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Dan Fullerton

Dan Fullerton

1998 AP Practice Exam: Free Response Questions

Slide Duration:

Table of Contents

I. Electricity
Electric Charge & Coulomb's Law

30m 48s

Intro
0:00
Objective
0:15
Electric Charges
0:50
Matter is Made Up of Atoms
0:52
Most Atoms are Neutral
1:02
Ions
1:11
Coulomb
1:18
Elementary Charge
1:34
Law of Conservation of Charge
2:03
Example 1
2:39
Example 2
3:42
Conductors and Insulators
4:41
Conductors Allow Electric Charges to Move Freely
4:43
Insulators Do Not Allow Electric Charges to Move Freely
4:50
Resistivity
4:58
Charging by Conduction
5:32
Conduction
5:37
Balloon Example
5:40
Charged Conductor
6:14
Example 3
6:28
The Electroscope
7:16
Charging by Induction
7:57
Bring Positive Rod Near Electroscope
8:08
Ground the Electroscope
8:27
Sever Ground Path and Remove Positive Rod
9:07
Example 4
9:39
Polarization and Electric Dipole Moment
11:46
Polarization
11:54
Electric Dipole Moment
12:05
Coulomb's Law
12:38
Electrostatic Force, Also Known as Coulombic Force
12:48
How Force of Attraction or Repulsion Determined
12:55
Formula
13:08
Coulomb's Law: Vector Form
14:18
Example 5
16:05
Example 6
18:25
Example 7
19:14
Example 8
23:21
Electric Fields

1h 19m 22s

Intro
0:00
Objectives
0:09
Electric Fields
1:33
Property of Space That Allows a Charged Object to Feel a Force
1:40
Detect the Presence of an Electric Field
1:51
Electric Field Strength Vector
2:03
Direction of the Electric Field Vector
2:21
Example 1
3:00
Visualizing the Electric Field
4:13
Electric Field Lines
4:56
E Field Due to a Point Charge
7:19
Derived from the Definition of the Electric Field and Coulomb's Law
7:24
Finding the Electric Field Due to Multiple Point Charges
8:37
Comparing Electricity to Gravity
8:51
Force
8:54
Field Strength
9:09
Constant
9:19
Charge Units vs. Mass Units
9:35
Attracts vs. Repel
9:44
Example 2
10:06
Example 3
17:25
Example 4
24:29
Example 5
25:23
Charge Densities
26:09
Linear Charge Density
26:26
Surface Charge Density
26:30
Volume Charge Density
26:47
Example 6
27:26
Example 7
37:07
Example 8
50:13
Example 9
54:01
Example 10
03:10
Example 11
13:58
Gauss's Law

52m 53s

Intro
0:00
Objectives
0:07
Electric Flux
1:16
Amount of Electric Field Penetrating a Surface
1:19
Symbol
1:23
Point Charge Inside a Hollow Sphere
4:31
Place a Point Charge Inside a Hollow Sphere of Radius R
4:39
Determine the Flux Through the Sphere
5:09
Gauss's Law
8:39
Total Flux
8:59
Gauss's Law
9:10
Example 1
9:53
Example 2
17:28
Example 3
22:37
Example 4
25:40
Example 5
30:49
Example 6
45:06
Electric Potential & Electric Potential Energy

1h 14m 3s

Intro
0:00
Objectives
0:08
Electric Potential Energy
0:58
Gravitational Potential Energy
1:02
Electric Potential Energy
1:11
Electric Potential
1:19
Example 1
1:59
Example 2
3:08
The Electron-Volt
4:02
Electronvolt
4:16
1 eV is the Amount of Work Done in Moving an Elementary Charge Through a Potential Difference of 1 Volt
4:26
Conversion Ratio
4:41
Example 3
4:52
Equipotential Lines
5:35
Topographic Maps
5:36
Lines Connecting Points of Equal Electrical Potential
5:47
Always Cross Electrical Field Lines at Right Angles
5:57
Gradient of Potential Increases As Equipotential Lines Get Closer
6:02
Electric Field Points from High to Low Potential
6:27
Drawing Equipotential Lines
6:49
E Potential Energy Due to a Point Charge
8:20
Electric Force from Electric Potential Energy
11:59
E Potential Due to a Point Charge
13:07
Example 4
14:42
Example 5
15:59
Finding Electric Field From Electric Potential
19:06
Example 6
23:41
Example 7
25:08
Example 8
26:33
Example 9
29:01
Example 10
31:26
Example 11
43:23
Example 12
51:51
Example 13
58:12
Electric Potential Due to Continuous Charge Distributions

1h 1m 28s

Intro
0:00
Objectives
0:10
Potential Due to a Charged Ring
0:27
Potential Due to a Uniformly Charged Desk
3:38
Potential Due to a Spherical Shell of Charge
11:21
Potential Due to a Uniform Solid Sphere
14:50
Example 1
23:08
Example 2
30:43
Example 3
41:58
Example 4
51:41
Conductors

20m 35s

Intro
0:00
Objectives
0:08
Charges in a Conductor
0:32
Charge is Free to Move Until the
0:36
All Charge Resides at Surface
2:18
Field Lines are Perpendicular to Surface
2:34
Electric Field at the Surface of a Conductor
3:04
Looking at Just the Outer Surface
3:08
Large Electric Field Where You Have the Largest Charge Density
3:59
Hollow Conductors
4:22
Draw Hollow Conductor and Gaussian Surface
4:36
Applying Gaussian Law
4:53
Any Hollow Conductor Has Zero Electric Field in Its Interior
5:24
Faraday Cage
5:35
Electric Field and Potential Due to a Conducting Sphere
6:03
Example 1
7:31
Example 2
12:39
Capacitors

41m 23s

Intro
0:00
Objectives
0:08
What is a Capacitor?
0:42
Electric Device Used to Store Electrical Energy
0:44
Place Opposite Charges on Each Plate
1:10
Develop a Potential Difference Across the Plates
1:14
Energy is Stored in the Electric Field Between the Plates
1:17
Capacitance
1:22
Ratio of the Charge Separated on the Plates of a Capacitor to the Potential Difference Between the Plates
1:25
Units of Capacitance
1:32
Farad
1:37
Formula
1:52
Calculating Capacitance
1:59
Assume Charge on Each Conductor
2:05
Find the Electric Field
2:11
Calculate V by Integrating the Electric Field
2:21
Utilize C=Q/V to Solve for Capitance
2:33
Example 1
2:44
Example 2
5:30
Example 3
10:46
Energy Stored in a Capacitor
15:25
Work is Done Charging a Capacitor
15:28
Solve For That
15:55
Field Energy Density
18:09
Amount of Energy Stored Between the Plates of a Capacitor
18:11
Example
18:25
Dielectrics
20:44
Insulating Materials Place Between Plates of Capacitor to Increase The Devices' Capacitance
20:47
Electric Field is Weakened
21:00
The Greater the Amount of Polarization The Greater the Reduction in Electric Field Strength
21:58
Dielectric Constant (K)
22:30
Formula
23:00
Net Electric Field
23:35
Key Take Away Point
23:50
Example 4
24:00
Example 5
25:50
Example 6
26:50
Example 7
28:53
Example 8
30:57
Example 9
32:55
Example 10
34:59
Example 11
37:35
Example 12
39:57
II. Current Electricity
Current & Resistance

17m 59s

Intro
0:00
Objectives
0:08
Electric Current
0:44
Flow Rate of Electric Charge
0:45
Amperes
0:49
Positive Current Flow
1:01
Current Formula
1:19
Drift Velocity
1:35
Constant Thermal Motion
1:39
Net Electron Flow
1:43
When Electric Field is Applied
1:49
Electron Drift Velocity
1:55
Derivation of Current Flow
2:12
Apply Electric Field E
2:20
Define N as the Volume Density of Charge Carriers
2:27
Current Density
4:33
Current Per Area
4:36
Formula
4:44
Resistance
5:14
Ratio of the Potential Drop Across an Object to the Current Flowing Through the Object
5:19
Ohmic Materials Follow Ohm's Law
5:23
Resistance of a Wire
6:05
Depends on Resistivity
6:09
Resistivity Relates to the Ability of a Material to Resist the Flow of Electrons
6:25
Refining Ohm's Law
7:22
Conversion of Electric Energy to Thermal Energy
8:23
Example 1
9:54
Example 2
10:54
Example 3
11:26
Example 4
14:41
Example 5
15:24
Circuits I: Series Circuits

29m 8s

Intro
0:00
Objectives
0:08
Ohm's Law Revisited
0:39
Relates Resistance, Potential Difference, and Current Flow
0:39
Formula
0:44
Example 1
1:09
Example 2
1:44
Example 3
2:15
Example 4
2:56
Electrical Power
3:26
Transfer of Energy Into Different Types
3:28
Light Bulb
3:37
Television
3:41
Example 5
3:49
Example 6
4:27
Example 7
5:12
Electrical Circuits
5:42
Closed-Loop Path Which Current Can Flow
5:43
Typically Comprised of Electrical Devices
5:52
Conventional Current Flows from High Potential to Low Potential
6:04
Circuit Schematics
6:26
Three-dimensional Electrical Circuits
6:37
Source of Potential Difference Required for Current to Flow
7:29
Complete Conducting Paths
7:42
Current Only Flows in Complete Paths
7:43
Left Image
7:46
Right Image
7:56
Voltmeters
8:25
Measure the Potential Difference Between Two Points in a Circuit
8:29
Can Remove Voltmeter from Circuit Without Breaking the Circuit
8:47
Very High Resistance
8:53
Ammeters
9:31
Measure the Current Flowing Through an Element of a Circuit
9:32
Very Low Resistance
9:46
Put Ammeter in Correctly
10:00
Example 8
10:24
Example 9
11:39
Example 10
12:59
Example 11
13:16
Series Circuits
13:46
Single Current Path
13:49
Removal of Any Circuit Element Causes an Open Circuit
13:54
Kirchhoff's Laws
15:48
Utilized in Analyzing Circuits
15:54
Kirchhoff's Current Law
15:58
Junction Rule
16:02
Kirchhoff's Voltage Law
16:30
Loop Rule
16:49
Example 12
16:58
Example 13
17:32
Basic Series Circuit Analysis
18:36
Example 14
22:06
Example 15
22:29
Example 16
24:02
Example 17
26:47
Circuits II: Parallel Circuits

