Call Now to Set Up Tutoring:

(888) 888-0446

AP Physics 2 : Electric Force Between Point Charges

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 #21 : Electric Force Between Point Charges

There are two point charges suspended in space. Charge A has a charge of $-5\ \mu \textup{C}$ . Charge B has a charge of $4\ \mu \textup{C}$ . If they are 3 meters apart, what is the magnitude of the force between them? Is the force attractive or repulsive?

Possible Answers:

$0.02 \textup{ N}$ ; repulsive

None of these answers are correct.

$0.02 \textup{ N}$ ; attractive

$0.06 \textup{ N}$ ; repulsive

$0.06 \textup{ N}$ ; attractive

Correct answer:

$0.02 \textup{ N}$ ; attractive

Explanation:

To determine the force between two point charges, you use Coulomb's Law.

$F=\frac{k*q_1*q_2}{r^2}$

We have the values of q1, q2, and r. We already know the value of k: 9 x 10⁹. To find the force, we can plug in each of the values.

$\begin{align*} F&=\frac{(9E9)*(-5E{-6})*(4E{-6})}{3^2} \\ &= -0.02 \end{align*}$

Therefore, the magnitude of the force is 0.02. Now, to determine the direction of the force, we examine the sign. If the sign is negative, then the force is attractive. As our force value is negative, it therefore is an attractive force.

Report an Error

Example Question #21 : Electric Force Between Point Charges

A -9nC charge is located at the origin. A -7nC charge is located at (5,0)mm . Where could an electron be placed where it would experience no net force?

Possible Answers:

None of these

(0,2.65)mm

(2.35,0)mm

(2.65,0)mm

(4.25,0)mm

Correct answer:

(2.65,0)mm

Explanation:

Since both charges are located on the x-axis, this is a one dimensional problem.

The electron, which has a negative charge, will be repelled by both charges.

Placing the electron somewhere "in the middle" will allow the forces to balance out.

$F_{net}=0=k\frac{q_1q_e}{r_1^2}-k\frac{q_2q_e}{r_2^2}$

Where

is the constant $9.0*10^{9}\frac{N\cdot m^2}{C^2}$

q_1 is the value of the first charge

q_2 is the value of the second charge

q_e is the charge of the electron

r_1 is the distance from the first charge to the electron

r_2 is the distance from the second charge to the electron

q_1 is 5mm away from q_2 , so:

r_1+r_2=5mm

Combining equations and plugging in values:

$F_{net}=0=9*10^9\frac{-9*10^{-9}*-1.6*10^{-19}}{r_1^2}-9*10^9\frac{-7*10^{-9}*-1.6*10^{-19}}{(.005-r_1)^2}$

$F_{net}=0=9*10^9\frac{-9*10^{-9}*-1.6*10^{-19}}{r_1^2}-9*10^9\frac{-7*10^{-9}*-1.6*10^{-19}}{r_2^2}$

$\frac{1.296*10^{-17}}{r_1^2}=\frac{1.008*10^{-17}}{r_2^2}$

$\frac{r_1^2}{1.296*10^{-17}}=\frac{r_2^2}{1.008*10^{-17}}$

$\frac{r_1^2}{1.296}=\frac{r_2^2}{1.008}$

$r_1=\sqrt{\frac{1.296r_2^2}{1.008}}$

r_1=1.13r_2

1.13r_2+r_2=5mm

r_2=2.35mm

r_1=2.65mm

Thus, the location of balanced forces is at:

(2.65,0)mm

Report an Error

Example Question #71 : Electrostatics

A 30nC charge is located at the origin. A 50nC charge is located at (10,0)mm . Where could a proton be placed where it would experience no net force?

$k=9.0*10^{9}\frac{N\cdot m^2}{C^2}$

Possible Answers:

(8.74,0)mm

None of these

(2.19,0)mm

(4.37,0)mm

(0,4.37)mm

Correct answer:

(4.37,0)mm

Explanation:

Since both point charges are located on the x-axis, this is a one dimensional problem.

The proton, which has a positive charge, will be repelled by both charges.

Placing the proton somewhere "in the middle" will allow the forces to balance out.

$F_{net}=0=k\frac{q_1q_e}{r_1^2}-k\frac{q_2q_e}{r_2^2}$

Where

$k=9.0*10^{9}\frac{N\cdot m^2}{C^2}$

q_1 is the value of the first charge

q_2 is the value of the second charge

q_e is the charge of the proton

r_1 is the distance from the first charge to the proton

r_2 is the distance from the second charge to the proton

q_1 is 5mm away from q_2 , so:

r_1+r_2=10mm

Combining equations and plugging in values:

$F_{net}=0=9*10^9\frac{30*10^{-9}*1.6*10^{-19}}{r_1^2}-9*10^9\frac{50*10^{-9}*1.6*10^{-19}}{r_2^2}$

$\frac{4.32*10^{-17}}{r_1^2}=\frac{7.2*10^{-17}}{r_2^2}$

$\frac{r_1^2}{4.32*10^{-17}}=\frac{r_2^2}{7.2*10^{-17}}$

$\frac{r_1^2}{4.32}=\frac{r_2^2}{7.2}$

$r_1=\sqrt{\frac{4.32r_2^2}{7.2}}$

r_1=.775r_2

.775r_2+r_2=10mm

r_2=5.63mm

r_1=4.37mm

Thus, the location of balanced forces is at:

(4.37,0)mm

Report an Error

Example Question #72 : Electrostatics

An electron is 500mm away from a point charge. It experiences a force of $2.6*10^{-16}N$ towards the point charge. Determine the value of the point charge.

