Call Now to Set Up Tutoring:

(888) 888-0446

AP Physics 2 : AP Physics 2

Study concepts, example questions & explanations for AP Physics 2

Create An Account

All AP Physics 2 Resources

6 Diagnostic Tests 149 Practice Tests Question of the Day Flashcards Learn by Concept

Example Questions

Example Question #1 : Point Charges

An object of mass 25kg accelerates at $2.1\frac{m}{s^2}$ in an electric field of $4.5\frac{N}{C}$ . Determine the charge of the object.

Possible Answers:

13.55C

11.66C

3.45C

7.75C

8.89C

Correct answer:

11.66C

Explanation:

Combine Newton's second law with the equation for electric force due to an electric field:

$F_{net}=ma=F_E=qE$

Plug in values:

25*2.1=4.5q

q=11.66C

Report an Error

Example Question #1 : Electrostatics

At $2\textup{ m}$ away from a point charge, the electric field is $15\ \frac{\textup{N}}{\textup{C}}$ , pointing towards the charge. Determine the value of the point charge.

Possible Answers:

$q=-8.85\textup{ nC}$

None of these

$q=-6.67\textup{ nC}$

$q=8.85\textup{ nC}$

$q=6.67\textup{ nC}$

Correct answer:

$q=-6.67\textup{ nC}$

Explanation:

Since the electric field is pointing towards the charge, it is known that the charge has a negative value.

Using electric field formula:

$|E|=k\frac{|q|}{r^2}$

Solving for

$|q|=|E|\frac{r^2}{k}$

Plugging in values:

$q=15*\frac{2^2}{9*10^9}$

|q|=6.67nC

Since the charge must have a negative value:

q=-6.67nC

Report an Error

Example Question #9 : Point Charges

Imagine two point charges separated by 5 meters. One has a charge of $5 \ \mu \textup{C}$ and the other has a charge of $-2 \ \mu \textup{C}$ . What is the magnitude of the force between them? Is it attractive or repulsive?

Possible Answers:

$9.0 \times 10^{-2} \textup{ N}$

Repulsive

There is no force felt by the two charges.

$9.0 \times 10^{-2} \textup{ N}$

Attractive

$3.6 \times 10^{-3} \textup{ N}$

Attractive

$3.6 \times 10^{-3} \textup{ N}$

Repulsive

Correct answer:

$3.6 \times 10^{-3} \textup{ N}$

Attractive

Explanation:

The equation for the force experienced by two point charges is known as Coulomb's Law, and is as follows.

$F=\frac{kq_1q_2}{r^2}$

The value 'k' is known as Coulomb's constant, and has a value of approximately $9.0 \times 10^{9} N$ .

We have all of the numbers necessary to use this equation, so we can just plug them in.

$\begin{align*} F&= \frac{kq_1q_2}{r^2}\\ &= \frac{(9\times 10^9)(5\times 10^{-6})(-2\times 10^{-6})}{(5^2)} \\ &= -0.0036\\ &= -3.6\times 10^{-3} \end{align*}$

Since we're given a negative number (and through our intuition: "opposites attract"), we can determine that the force is attractive. Because we're asked for the magnitude of the force, we take the absolute value, so our answer is

$3.6 \times 10^{-3} \textup{ N}$ , attractive force.

Report an Error

Example Question #2 : Point Charges

What is the value of the electric field 3 meters away from a point charge with a strength of $10\ \mu\textup{C}$ ?

Possible Answers:

$10000\ \frac{\textup{N}}{\textup{C}}$

None of the answers are correct.

$30000\ \frac{\textup{N}}{\textup{C}}$

$-30000\ \frac{\textup{N}}{\textup{C}}$

$-10000\ \frac{\textup{N}}{\textup{C}}$

Correct answer:

$10000\ \frac{\textup{N}}{\textup{C}}$

Explanation:

To find the strength of an electric field generated from a point charge, you apply the following equation.

$E=\frac{k*Q}{r^2}$

We know the value of Q and r (the charge and distance, respectively), so we can simply plug in the numbers we have to find the answer.