39m 9s

Intro
0:00
Objectives
0:16
Parallel Circuits
0:38
Multiple Current Paths
0:40
Removal of a Circuit Element May Allow Other Branches of the Circuit to Continue Operating
0:44
Draw a Simple Parallel Circuit
1:02
Basic Parallel Circuit Analysis
3:06
Example 1
5:58
Example 2
8:14
Example 3
9:05
Example 4
11:56
Combination Series-Parallel Circuits
14:08
Circuit Doesn't Have to be Completely Serial or Parallel
14:10
Look for Portions of the Circuit With Parallel Elements
14:15
Lead to Systems of Equations to Solve
14:42
Analysis of a Combination Circuit
14:51
Example 5
20:23
Batteries
28:49
Electromotive Force
28:50
Pump for Charge
29:04
Ideal Batteries Have No Resistance
29:10
Real Batteries and Internal Resistance
29:20
Terminal Voltage in Real Batteries
29:33
Ideal Battery
29:50
Real Battery
30:25
Example 6
31:10
Example 7
33:23
Example 8
35:49
Example 9
38:43
RC Circuits: Steady State

34m 3s

Intro
0:00
Objectives
0:17
Capacitors in Parallel
0:51
Store Charge on Plates
0:52
Can Be Replaced with an Equivalent Capacitor
0:56
Capacitors in Series
1:12
Must Be the Same
1:13
Can Be Replaced with an Equivalent Capacitor
1:15
RC Circuits
1:30
Comprised of a Source of Potential Difference, a Resistor Network, and Capacitor
1:31
RC Circuits from the Steady-State Perspective
1:37
Key to Understanding RC Circuit Performance
1:48
Charging an RC Circuit
2:08
Discharging an RC Circuit
6:18
The Time Constant
8:49
Time Constant
8:58
By 5 Time Constant
9:19
Example 1
9:45
Example 2
13:27
Example 3
16:35
Example 4
18:03
Example 5
19:39
Example 6
26:14
RC Circuits: Transient Analysis

1h 1m 7s

Intro
0:00
Objectives
0:13
Charging an RC Circuit
1:11
Basic RC Circuit
1:15
Graph of Current Circuit
1:29
Graph of Charge
2:17
Graph of Voltage
2:34
Mathematically Describe the Charts
2:56
Discharging an RC Circuit
13:29
Graph of Current
13:47
Graph of Charge
14:08
Graph of Voltage
14:15
Mathematically Describe the Charts
14:30
The Time Constant
20:03
Time Constant
20:04
By 5 Time Constant
20:14
Example 1
20:39
Example 2
28:53
Example 3
27:02
Example 4
44:29
Example 5
55:24
III. Magnetism
Magnets

8m 38s

Intro
0:00
Objectives
0:08
Magnetism
0:35
Force Caused by Moving Charges
0:36
Dipoles
0:40
Like Poles Repel, Opposite Poles Attract
0:53
Magnetic Domains
0:58
Random Domains
1:04
Net Magnetic Field
1:26
Example 1
1:40
Magnetic Fields
2:03
Magnetic Field Strength
2:04
Magnets are Polarized
2:16
Magnetic Field Lines
2:53
Show the Direction the North Pole of a Magnet Would Tend to Point if Placed on The Field
2:54
Direction
3:25
Magnetic Flux
3:41
The Compass
4:05
Earth is a Giant Magnet
4:07
Earth's Magnetic North Pole
4:10
Compass Lines Up with the Net Magnetic Field
4:48
Magnetic Permeability
5:00
Ratio of the magnetic Field Strength Induced in a Material to the Magnetic Field Strength of the Inducing Field
5:01
Free Space
5:13
Permeability of Matter
5:41
Highly Magnetic Materials
5:47
Magnetic Dipole Moment
5:54
The Force That a Magnet Can Exert on Moving Charges
5:59
Relative Strength of a Magnet
6:04
Example 2
6:26
Example 3
6:52
Example 4
7:32
Example 5
7:57
Moving Charges In Magnetic Fields

29m 7s

Intro
0:00
Objectives
0:08
Magnetic Fields
0:57
Vector Quantity
0:59
Tesla
1:08
Gauss
1:14
Forces on Moving Charges
1:30
Magnetic Force is Always Perpendicular to the Charged Objects Velocity
1:31
Magnetic Force Formula
2:04
Magnitude of That
2:20
Image
2:29
Direction of the Magnetic Force
3:54
Right-Hand Rule
3:57
Electron of Negative Charge
4:04
Example 1
4:51
Example 2
6:58
Path of Charged Particles in B Fields
8:07
Magnetic Force Cannot Perform Work on a Moving Charge
8:08
Magnetic Force Can Change Its Direction
8:11
Total Force on a Moving Charged Particle
9:40
E Field
9:50
B Field
9:54
Lorentz Force
9:57
Velocity Selector
10:33
Charged Particle in Crosses E and B Fields Can Undergo Constant Velocity Motion
10:37
Particle Can Travel Through the Selector Without Any Deflection
10:49
Mass Spectrometer
12:21
Magnetic Fields Accelerate Moving Charges So That They Travel in a Circle
12:26
Used to Determine the Mass of An Unknown Particle
12:32
Example 3
13:11
Example 4
15:01
Example 5
16:44
Example 6
17:33
Example 7
19:12
Example 8
19:50
Example 9
24:02
Example 10
25:21
Forces on Current-Carrying Wires

17m 52s

Intro
0:00
Objectives
0:08
Forces on Current-Carrying Wires
0:42
Moving Charges in Magnetic Fields Experience Forces
0:45
Current in a Wire is Just Flow of Charges
0:49
Direction of Force Given by RHR
4:04
Example 1
4:22
Electric Motors
5:59
Example 2
8:14
Example 3
8:53
Example 4
10:09
Example 5
11:04
Example 6
12:03
Magnetic Fields Due to Current-Carrying Wires

24m 43s

Intro
0:00
Objectives
0:08
Force on a Current-Carrying Wire
0:38
Magnetic Fields Cause a Force on Moving Charges
0:40
Current Carrying Wires
0:44
How to Find the Force
0:55
Direction Given by the Right Hand Rule
1:04
Example 1
1:17
Example 2
2:26
Magnetic Field Due to a Current-Carrying Wire
4:20
Moving Charges Create Magnetic Fields
4:24
Current-Carrying Wires Carry Moving Charges
4:27
Right Hand Rule
4:32
Multiple Wires
4:51
Current-Carrying Wires Can Exert Forces Upon Each Other
4:58
First Right Hand Rule
5:15
Example 3
6:46
Force Between Parallel Current Carrying Wires
8:01
Right Hand Rules to Determine Force Between Parallel Current Carrying Wires
8:03
Find Magnetic Field Due to First Wire, Then Find Direction of Force on 2nd Wire
8:08
Example
8:20
Gauss's Law for Magnetism
9:26
Example 4
10:35
Example 5
12:57
Example 6
14:19
Example 7
16:50
Example 8
18:15
Example 9
18:43
The Biot-Savart Law

21m 50s

Intro
0:00
Objectives
0:07
Biot-Savart Law
0:24
Brute Force Method
0:49
Draw It Out
0:54
Diagram
1:35
Example 1
3:43
Example 2
7:02
Example 3
14:31
Ampere's Law

26m 31s

Intro
0:00
Objectives
0:07
Ampere's Law
0:27
Finds the Magnetic Field Due to Current Flowing in a Wire in Situations of Planar and Cylindrical Symmetry
0:30
Formula
0:40
Example
1:00
Example 1
2:19
Example 2
4:08
Example 3
6:23
Example 4
8:06
Example 5
11:43
Example 6
13:40
Example 7
17:54
Magnetic Flux

7m 24s

Intro
0:00
Objectives
0:07
Magnetic Flux
0:31
Amount of Magnetic Field Penetrating a Surface
0:32
Webers
0:42
Flux
1:07
Total Magnetic Flux
1:27
Magnetic Flux Through Closed Surfaces
1:51
Gauss's Law for Magnetism
2:20
Total Flux Magnetic Flux Through Any Closed Surface is Zero
2:23
Formula
2:45
Example 1
3:02
Example 2
4:26
Faraday's Law & Lenz's Law

1h 4m 33s

Intro
0:00
Objectives
0:08
Faraday's Law
0:44
Faraday's Law
0:46
Direction of the Induced Current is Given by Lenz's Law
1:09
Formula
1:15
Lenz's Law
1:49
Lenz's Law
2:14
Lenz's Law
2:16
Example
2:30
Applying Lenz's Law
4:09
If B is Increasing
4:13
If B is Decreasing
4:30
Maxwell's Equations
4:55
Gauss's Law
4:59
Gauss's Law for Magnetism
5:16
Ampere's Law
5:26
Faraday's Law
5:39
Example 1
6:14
Example 2
9:36
Example 3
11:12
Example 4
19:33
Example 5
26:06
Example 6
31:55
Example 7
42:32
Example 8
48:08
Example 9
55:50
IV. Inductance, RL Circuits, and LC Circuits
Inductance

