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Dan Fullerton

Dan Fullerton

1998 AP Practice Exam: Multiple Choice 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
1:03:10
Example 11
1:13:58
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 (8)

1 answer

Last reply by: Professor Dan Fullerton
Sun Aug 26, 2018 1:43 PM

Post by Laura Darrow on August 25 at 08:43:20 PM

Hello Mr. Fullerton ~ I have a question about #70. If you put an insulator between the plates wouldn't the distance d be smaller? Smaller distance, larger capacitance. What does "without otherwise disturbing the system" mean? Many thanks.




2 answers

Last reply by: Laura Darrow
Mon Aug 27, 2018 4:33 PM

Post by Laura Darrow on August 25 at 05:17:22 PM

Hello Mr. Fullerton - I don't have a clear understanding of #67, could you clarify further? Thanks!

1 answer

Last reply by: Professor Dan Fullerton
Mon May 8, 2017 10:39 AM

Post by Stephen Saucedo on May 8, 2017

You're not really explaining anything that well. Also you're going super fast.

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: Multiple Choice Questions

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  • 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

Transcription: 1998 AP Practice Exam: Multiple Choice Questions

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

In this lesson, we are going to take an AP Practice Exam and go through the multiple choice portion.0003

The link of that is available right here.0011

I highly recommend that you take a minute, go through, and print it out, and try and take the test on your own first.0013

Then come back and we will go through the walk through this together.0020

As we take a look here, let us start on the E and M portion of this practice test from 1998.0025

In question 36 is where we are going to begin.0031

As we look at number 36, let us see.0036

We a resistor and capacitor acted in series to a battery.0040

I always like to draw these out.0044

There is V0 ±, we have got our resistor R, we got our capacitor C, and there is our circuit.0048

I will define the direction for positive current flow there I.0059

Which in the following questions relating the current and the circuit and the charge describes the circuit?0065

If we use Kirchhoff’s voltage law and we go around, let us start from here and go around clockwise.0071

I could have write my KVL that we have -V0 + IR + the voltage across our capacitor VC = 0.0076

But since capacitance is charge / voltage, voltage is charge/ capacitance so we could write this as - V0 + IR + Q/ C = 0.0091

Or V0 - IR -Q/ C = 0 and that will match the form of answer B.0108

37, which of those following combinations of 4 ohm resistors would dissipate 24 W connected to a 12V battery?0126

First, it would be nice know what that resistance is.0135

If power is V² / R, then resistance is V²/ our power which is going to be 12 V²/ 24W.0139

144/ 24 will be 6 ohms.0149

We need a combination of resistors that will give us 6 ohms.0152

If those are all 4 ohms resistors, it looks to me like E gives us a 4 ohm resistor in series with 2/ 4.0156

The 2/ 4 in parallel gives us 2 ohms.0164

4 + 2 is 6, correct answer must be E.0166

38, we have 2 uncharged conductors 1 and 2 on insulating stands, they are in contacts.0175

Let us draw those and we are going to bring a negatively charged rod near them, there it is.0181

When we do that, the electrons are closest to the rod and they are screaming away.0187

They are repelled so they are all going to hang out over here as much as possible, leaving this one slightly positively charged.0192

We separate those spheres, what is now true of conductor number 2, this one over here.0199

It is negatively charged so the answer must be C.0206

Taking a look at 39, we have a couple of charges in a test charge + Q,0212

what is the direction of force on the test charge due to the other two charges?0223

We are going to get a force from the top one that is down the force F.0227

We are going to get a force to the left F from the one on the right because they repel.0232

The net is going to be down into the left or E.0237

40, relates to the same question, if F is the magnitude of the force on the test charge due to 1, what is the net force acting on it?0244

We can figure this out with the Pythagorean Theorem.0253

That is going to be in that direction, the combination of those two.0257

The total is going to be √ F² + F² which is √ 2F² or √ 2 × F.0260

The answer is D.0274

Taking a look at 41, Gauss’s law provides a convenient way to calculate the electric field outside and each of the following accept what?0278

A large plate, we did that.0298

A sphere, we did that.0301

A cubed, it is going to be difficult with Gauss’s law because that is symmetry piece.0303

The long solid and a long hollow cylinder, our lectures on Gauss’s law we did A, B, D, and E.0307

The only one that is going to be difficult to use Gauss’s law for that purpose is going to be C.0314

Let us take a look, moving on here to 42, have a wire resistance R dissipating some power when the current passes through it.0321

The wire is replaced by another wire with resistance 3R, what is the power dissipated with the same current?0332

Power is I² R so our final power, once we triple our resistance is going to be I² × 3R, which is just going to be 3 I² R or 3 × our initial power.0338

The answer must be D, 3P.0354

Moving on 43, we got a narrow beam of protons producing that current of 1.6 ma.0361

They are 10⁹ proton in each meter, which is the best estimate of the average speed of the protons in the beam?0369

I like it, a little more challenging.0375

Current is 1.6 × 10⁻³ amps.0378

We have got 10⁹ protons / m and we are trying to find the average speed.0383

The way I might go about this is, I’m first going to recognize the current as charge per unit time.0394

We need to also somehow get velocity in here so I'm going to look at velocity being distance /time.0400

Therefore, time is distance / velocity and I'm going to plug that in for T here to get that my current is going to be Q/ D × V.0406

Let us also look and see if we can figure out the charge / m, since we got a Q/ D factor here.0421

We have 10⁹ elementary charges, the charge on each proton / m.0427

Let us convert that into Coulomb’s, that is 1.6 × 10 ⁻19 C/ elementary charge, which gives us 1.6 × 10 ⁻10 C / m.0433

Putting all of this together now, we said current is Q/ D × V.0447

Therefore, V must be current × D/ Q which is going to be our current was 1.6 × 10⁻³ amps Q/ D, that is what we have here C/ m.0454

Q/ D in the denominator would be 1.6 × 10 ⁻10 C/ m, which is going to leave us with 10⁷ m/ s.0469

Is it one of our answer? It is, answer D.0482

Very good, let us take a look at 44.0487

Which of the following describes the line of magnetic field in the vicinity of the beam due to the beam's current?0493

Either way, concentric circles around the beam.0499

Absolutely, that is got to be at 44 A.0502

45, we have got a couple of charges on X axis and we want to know where the electric field strength could be 0.0508

That is got to be between the two so that they cancel out.0518

And where is it going to be between the two?0525

That has got to be a little bit closer to that + 2Q charge so that the distance factor0526

to make up for the fact that the charge on the right has less charge, so less force.0532

I would say it has to be between 2 and 3, answer C.0536

And 46, relating to the same problem.0544

The electric potential is negative at some points on the line in which of the following ranges?0546

That does not make sense, with only positive charges we are not going to have negative electric potentials.0551

I'm going to go with E on that one.0557

There are no places where you are going to have a negative potential at least in that current charge configuration.0560

Let us take a look here at number 47, we have got a graph showing the electric potential and region of space as a function of position.0570

Where would the charged particle experience the force of greatest magnitude?0580

You are going to have the greatest force where you have the greatest electric field strength.0584

Remember that the electric field is - DV DR so if that relates to the slope,0592

we are going to have the strongest field where we have the strongest, the greatest slope on our voltage position graph.0598

That to me looks like it is going to be position D, the greatest slope is going to have the greatest electric field.0604

Therefore, the greatest force on a charged particle.0612

48, work that must be done by an external agent to move a point charge of 2 µc from the origin 2, 3 m away as 5 J.0619

Find the potential difference.0628

Potential difference is work per unit charge, that is going to be 5 J / 2 mc or 0.002 C.0631

It is just 2500 V, the answer is C.0639

49, we have got a wire loop, we have got a wire, and we are looking for net force on the loop.0648

We have done similar things here in our lectures.0655

Here is our current I2, let us draw our loop up here.0658

Where this is I1 going that direction and the first thing I'm going to do if I want to know the net force on the loop is I'm going to figure out the direction of the magnetic field due to I2.0667

From the right hand rule, it looks like that is going to be into the plane of the page0677

so I'm just going to draw that with some axis up here to help me remember that.0680

They should be uniform but I have poor drawing skills here.0687

The direction of the force, by the right hand rule, if we have positive current,0695

point the fingers of my right hand in the direction current is flowing, bend in the direction of the magnetic field.0700