$\begin{align*} E&= \frac{9E9*10E{-6}}{3^2}\\ &= 10000\ \frac{N}{C} \end{align*}$

While this might seem like a very large number coming from such a small charge, remember that the typical charges interacting with it will be in the same magnitude of strength, roughly. This yields a force much smaller than 10,000 Newtons.

Report an Error

Example Question #1 : Electric Fields

Physics2set1q8

What is the value of the electric field at point C?

Points A and B are point charges.

Possible Answers:

$8.99 \frac{N}{C}$ in the direction

$22.5 \frac{N}{C}$ in the direction

$8.99 \frac{N}{C}$ in the direction

$5.24 \frac{N}{C}$ in the direction

Correct answer:

$22.5 \frac{N}{C}$ in the direction

Explanation:

First, let's calculate the electric field at C due to point A.

$E_{1}=k\frac{Q_{1}}{R^2}$

$E_{2}=k\frac{Q_{2}}{R^2}$

We can tell that the net electric field will be in the direction.

$E_{net}=E_{2}-E_{1}$

$E_{net}=\frac{k\cdot 10^{-9}}{1^2}(3-\frac{1}{2})$

$E_{net}= 22.5 \frac{N}{C}$ in the direction.

Report an Error

Example Question #1 : Electric Fields

What is the electric field 15m away from a particle with a charge of 15mC ?

Possible Answers:

$6\cdot 10^5\frac{N}{C}$ away from the charged particle

$6\cdot 10^5\frac{N}{C}$ towards the charged particle

$6\frac{N}{C}$ away from the charged particle

$6\cdot 10^3\frac{N}{C}$ away from the charged particle

$6\cdot 10^8\frac{N}{C}$ towards the charged particle

Correct answer:

$6\cdot 10^5\frac{N}{C}$ away from the charged particle

Explanation:

Use the equation to find the magnitude of an electric field at a point.

$E_{field}=\frac{kq}{r^2}$

$E_{field}=9\cdot10^{9}\frac{N\cdot m^{2}}{C^{2}}\cdot\frac{15mC}{(15m)^{2}}$

Solve.

$E_{field}=600000=6\cdot 10^5\frac{N}{C}$

Since it is a positive charge, the electric field lines will be pointing away from the charged particle.

Report an Error

Example Question #2 : Electric Fields

You are at point (0,5). A charge of -50nC is placed at the origin. What charge would you need to place at (0,-3) to cause there to be no net electric field at your location.

Possible Answers:

128nC

64nC

-128nC

-64nC

None of these

Correct answer:

128nC

Explanation:

We will need to use the electric field equation, twice. Because we are given coordinates, we will need to use vector notation.

$E_{total}=E_{1}+E_{2}$

$E_{1}=k\frac{q_{1}}{r_{1}^2} \widehat{r_1}$

$E_{2}=k\frac{q_{2}}{r_{2}^2} \widehat{r_{2}}$

Combine the two equations.

$E_{total}=k\frac{q_{1}}{r_{1}^2} \widehat{r_1} + k\frac{q_{2}}{r_{2}^2} \widehat{r_2}$

Plug in known values.

$0 = \frac{50 nC}{5^2}\frac{<0, 5>}{5}+ \frac{q_2}{8^2}\frac{<0, 8>}{8}$

q_2 = 128 nC

Note that the charge is positive. This is because the electric field lines point towards the negative charge at the origin, and in order to balance this at your location, the electric field lines of the charge at (0,-3) must be pointing away from the charge.

Report an Error

Example Question #2 : Electric Fields

Potential problem

In the diagram above where along the line connecting the two charges is the electric potential (V) due to the two charges zero?

Possible Answers:

1.2 m to the right of Q_B

There is no point on the line where the electric potential is zero

1.2 m to the left from Q_A

1.2 m to the right of Q_A

1.2 m to the left of Q_B

Correct answer:

1.2 m to the right of Q_A

Explanation:

Potential is not a vector, so we just add up the two potentials and set them to each other. The equation for electric potential is:

$V=\frac{1}{4\pi \epsilon _{0}}\frac{q}{r}$

$\frac{1}{4\pi \epsilon _{0}}\frac{Q_A}{d}=\frac{1}{4\pi \epsilon _{0}}\frac{Q_B}{(2-d)}$

If the point we are looking for is distance from Q_A , it's (2-d) from Q_B . Cancel all the common terms, then cross-multiply:

$\frac{1}{4\pi \epsilon _{0}}\frac{3\times10^{-9}}{d}=\frac{1}{4\pi \epsilon _{0}}\frac{2\times10^{-9}}{(2-d)}$

$\frac{3}{d}=\frac{2}{(2-d)}$

3(2-d)=2d

d=1.2 m

Since we had associated with Q_A , it's from that charge toward the weaker charge.

Report an Error

Example Question #1 : Electric Fields

Electric field problem

In the diagram above, where is the electric field due to the two charges zero?

Possible Answers:

0.56 m to the right of Q_B

0.56 m to the right of Q_A

0.56 m to the left of Q_B

There is no point where the electric field is zero

0.56 m to the left of Q_A

Correct answer:

0.56 m to the right of Q_A

Explanation:

Electric field is a vector. In between the charges is where Q_A 's field points right and Q_B 's field points left, so somewhere in between, the two vectors will add to zero. It will be closer to the weaker charge, Q_B , but since field depends on the inverse-square of the distance, it will not be linear, and we'll have to do some math.

First set the magnitudes of the two fields equal to each other. The vectors point in opposite directions, so when their magnitudes are equal, the vector sum is zero.

$E_A=E_B \newline \frac{1}{4\pi _0}\frac{Q_A}{d_A^2}=\frac{1}{4\pi _0}\frac{Q_B}{d_B^2}\newline \frac{1}{4\pi _0}\frac{5\times10^{-9}}{d_A^2}=\frac{1}{4\pi _0}\frac{3\times10^{-9}}{(1-d_A)^{2}} \newline \frac{5}{d_A^{2}}=\frac{3}{(1-d_A)^{2}}$

Many of the terms cancel, making it a bit easier. Now cross multiply and solve the quadratic:

$2d_A^{2}-10d_A +5=0$

d_A=0.56 m

Report an Error

Example Question #1 : Electric Fields

Coloumbs law for varsity

If charge has a value of $-9*10^{-13}C$ , charge has a value of $8*10^{-13}C$ , and is equal to 50 cm , what will be the magnitude of the force experienced by charge ?

Possible Answers:

None of these

$|F|=2.59*10^{-14}N$

$|F|=5.55*10^{-14}N$

$|F|=6.75*10^{-14}N$

$|F|=1.11*10^{-14}N$

Correct answer:

$|F|=2.59*10^{-14}N$

Explanation:

Using coulombs law to solve

$F=k\frac{q_1q_2}{r^2}$

Where:

q_1 it the first charge, in coulombs.

q_2 is the second charge, in coulombs.

is the distance between them, in meters

is the constant of $9*10^9\frac{N*m^2}{C^2}$

Converting into

$50cm*\frac{1m}{100cm}=.50m$

Plugging values into coulombs law

$F=9*10^{9}\frac{-9*10^{-13}8*10^{-13}}{.50^2}$

$F=-2.59*10^{-14}N$

Magnitude will be the absolute value

$|F|=2.59*10^{-14}N$

Report an Error

View Tutors

Sylvanus
Certified Tutor

university of Yaounde I, Bachelor of Science, Psychology. university of Yaounde I, Master of Science, Education.

View Tutors

Brandon
Certified Tutor

Upstate Medical University, Doctor of Philosophy, Cellular and Molecular Biology.

View Tutors

Nancy
Certified Tutor

Rhodes College, Bachelor in Arts, Economics. Harvard Business School, Masters in Business Administration, Finance.