6m 41s

Intro
0:00
Objectives
0:08
Self Inductance
0:25
Ability of a Circuit to Oppose the Magnetic Flux That is Produced by the Circuit Itself
0:27
Changing Magnetic Field Creates an Induced EMF That Fights the Change
0:37
Henrys
0:44
Function of the Circuit's Geometry
0:53
Calculating Self Inductance
1:10
Example 1
3:40
Example 2
5:23
RL Circuits

42m 17s

Intro
0:00
Objectives
0:11
Inductors in Circuits
0:49
Inductor Opposes Current Flow and Acts Like an Open Circuit When Circuit is First Turned On
0:52
Inductor Keeps Current Going and Acts as a Short
1:04
If the Battery is Removed After a Long Time
1:16
Resister Dissipates Power, Current Will Decay
1:36
Current in RL Circuits
2:00
Define the Diagram
2:03
Mathematically Solve
3:07
Voltage in RL Circuits
7:51
Voltage Formula
7:52
Solve
8:17
Rate of Change of Current in RL Circuits
9:42
Current and Voltage Graphs
10:54
Current Graph
10:57
Voltage Graph
11:34
Example 1
12:25
Example 2
23:44
Example 3
34:44
LC Circuits

9m 47s

Intro
0:00
Objectives
0:08
LC Circuits
0:30
Assume Capacitor is Fully Charged When Circuit is First Turned On
0:38
Interplay of Capacitor and Inductor Creates an Oscillating System
0:42
Charge in LC Circuit
0:57
Current and Potential in LC Circuits
7:14
Graphs of LC Circuits
8:27
V. Maxwell's Equations
Maxwell's Equations

3m 38s

Intro
0:00
Objectives
0:07
Maxwell's Equations
0:19
Gauss's Law
0:20
Gauss's Law for Magnetism
0:44
Faraday's Law
1:00
Ampere's Law
1:18
Revising Ampere's Law
1:49
Allows Us to Calculate the Magnetic Field Due to an Electric Current
1:50
Changing Electric Field Produces a Magnetic Field
1:58
Conduction Current
2:33
Displacement Current
2:44
Maxwell's Equations (Complete)
2:58
VI. Sample AP Exams
1998 AP Practice Exam: Multiple Choice Questions

32m 33s

Intro
0:00
1998 AP Practice Exam Link
0:11
Multiple Choice 36
0:36
Multiple Choice 37
2:07
Multiple Choice 38
2:53
Multiple Choice 39
3:32
Multiple Choice 40
4:37
Multiple Choice 41
4:43
Multiple Choice 42
5:22
Multiple Choice 43
6:00
Multiple Choice 44
8:09
Multiple Choice 45
8:27
Multiple Choice 46
9:03
Multiple Choice 47
9:30
Multiple Choice 48
10:19
Multiple Choice 49
10:47
Multiple Choice 50
12:25
Multiple Choice 51
13:10
Multiple Choice 52
15:06
Multiple Choice 53
16:01
Multiple Choice 54
16:44
Multiple Choice 55
17:10
Multiple Choice 56
19:08
Multiple Choice 57
20:39
Multiple Choice 58
22:24
Multiple Choice 59
22:52
Multiple Choice 60
23:34
Multiple Choice 61
24:09
Multiple Choice 62
24:40
Multiple Choice 63
25:06
Multiple Choice 64
26:07
Multiple Choice 65
27:26
Multiple Choice 66
28:32
Multiple Choice 67
29:14
Multiple Choice 68
29:41
Multiple Choice 69
31:23
Multiple Choice 70
31:49
1998 AP Practice Exam: Free Response Questions

29m 55s

Intro
0:00
1998 AP Practice Exam Link
0:14
Free Response 1
0:22
Free Response 2
10:04
Free Response 3
16:22
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Lecture Comments (1)

0 answers

Post by Professor Dan Fullerton on March 27, 2015

Correct link: http://apcentral.collegeboard.com/apc/public/courses/211623.html

1998 AP Practice Exam: Free Response Questions

Lecture Slides are screen-captured images of important points in the lecture. Students can download and print out these lecture slide images to do practice problems as well as take notes while watching the lecture.

  • Intro 0:00
  • 1998 AP Practice Exam Link 0:14
  • Free Response 1 0:22
  • Free Response 2 10:04
  • Free Response 3 16:22

Transcription: 1998 AP Practice Exam: Free Response Questions

Hello, everyone, and welcome back to www.educator.com.0000

I'm Dan Fullerton, and in this final lesson we are going to continue our work on the 1998 practice exam,0003

this time focusing on the free response questions.0010

Let us dive right in and make sure you have got the test printed out.0014

I highly recommend that you take a shot at these questions before coming back to the video.0017

Starting with number 1, let us look at part A.0022

We have that sphere B1 and a point/ A and that B1 sphere is held in place.0027

We area asked to find the charge on sphere B1.0034

As I look at this, the first thing I think I'm going to do is draw a free body diagram for B1 to see all the forces that are on it.0038

Let us take a look here.0046

If there is our sphere, we will put our Y axis and X axis.0048

And labeling their forces, we have an electrical force repulsion to the right.0057

We have a force of tension from our string, we will call that T.0062

We have the weight of the sphere MG.0069

Using Newton’s second law, we can write that the net force in the Y direction is going to be T sin 70°.0075

If 70° here, if that is 20 to match what they have on our diagram would give us 4 θ, - MG = 0,0088

which implies then that T must equal MG/ sin 70° which is our mass 0.025 × the acceleration0094

due to gravity on the surface of the Earth 9.8 m / s² / the sin of 70° or about 0.26 N.0111

We try the same thing in the X direction.0123

The net force in the X direction, we have the electrical force to the right - T cos 70, the X component of that tension.0127

Those all have to be = to 0 because it is in equilibrium, it is not accelerating.0139

Which implies that the electrical force = T cos 70° is going to be, we just found T was 0.26 N cos 70° or 0.089 N.0144

If we know the electrical force, we should be able to back out that charge using in Coulomb’s law.0162

Our electrical force is K QA QB / R² which implies that the charge on sphere B is just going to be our electrical force × R² ÷ K QA.0167

Our electrical force we just found was 0.089 N.0187

The distance between the centers of our charged particles is 1.5 m, K 9 × 10⁹ N meters²/ C² which is the same as 1/ 4 π ε₀.0196

The charge on A is 120 µc.0212

Put that all into my calculator and I find the value for QB of 1.86 × 10⁻⁷ C.0218

Part A, check.0231

Part B asks, suppose this b1 is replaced by a second suspended sphere B2 that has the same mass but this one is conducting.0235

If we establish equilibrium again, what happens to that equilibrium angle?0246

That angle has going to have to be less, that angle θ is going to be less than 20°0251

because the charges are going to move in the conductor leaving more positive charge on the far side of your B2 sphere0257

and less on the near side near A.0264

Effectively, that distance between charges increases so the electrical force between them is going to decrease by Coulomb’s law.0266

Electrical force decreases, it is going to come down a little bit.0274

I would say that that is going to be less than 20°.0277

Here for part C, the sphere B2 is now replaced by a very long horizontal none conducting tube.0282

The tube is hollow with thin walls, gives us the radius and the uniform positive charge per unit length λ.0290

Use Gauss’s law to show that the electric field is given by that expression.0298

I’m really good at Gauss’s law, so we will try that.0303

Here is our tube and the Gaussian surface I'm going to choose is a cylinder around that.0306

Choose a Gaussian surface over here at some distance R from its center.0317

The radius of the little one inside this R, the radius of our Gaussian surface is r.0324

We can write Gauss’s law, integral / the closed surface of E ⋅ DA is equal to the total enclosed charge ÷ ε 0.0333

The left hand side electric field should be constant because we chose this with symmetry, that in mind.0346

And the area of our close surface is going to be 2 π R × its length L and0353

that is going to be equal to the enclosed charge which is the linear charge density × our length ÷ ε₀.0361

Therefore, our electric field is going to be λ L / 2 π ε₀ RL.0370

Our L's make a ratio of 1 so that is just going to the λ / 2 π ε₀ R.0379

And if we start putting our values in here, λ is 0.1 × 10⁻⁶ C / m, 2 π ε₀ is 8.54 × 10 ⁻12 C²/ N-m².0386

And we also have our R down here.0406

This implies then that the electric field is going to be, I come up with about 1797/ R N-m/ C,0410

which is approximately 1800/ assuming R is in meters, N/ C with R in meters.0422

We prove that 1800/ R N/ C.0437

Very good, moving on to part B.0443

A small sphere A with charged 120 µc is now brought in the vicinity of the tube and held at distance of 1.5 m.0448

Find the force the tube exerts on the sphere.0455

That should be pretty easy now that we have the electric field strength.0458

The force is just charge × electric field which is going to be our 120 × 10⁻⁶ C × our 1800 N/ C / R 1.5 m or 0.144 N.0463

And part E, let us take a look at E here.0486

Calculate the work done against the electrostatic repulsion to move sphere A toward0494

the tube from a distance R = 1.5 m to a distance R= 0.3 m from the tube.0500

That should be a pretty straightforward calculation of work.0506

Work = the integral of F ⋅ DR, which is going to be the opposite of the integral as we go from R = 1.5 m to 0.3 m0509

throughout the work done against the electrostatic repulsion of, we have our 120 × 10⁻⁶ our charge × our 1800 N/ C ÷ R0521

to give us our force QE DR, which implies then that the work is going to be equal to -0.216, when I pull the constants out.0539

Integral from R = 1.5 to 0.3 of DR/ R.0552

Integral of DR/ R is nat log of R.0559

W = -0.216 log of R evaluated from 1.5 to 0.3, which is -0.216 log of 0.3 - log of 1.5,0563

which is when I plugged into my calculator about 0.348 J.0583

That finishes up free response problem number 1.0595

Let us move on to number 2, a circuit problem with the capacitor and inductor.0598