You are going to see a force that is up from that section of wire.0705

Let me use red here.0708

Up from that section of wire, over here it is going to be to the right.0710

Here it is going to be down and here it is going to be the left.0715

To where it is going to be that these two are going to cancel, these are not however0719

because the closest wire is going to be in the stronger magnetic field because it is closer to the source.0723

The smaller distance from the source of that magnetic fields.0728

This magnetic field is stronger here than here.0731

We will have a greater force down so the answer must be A, toward the wire.0733

Number 50, one of the trickier problems on the test.0746

A uniform magnetic field B is parallel to the XY plane as shown and we have got a proton initially moving with velocity V at that angle θ.0750

How will that follow that kind of path?0758

In here you have really got to think and visualize in 3 dimensions.0760

How the right hand rule is going to work?0764

Point your fingers in the direction it is moving, bend in the direction of the magnetic field,0766

and you are going to see an evolving, twisting path as you get a force but keep bringing it around in the helical path.0772

That is going to be D, A helical path with its axis parallel to the Y axis in the direction of the magnetic field.0780

51, taking a look here, we have a parallel plate capacitor.0791

I like drawing parallel plate capacitors, let us do that.0798

Parallel plate capacitor + Q on one plate -Q on the other and that is separated by some distance D.0805

Good so far.0816

Let us see, if each plate has an area A.0818

A single proton of charge + E is released from rest of the surface of the positively charged plate.0822

There is our proton charge + E.0829

What kinetic energy be or what will it be proportional to when it hits the other plate?0831

Let us go back to my definition of potential is work / charge, which implies that the work done which is QV.0841

When we get down here that is going to be the kinetic energy, 100% efficient in our problem.0851

But we also know that if C = Q/ V, then V = Q/ C.0857

We can write this as our charge of our proton × a charge on one of our plates ÷ the capacitance is equal to QV.0862

Since capacitance is proportional to area/ the separation of the plates,0876

we can say that the change in kinetic energy is going to be, not equal to, proportional to EQ/ C, which is A/ D.0882

DQ D/ A, do we have anything that looks like that, the answer is A.0896

Moving on to 52, in which of the following case does there exist a non 0 magnetic field that can be conveniently determined using Ampere’s law.0906

Outside the point charge if it is at rest.0919

It cannot be A, that is not going to work.0921

You do not get a magnetic field from a charge at rest.0923

Inside a stationary cylinder carrying current, no it is not moving again.0926

No charge, no magnetic field.0929

How about C, inside a very long current carrying solenoid?0931

We actually did that in the Ampere’s law video.0935

We know the answer is C, but let us take a look at the next 2 just for kicks.0938

At the center of the current carrying loop of wire, we did that but not with Ampere’s law.0942

That was pretty good, be pretty tough to integrate.0947

We did that with the Biot-Savart law and outside the square current carrying loop of wire.0950

That would be pretty tough to integrate too.0954

C is going to be the nice simple easy one.0956

53, positive beam of protons moves parallel to the X axis in the positive X direction as shown.0961

The magnetic field is pointed in the positive Y direction, in what direction must the electric field be pointed?0970

But let us se, the magnetic field is pointing in the positive Y and it is going to go with 0 deflection.0976

Then the magnetic force has to exactly balance the electrical force.0983

By the right hand rule then, that means that the direction of the electric field must be in the -Z direction0986

in order to balance the force from the magnetic field.0994

That is got a be E, negative Z direction.0998

Taking a look at 54, a vertical length of copper wire moves to the right with a steady velocity V1005

in the direction of constant horizontal magnetic field B.1012

To the following describes the induce charges on the ends of the wire.1016

54, V cross B is going to be 0, they are in the same direction.1020

I'm going to go with E on that one.1025

55, suppose an electron charge -E orbit a proton in a circular orbit of constant radius R.1031

If we assume the proton stationary, we are only worried about the electrostatic forces.1043

Find the kinetic energy of the two particle system which is going to be the kinetic energy of the electron.1048

The force on the electron, the electrical force is going to be the centripetal force,1056

which implies then our electrical force by Coulomb’s law is 1/ 4 π ε₀, the charge on each of these proton and electron.1062

The magnitude is E so that is going to be E² ÷ R and that has to equal our centripetal force MV²/ R.1070

Q1 Q2/ R, get rid of square there.1087

We got to solve then for MV² is going to be equal to, we multiply that by R.1093

We are going to get a E²/ 4 π ε₀ R.1101

And MV² is awfully close to the kinetic energy.1111

If we want the kinetic energy, that is kinetic energy so I just divide that by 2.1114

Our kinetic energy then is going to be E²/ 8 π ε₀ R.1120

Because that is one of our choices, 1/ 8 π ε₀ R.1133

Choice B gives us the correct answer.1139

56, we have got a square loop of wire with side L and resistance R.1148

It is held at rest in a uniform field but the field decreases with time according to that formula.1154

Find the induced current in the loop.1161

Faraday’s law, E = –D φ BDT which is - the derivative of BA because our area is not changing.1164

Which is going to be - the derivative with respect to time of, our magnetic field strength B is given by the formula A – BT.1175

And our area is just L², it is a square.1188

L² can come out, it is going to be L² × the derivative of A – BT, which implies that induced EMF then is just going to be BL².1191

The current then is E/ R just going to be BL² / R.1207

And how about direction, it looks like with time but that magnetic field is getting weaker.1218

Dense law as we wanted to oppose that change that will give us a counterclockwise current by the right hand rule.1224

BL²/ R counterclockwise, looks like that is going to be choice E.1230

57, a negatively charged particle in a uniform magnetic field moves in a circular path as shown,1240

which of the following graphs to picks have the frequency of revolution F depends on the radius R.1249

Let us start, we know that the velocity is distance /time or in this case 2 π R ÷ T, the time for once around is the period.1258

We can also look at this from a centripetal force perspective.1268

The centripetal force is MV² / R which has to be equal to whatever is causing that force.1272

What is causing the centripetal force is the magnetic force here, which is going to be Q VB.1281

Therefore, we could solve for V and say that that is going to be Q BR/ M.1287

Putting these together, Q BR/ M is our velocity has to equal 2 π R / T.1297

But by the way, 1/ the period is the frequency so I'm going to write that as 2 π R F.1309

And just solve for frequency is going to be Q BR/ M × 2 π R or Q B/ 2 π M.1315

Notice here that we have no dependence on the radius.1331

The correct answer must be A, does not depend on radius at all.1334

Let us take a look at 58, the only force acting on an electron is due to a uniform electric field.1343

The electron moves at constant.1353

It has got to be a constant acceleration because it is a constant force.1356

The direction is opposite that of the field because the field points in the direction of the force a positive charge would feel.1363

58 right away, A.1369

59, in a region of space we got a spherical symmetric electric potential given by the function V of R = KR², where K is a constant.1375

What is the magnitude of the electric field when your distance R from the origin?1387

The electric field from potential is - DV DR which is just going to be -2 KR.1392

Therefore, the magnitude of the electric field is just going to be 2K and we are looking at R₀ instead of R.1400

2 K R0 is going to be our answer C.1408

In the following question 60, what is the direction of the electric field at RO and the direction of the force on electron.1414

Notice we had negative here, so that means that the electric field is going to be toward the origin and1422

because it is a force on the electron, it is going to the opposite direction of the field.1428

The answer there B, it is toward the origin with the force, the electric field toward the origin of the force in the opposite direction.1433

The answer is B because of this -2 KR that we just found.1440

61, 2 charged particles each with a charge of +Q are located along the X axis at X = 2 and X =4.1450

Which of the following shows the graph of the magnitude of the electric field along the X axis?1459

As I look at 61, it is pretty obvious right in between that has to be 0 so that gets rid of choice C and choice B.1464

D and E do not make any sense, it is going to be A.1475

62, positive electric charge moved at constant speed between two locations in an electric field1481

with no work done by or against the field, how can that occur?1490

Right away, I’m thinking that the only way that happens is if you are on equal potential.1493

D, the charge is moved along the equal potential line.1498

There it is, nice and simple.1501

63, we got a non conducting hollow sphere of radius R carrying a large charge + Q.1507

A small charge + Q is located at point P and what must be the work done in moving the charge +Q1513

from P through the hole to the center of the sphere?1521

We are only going to have to do work to get to the surface and that amount of work is going to be charge × the potential difference,1525

which is going to be charge × the final potential - the initial potential, which is going to be our charge ×1534

our final K Q/ R the radius - K Q/ r, we can do some factoring Kq Q 1 / R -1 / r.1543

It looks like it is going to be the choice E.1561

64, we have got a couple capacitors altering µf connected to the circuit with a 12 V battery.1568

Find the equivalent capacitance between X and Z.1577

The first thing I'm going to do is see that those 2, 3 µf capacitor is in parallel.1580

Those add up so that is going to be 6 µf.1585

Then we have got to go through A 3 µf capacitor.1590

Those two in series, if we found that equivalent should tell us our answer.1596

We can do that a bunch of ways but right away we know that that is going to be less than the smallest so less than 3.1601

As I do that, 1/ C equivalent is 1/6 µf + 1/3 µf is going to be, 1/ C equivalent is going to being 1/6 µf + 2/6 micro F is 3/6 µf.1615

Therefore, C equivalent is 2 µf.1634

Our choice there for 64 must be B.1640

And 65, the follow on question.1646

Potential difference between Y and Z is going to be what?1650

We are looking for the potential difference if we can put a vault meter right there between Y and Z.1654

If C = Q/ V then potential = Q/ C, we know our charge is going to be VC, is going to be 12V × 2 µf or 24 µc.1662

The charge on each plate is 24 µc.1685

Once we know that, the potential between Y and Z is just going to be Q / C or we will have 24 µc on the plate ÷ that capacitance 3 µf is 8V.1689

65 should be D.1705

66, in the figure contains two identical light bulbs in series with a battery.1713

At first they are at equal brightness when switch S is closed, which of the following occurs to the bulbs?1720

Once switch S is closed, right away bulb 2 is not going to have any electricity going through.1727

All the electricity is going to go right through S, it has shorted out.1734

Our choice is either B or E.1737

If we short out bulb 2, all of that potential is now going to go through bulb 1 so it is going to expand more power,1740

the answer has to be B, bulb 1 gets brighter and bulb 2 goes out.1748

67, the bar magnet and a wire loop current carrying I are arranged in which direction is the force on the current loop due to the magnet?1755

As I look at that one, let us say that that current loop creates a south pole1765

but nears the magnet due to the right hand rule so that is going to attract it.1770

Your force then has to be A, toward the magnet.1774

68, a wire loop of area A is placed in a time varying that is spatially uniform magnetic field perpendicular to the plane of the loop.1782

The induced EMF was given by that function.1791

The time varying magnetic field could be given by, EMF is – D φ B DT is going to be B AT ^½.1794

Or which implies that E = - the derivative of the integral of B ⋅ DA which is - D/ DT of BA or - A DB DT.1807

Which implies then, we have – B T ^½ must equal DB DT, which implies that the integral of -BT ^½ DT must equal the integral of DB.1828

Or integral of DB is just B integral of - BT ^½ is going to be – B T³/2 / 3/2 + some constant,1837

implies then that B is going to be equal to -2/3 B T³/2, which is we are looking for that and not worry about direction.1859

2/3 BT³/2, the answer choice E.1875

That is more involved multiple choice in there.1880

69, talking about a capacitor with two identical conducting plates parallel to each other separated by a distance D,1884

the stand is connected if you ignore effects what is the electric field between the plates?1892

Electric field between the plates are parallel capacitor is constant as long as you are not near the edges.1897

Just B/ D, the answer is B.1904

70, an insulating plastic material is inserted between the plates without otherwise disturbing the system.1909

What does that do to the capacitance?1916

Capacitance is ε A/ D.1919

If you increase ε by putting in an insulator, your capacitance has to go up because A and D are changing.1923

Capacitances must increase, the answer has to be A.1930

Alright that is the end of the multiple choice section of this practice test.1937

Hopefully that gives you a good feel for where you are strong and some opportunities for some more work.1940

Thank you so much for watching www.educator.com.1946

Come on back in the next lesson and we will do the free response portion of this test.1948

Make it a great day everyone.1952

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