All AP Physics 2 Resources

6 Diagnostic Tests 149 Practice Tests Question of the Day Flashcards Learn by Concept

Popular Subjects

Physics Tutors in Seattle, Reading Tutors in Miami, MCAT Tutors in Atlanta, Calculus Tutors in Phoenix, Math Tutors in Washington DC, Biology Tutors in New York City, Computer Science Tutors in Atlanta, LSAT Tutors in Dallas Fort Worth, SAT Tutors in Phoenix, MCAT Tutors in Chicago

Popular Courses & Classes

Spanish Courses & Classes in Boston, ACT Courses & Classes in San Diego, SAT Courses & Classes in Los Angeles, SSAT Courses & Classes in Houston, ISEE Courses & Classes in Houston, SAT Courses & Classes in Seattle, Spanish Courses & Classes in Houston, SAT Courses & Classes in Denver, SSAT Courses & Classes in Atlanta, GMAT Courses & Classes in Seattle

Popular Test Prep

MCAT Test Prep in Miami, ISEE Test Prep in New York City, ACT Test Prep in Houston, GMAT Test Prep in Houston, SAT Test Prep in Boston, GRE Test Prep in Denver, GRE Test Prep in Phoenix, ACT Test Prep in Los Angeles, SSAT Test Prep in Miami, SSAT Test Prep in Dallas Fort Worth

DMCA Complaint

If you believe that content available by means of the Website (as defined in our Terms of Service) infringes one or more of your copyrights, please notify us by providing a written notice (“Infringement Notice”) containing the information described below to the designated agent listed below. If Varsity Tutors takes action in response to an Infringement Notice, it will make a good faith attempt to contact the party that made such content available by means of the most recent email address, if any, provided by such party to Varsity Tutors.

Your Infringement Notice may be forwarded to the party that made the content available or to third parties such as ChillingEffects.org.

Please be advised that you will be liable for damages (including costs and attorneys’ fees) if you materially misrepresent that a product or activity is infringing your copyrights. Thus, if you are not sure content located on or linked-to by the Website infringes your copyright, you should consider first contacting an attorney.

Please follow these steps to file a notice:

You must include the following:

A physical or electronic signature of the copyright owner or a person authorized to act on their behalf; An identification of the copyright claimed to have been infringed; A description of the nature and exact location of the content that you claim to infringe your copyright, in \ sufficient detail to permit Varsity Tutors to find and positively identify that content; for example we require a link to the specific question (not just the name of the question) that contains the content and a description of which specific portion of the question – an image, a link, the text, etc – your complaint refers to; Your name, address, telephone number and email address; and A statement by you: (a) that you believe in good faith that the use of the content that you claim to infringe your copyright is not authorized by law, or by the copyright owner or such owner’s agent; (b) that all of the information contained in your Infringement Notice is accurate, and (c) under penalty of perjury, that you are either the copyright owner or a person authorized to act on their behalf.

Send your complaint to our designated agent at:

Charles Cohn Varsity Tutors LLC
101 S. Hanley Rd, Suite 300
St. Louis, MO 63105

Or fill out the form below:

Do not fill in this field

Contact Information

* Full legal name

Company name

* Phone number

* Email address

Address

* Address1

Address2

* City

* State

* Zipcode

* Country

Complaint Details

* URL of copyrighted work (where your original material is located)

* Please describe the copyrighted work so that it may be easily identified

I am the owner, or an agent authorized to act on behalf of the owner of an exclusive right that is allegedly infringed.

I have a good faith belief that the use of the material in the manner complained of is not authorized by the copyright owner, its agent, or the law.

This notification is accurate.

I acknowledge that there may be adverse legal consequences for making false or bad faith allegations of copyright infringement by using this process.

* Signature (your digital signature is legally binding)

HIGH SCHOOL

GRADUATE SCHOOL

K-8

Search 50+ Tests

math tutoring

science tutoring

foreign languages

elementary tutoring

other

Search 350+ Subjects

AP Physics 2 : AP Physics 2

Study concepts, example questions & explanations for AP Physics 2

All AP Physics 2 Resources

Example Questions

Example Question #1 : Point Charges

Example Question #1 : Electrostatics

Example Question #9 : Point Charges

Example Question #2 : Point Charges

Example Question #1 : Electric Fields

Example Question #1 : Electric Fields

Example Question #2 : Electric Fields

Example Question #2 : Electric Fields

Example Question #1 : Electric Fields

Example Question #1 : Electric Fields

All AP Physics 2 Resources

Report an issue with this question

DMCA Complaint

Contact Information

Address

Complaint Details

Email address:
Your name:
Feedback: