of the current in the capacitor. Express your answer in of any or all of ANSWER:

,

,

, and

.

= Correct

Part B What is the amplitude Hint B.1

Relating

of the voltage across the resistor? to Hint not displayed

Express your answer in of ANSWER:

,

,

, and

.

= 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 your calculation. Express your answer numerically, in millivolts, to the nearest integer. ANSWER:

= 3

Correct

mV

Energy Flow and AC Phasor Diagrams Conceptual Question In each of the phasor diagrams shown in the figure, the phasors rotate counterclockwise with angular frequency . 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 circuit consisting of _________ Hint A.1

Any easy way to 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



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 circuit consisting of _________ ANSWER: Correct

Part D At the instant shown, which of the following statements is correct? 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 At the instant shown, which of the following statements is correct? Hint E.1

Find the algebraic signs of


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 circuit consisting of ________ ANSWER: Correct

Part G At the instant shown, which of the following statements is correct? 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 At the instant shown, which of the following statements is correct? ANSWER:

Energy is flowing from the AC source to the inductor. Energy is flowing from the inductor to the AC source. Correct

Reactance and Current Consider the two circuits shown in the figure. The current in circuit 1, containing an inductor of self-inductance , has an angular frequency , while the current in circuit 2, containing a capacitor of capacitance , has an angular frequency . If we increase dimmer.

and decrease

, both bulbs grow

Part A If we keep and constant, we can achieve the exact same effect of decreasing the brightness of each bulb by performing which of the following sets of actions? Hint A.1


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 decreasing

and

and increasing

decreasing both

and

Correct

As you found out, the reactance of these circuits can be changed not only by varying the frequency of the current, but also by changing the characteristics of the elements in them, i.e., by changing the inductance 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. What happens to the brightness of each bulb if you increase the frequency of the current in the new circuit while keeping and constant?

Hint B.1


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 will have a much smaller amplitude through the inductor than through the capacitor. That is, the inductor tends 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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