As we look here at number 2, we are going to start off with the circuit, the switches initially open,0605

the capacitor is uncharged, and there is a voltmeter but they are not showing the measure of potential difference across R1.0612

On the diagram above, draw the voltmeter with a proper connections for measuring the potential difference.0618

That was nice and simple, what you will only do is you take your drawing that you have there, there is your 20V.0625

We have R1 just put your voltmeter in parallel V with R1.0629

Taking a look at part B, at time T = 0, the switch is moved to position A so we now have the capacitor in the circuit.0640

Find the voltmeter reading for the time right after you do.0649

The trick here is the moment you close that, that capacitor initially is going to act like a wire.0653

Effectively, you have a circuit that looks like this.0658

There is R1, there is R2.0662

R1 is 10 ohms, R 2 is 20 ohms, and if we want to know the voltmeter reading,0670

we could do this as a voltage divider and probably not too tough to say that it is going to drop 1/3 of the voltage across,0677

1/3 of the resistance to 10 ohms which will be 1/3 of 20 or 6.67V.0685

But more formally, we can do this by saying the total current in the circuit is E/ R which is going to be 20 V/ your total resistance 30 ohms or 0.667 amps.0690

The voltage drop across R1 is just going to be current × resistor 1 which is 0.667 amps × our 10 ohms which is about 6.67 V.0704

Moving on to part C, after a long time the measurement of potential difference across R1 is again taken,0721

determine for this later time the voltmeter reading.0731

After a long time, the capacitor is going to act like an open.0736

That is going to break your circuit, you are not going to have any current flowing through R1.0741

Therefore, there is no voltage drop across R1 so the voltmeter reading is going to be equal to 0.0747

And C2, the charge on the capacitor.0757

The entire voltage will be across the capacitor at that point.0759

If C = Q/ V, that implies that Q = CV, which is going to be your capacitance 15 µf or 15 × 10⁻⁶ F ×0763

your potential difference across your capacitor 20 V or 3 × 10⁻⁴ C.0775

There is part C, let us take a look at D.0787

At a still later time t = T, the switch is moved to position B.0795

Determine the voltmeter reading right after that happens.0800

Now we have got a circuit that looks kind of like this, for 2D we got out 20V.0803

At that next position for our switch, there is R1.0812

Here is our inductor and we have R2 in the mix, that is a 2 H conductor.0816

Initially, the inductor poses current flowing and acts like an open so potential across R1 is going to be 0.0827

Let us take a look at 2 E, a long time after t = T, the current in R1 reaches some constant final value I final.0839

Determine what that is going to be.0848

After a long time, that inductor just acts like a wire so I final is going to be equal to the potential ÷ the resistance in our circuit.0851

Which is again, 20 V/ 30 ohms or 0.667 amps.0861

Part 2, determine the energy stored in the inductor.0872

The energy in the inductor is given by ½ LI² and it is going to be ½ × our inductance 2 H × the square of our current 0.667 amps² or about 0.445 J.0875

We have got a part F, write but do not solve the differential equation for the current in R1 as a function of time.0896

We can use Faraday’s law to do this, the integral/ the closed loop of E ⋅ DL = - the time rate of change of the magnetic flux.0908

As we go around here, remember no electric field in there so we have got - E + I R1.0921

We have got the voltage drop across I R2 and all of that has to equal their change in flux - L DI DT.0932

There is no electric field here so no contribution to the left hand side from your inductor.0944

We got ID IDT so I suppose we have already written that differential equation.0949

But just to clean it up a touch, let us call that E – I R1 + R2 - L DI DT = 0.0954

There is the differential equation and we do not have to solve it this time.0971

That is a straightforward number 2 and I think we are going to make up for here in number 3.0976

This is one of the trickier free response problems, more involved free response problems I have seen on AP exam.0982

Here we have a conducting bar of mass M that is placed on some conducting rails0992

at distance L apart that is at an angle with respect to the horizontal.0997

We have got a magnetic field coming out of that ramp.1001

We are asked to determine the current on the circuit when the bar has reached the constant final speed, so 3A.1005

First thing I'm going to do is I'm going to make a free body diagram for that bar and I'm going to tilt my axis1013

so it is a little easier for me to see what is going.1021

On our bar we are going to have a magnetic force up the ramp and we are going to have gravitational force pulling it down.1029

We know that that force is going to be up the ramp, we can use the right hand rule Q × V × B1040

to find the force on the charges in the bar creating the current.1046

And then the charges in the bar move to find the force on the bar which is going to be up the incline.1049

As I do this, as I take a look here, determine the current in the circuit when the bar is reached a constant final speed.1054

Let us see, net force in the X direction, Newton’s second law point our free body diagram1065

is going to be the X component of its weight MG sin θ - FB the magnetic force, all has to equal MA.1072

Since V is constant, A = 0.1085

Therefore, we can say that the magnetic force has to equal MG sin θ, its constant final speed.1089

The magnetic force, my current flowing through the wire is going to be I LB.1100

Therefore, we can state that MG sin θ must equal I LB or the current must be equal to MG sin θ/ LB.1107

For part B, we are asked to determine the constant final speed of the bar.1130

For B, if current is potential / resistance which is – D φ B DT/ R.1138

Let us see here, we could do that and all of that must equal MG sin θ/ LB.1151

It would be helpful to know what this D φ B term is.1162

Φ B is going to be B × A or in this case B LX are geometry.1167

D φ B with respect to time is just going to be BL and X is the only thing that changes with time.1177

That is DX DT which is going to B BL × its velocity.1185

I can put that in up there for that.1189

This implies then that, we have B LV/ R = MG sin θ/ LB.1194

And solving just for B 4V, V would then equal, we will have MG R sin θ/ L² B².1212

Plugging away through this, part C.1234

Part C says, determine the rate at which energy is being dissipated when its reach its constant final speed.1241

Varying at which energy is dissipated.1247

A lots of fancy way of saying find the power.1249

We know I, we know R, so power = I² R is just going to be this² multiplied by R.1251

We will have M² G² R sin² θ / L² B².1260

Not too bad on C.1279

D, express the speed of the bar as a function of time T from the time it is released at T equal 0.1283

This one looks a bit more involved.1292

Let us see, we know MG sin θ - the magnetic force = MA Newton’s second law.1297

We said the magnetic force was I LB, we also know that A is the time rate of change of velocity DV DT.1310

MG sin θ – I LB must equal M DV DT.1320

But we also know that I = B LV / R, therefore, MG sin θ -, substituting in here for FB, B LV/ R.1332

We have still got our LB again, LV LB = M DV DT.1355

Let us move that M by all sides by M.1370

I get on the left hand side G sin θ - L² B² V/ MR = DV DT.1375

Which implies then, it is time to do our separation of variables.1394

This might get a little bit ugly, DV/ L² B² V/ MR - G sin θ = – DT.1398

We are going to try and integrate this.1419

We are going to integrate, the composite.1422

The integral from V equal 0 to some final value of V of, we have got DV / L² B² V / MR -G sin θ.1425

If I want that = the integral from t equal 0 to T of DT with a negative sign, I want this to fit the formula DU/ U.1445

I have got DV, I got V here.1459

I have got all this other stuff there as well.1461

I’m going to have to have that at the top to make that fit the form so that is L² B² / MR.1464

If I multiplying the left by L² B²/ MR, I have to multiply the right by L² B²/ MR × negative integral from 0 to T of DT.1473

That fits the form of DU/ U.1486

This implies then that this should be a nat log of our U which is down here,1491

L² B² V/ MR -G sin θ all evaluated from V = 0 to V, must equal.1499

This thankfully is little bit easier to integrate, - L² D² T/ MR.1514

We can expand this in the left as we substitute in our 0 and V.1525

The left hand side will be, we get L² B² V/ MR - G sin θ - the log of - G sin θ = - L² B² T/ MR.1529

We can compress this right hand side.1557

The difference of logs is the log of the quotient.1558

That is going to imply that the log of, we have L² B² V/ M R - G sin θ/ -G sin θ = - L² B² T/ MR.1562

Where do we go from here?1594

Let us see, it looks like we are going to have to do some simplifications.1597

I think we can get rid of that G sin θ or at least incorporate it a little bit better.1601

Which implies then that the log of - L² B² V/ MR G sin θ + 1 = - L² B² T/ MR, raising both sides to E1606

to get rid of that natural log, which implies in the left hand side becomes - L² B² V/ M R G sin θ +11632

must equal E ^- L² B² T/ MR, which implies then, moving on to the next page.1646

L² B² V/ MR G sin θ = 1 - E ^- L² B² T/ MR.1660

Since we are solving for V, we can do one last step.1678

Get VL by itself and say that V =, we will have MR G sin θ/ L² B² × (1 - E ^- L² B² T/ MR).1683

Let us get that a nice big 3D box because good heaven knows we certainly earned that on that problem.1714

A lot involved on part D there.1721

Quite a tricky bit of work and a lot of math.1725

Let us finish off with part E, so 3 even.1729

Suppose the experiment is performed again, this time with the second identical resistor connecting the rails at the bottom.1735

Will this affect the final speed and how? Justify your answer.1741

If you have 2 resistors in parallel, your total equivalent resistance is going to go down.1747

We just said that the final velocity is MG R sin θ/ B² L².1758

If R decreases, our final velocity must go down.1770

I would say that the final speed increases because R is decreasing there.1774

Hopefully that gets you a great start to AP Physics C Electricity and Magnetism.1785

Thank you so much for joining us at www.educator.com.1792

Make it a great day everybody.1794