Possible Answers:

$4.5*10^{-8}C$

$-1.6*10^{-19}C$

$9*10^{-8}C$

$4.5*10^{-1}C$

None of these

Correct answer:

$4.5*10^{-8}C$

Explanation:

Using

$F=k\frac{q_1q_e}{r^2}$

Solving for q_1

$F\frac{r^2}{k*q_e}=q_1$

Converting to and plugging in values

$-2.6*10^{-16}\frac{.500^2}{(9*10^{9})*-1.6*10^{-19}}=q_1$

*Note: a negative sign is used for the force because it is an attractive force, if it was a repulsive force, the opposite sign would be used.

$q=4.5*10^{-8}C$

Report an Error

Example Question #21 : Electric Force Between Point Charges

Which of the following will cut the magnitude of the electric force between two charged particles in half?

Possible Answers:

Placing the two particles twice as far apart

Double the charge of only one of the particles

Reducing the charge of only one of the particles by half

Doubling the charge of each particle

Reducing the charge of each particle by half

Correct answer:

Reducing the charge of only one of the particles by half

Explanation:

To answer this question, it's necessary to understand the factors that affect the electric force. To show this, we can write the electrical force expression.

$F_{e}=k\frac{q_{1}q_{2}}{r^{2}}$

Using the above expression, we can look at how each answer choice would change the electric force.

If we place the two particles twice as far apart, the magnitude of the electric force will be reduced by a factor of .

Doubling the charge of one particle would double the electric force. Doubling the charge on both particles would cause the electric force to become times as great.

Halving the charge of both particles would cause the electric force to decrease by a factor of .

If the charge on only one of the particles is cut in half, then the electric force would be cut in half as well. Thus, this is the correct answer.

Report an Error

Example Question #22 : Electric Force Between Point Charges

Two charges are a fixed distance apart. Both charges have charge . If another charge of charge and mass is placed a distance $\frac{d}{4}$ from one of the charges and $\frac{3d}{4}$ from the other, what will be the magnitude of its acceleration the moment it's released?

Possible Answers:

$\frac{16kqQ}{md^2}$

$\frac{8kqQ}{3md}$

$\frac{8kqQ}{9md}$

$\frac{128kqQ}{9md^2}$

$\frac{16kqQ}{9md^2}$

Correct answer:

$\frac{128kqQ}{9md^2}$

Explanation:

All we have to do is find the sum of the forces on the charge and divide by its mass. To find the force from each charge, we can use Coulomb's law:

$F=\frac{kq_{1}q_{2}}{d^2}$

Let's let the force from the charge a distance $\frac{d}{4}$ away be positive. That force is $\frac{kqQ}{(\frac{d}{4})^2}=\frac{16kqQ}{d^2}$ . The other force will be negative because it's acting in the opposite direction. This force is $\frac{-kqQ}{(\frac{3d}{4})^2}=\frac{-16kqQ}{9d^2}$ . Adding these two together we get

$\frac{128kqQ}{9d^2}$

This is the magnitude of the net force. To find acceleration, we divide by the mass to get

$\frac{128kqQ}{9md^2}$

Report an Error

View Tutors

Pamela
Certified Tutor

Mount Holyoke College, Bachelor in Arts, English. Columbia University in the City of New York, Master of Public Policy, Inter...

View Tutors

Anthony
Certified Tutor

CUNY New York City College of Technology, Bachelor of Science, Civil Engineering.

View Tutors

Chiquita
Certified Tutor

Troy University, Bachelor of Science, Education. University of Phoenix-Atlanta Campus, Master of Arts, Adult and Continuing E...

All AP Physics 2 Resources

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

Popular Subjects

GRE Tutors in Phoenix, Statistics Tutors in Dallas Fort Worth, French Tutors in Miami, French Tutors in Washington DC, Computer Science Tutors in Los Angeles, French Tutors in New York City, Algebra Tutors in Atlanta, English Tutors in Philadelphia, Statistics Tutors in Denver, English Tutors in Houston

Popular Courses & Classes

LSAT Courses & Classes in San Francisco-Bay Area, ACT Courses & Classes in Atlanta, SAT Courses & Classes in Seattle, MCAT Courses & Classes in Denver, GRE Courses & Classes in Boston, SAT Courses & Classes in New York City, ACT Courses & Classes in Houston, GRE Courses & Classes in Seattle, GMAT Courses & Classes in Miami, SSAT Courses & Classes in Chicago

Popular Test Prep

SAT Test Prep in Los Angeles, ACT Test Prep in Atlanta, GMAT Test Prep in New York City, SAT Test Prep in Dallas Fort Worth, LSAT Test Prep in Washington DC, SSAT Test Prep in Chicago, MCAT Test Prep in New York City, SSAT Test Prep in Phoenix, MCAT Test Prep in Atlanta, ACT Test Prep in Washington DC

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 : Electric Force Between Point Charges

Study concepts, example questions & explanations for AP Physics 2

All AP Physics 2 Resources

Example Questions

Example Question #21 : Electric Force Between Point Charges

Example Question #21 : Electric Force Between Point Charges

Example Question #71 : Electrostatics

Example Question #72 : Electrostatics

Example Question #21 : Electric Force Between Point Charges

Example Question #22 : Electric Force Between Point Charges

All AP Physics 2 Resources

Report an issue with this question

DMCA Complaint

Contact Information

Address

Complaint Details

Email address:
Your name:
Feedback: