We are open Saturday and Sunday!

Call Now to Set Up Tutoring:

(888) 888-0446

AP Physics C: Mechanics : Electricity

Study concepts, example questions & explanations for AP Physics C: Mechanics

Create An Account

All AP Physics C: Mechanics Resources

2 Diagnostic Tests 92 Practice Tests Question of the Day Flashcards Learn by Concept

Example Questions

Example Question #21 : Electricity And Magnetism Exam

Three identical point charges with $q=1\mu C$ are placed so that they form an equilateral triangle as shown in the figure. Find the electric potential at the center point (black dot) of that equilateral triangle, where this point is at a equal distance, d=3mm , away from the three charges.

3_charges

$k=8.99*10^9\frac{Nm^2}{C^2}$

Possible Answers:

8.99*10^6V

7.25*10^5V

5.43*10^8V

8.99*10^3V

1.90*10^6V

Correct answer:

8.99*10^6V

Explanation:

The electric potential from point charges is $V=\frac{kq}{d}$ .

Knowing that all three charges are identical, and knowing that the center point at which we are calculating the electric potential is equal distance from the charges, we can multiply the electric potential equation by three.

$V=3\frac{kq}{d}$

Plug in the given values and solve for .

$V=3\frac{(8.99*10^9\ \frac{\text{Nm}^2}{\text{C}^2})(1* 10^{-6}\ \text{C})}{3*10^{-3}\ \text{m}}$

$V=8.99* 10^6\ \text{V}$

Report an Error

Example Question #21 : Electricity

Ap physics c e m potential problems 2 6 16 1

Eight point charges of equal magnitude are located at the vertices of a cube of side length . Calculate the potential at the center of the cube.

Possible Answers:

$\frac{8Q}{3\pi\epsilon_0a}$

$\frac{2\sqrt{2}Q}{\pi\epsilon_0a}$

$\frac{2Q}{\pi\epsilon_0a\sqrt{3}}$

$\frac{4Q}{\pi\epsilon_0a\sqrt{3}}$

$\frac{\sqrt{3}Q}{6\pi\epsilon_0a}$

Correct answer:

$\frac{4Q}{\pi\epsilon_0a\sqrt{3}}$

Explanation:

By the Pythagorean theorem, each charge is a distance

$d = \sqrt{\left (\frac{a}{2} \right )^2 + \left (\frac{a}{2} \right )^2+\left (\frac{a}{2} \right )^2} = \frac{\sqrt{3}}{2}a$

from the center of the cube, so the potential is

$V=8\times\left (\frac{Q}{4\pi\epsilon_0\left (\frac{\sqrt{3}}{2}a \right )} \right )$

$=\frac{4Q}{\pi\epsilon_0a\sqrt{3}}$

Report an Error

Example Question #1 : Calculating Electric Potential

Ap physics c e m potential problems 2 6 16 1 4

An infinite plane has a nonuniform charge density given by $\sigma(r)=\frac{qa}{2\pi(r^2+a^2)^{3/2}}$ . Calculate the potential at a distance above the origin.

You may wish to use the integral:

$\int\frac{x}{(x^2+a^2)^2}dx=\frac{1}{2a^2}\left(\frac{x^2}{x^2+a^2} \right )+C$

Possible Answers:

$\frac{q}{8\pi\epsilon_0a}$

$\frac{q}{\pi\epsilon_0a}$

$\frac{q}{16\pi\epsilon_0a}$

$\frac{q}{2\pi\epsilon_0a}$

$\frac{q}{4\pi\epsilon_0a}$

Correct answer:

$\frac{q}{8\pi\epsilon_0a}$

Explanation:

Use polar coordinates with the given surface charge density, $\sigma(r)=\frac{qa}{2\pi(r^2+a^2)^{3/2}}$ and area element $rd\theta d\phi$ . Noting that a point from the origin is a distance $\sqrt{r^2+a^2}$ from the point of interest, we calculate the potential as follows, integrating with respect to from to $\infty$ .

$V=\frac{1}{4\pi\epsilon_0} \int_{0}^{2\pi} \int_{0}^{\infty}\left (\frac{qa}{2\pi(r^2+a^2)^{3/2}} \right ) \frac{rdrd\theta}{\sqrt{r^2+a^2}}$

$=\frac{1}{4\pi\epsilon_0} (2\pi) \left(\frac{qa}{2\pi} \right )\int_{0}^{\infty}\left (\frac{1}{(r^2+a^2)^{3/2}} \right ) \frac{rdr}{\sqrt{r^2+a^2}}$

$=\frac{qa}{4\pi\epsilon_0} \int_{0}^{\infty}\frac{r}{(r^2+a^2)^2} dr$

$=\frac{qa}{4\pi\epsilon_0} \left [\frac{1}{2a^2}\left (\frac{x^2}{x^2+a^2} \right ) \right ]_{0}^{\infty}$

$=\frac{q}{8\pi\epsilon_0a} (1-0)$ (by limit)

$=\frac{q}{8\pi\epsilon_0a}$

Remark: This is exactly the charge distribution that would be induced on an infinite sheet of (grounded) metal if a negative charge were held a distance above it.

Report an Error

Example Question #2 : Calculating Electric Potential

Ap physics c e m potential problems 2 6 16 1 3

A nonuniformly charged hemispherical shell of radius (shown above) has surface charge density $\sigma(\theta)=b\cos^3\theta$ . Calculate the potential at the center of the opening of the hemisphere (the origin).

Possible Answers:

$\frac{Rb}{3\epsilon_0}$

$\frac{3Rb}{8\epsilon_0}$

$\frac{Rb}{2\epsilon_0}$

$\frac{Rb}{15\epsilon_0}$

$\frac{Rb}{8\epsilon_0}$

Correct answer:

$\frac{Rb}{8\epsilon_0}$

Explanation:

Use spherical coordinates with the given surface charge density $\sigma(\theta)=b\cos^3\theta$ , and area element $R^2\sin\theta d\theta d\phi$ . Every point on the hemispherical shell is a distance from the origin, so we calculate the potential as follows, noting the limits of integration for $\theta$ range from to $\pi/2$ .

$V=\frac{1}{4\pi\epsilon_0}\int_{0}^{2\pi}\int_{0}^{\pi/2}(b\cos^3\theta)\left (\frac{1}{R} \right )R^2\sin\theta d\theta d\phi$

$=\frac{Rb}{4\pi\epsilon_0} (2\pi) \int_{0}^{\pi/2}\cos^3\theta\sin\theta d\theta$

$=\frac{Rb}{2\epsilon_0} \left[\frac{-\cos^4\theta}{4} \right]_{0}^{\pi/2}$

$=\frac{Rb}{8\epsilon_0}$

Report an Error

Example Question #11 : Calculating Electric Potential

Ap physics c e m potential problems 2 6 16 1 2

Three equal point charges are placed at the vertices of an equilateral triangle of side length . Calculate the potential at the center of the triangle, labeled P.

Possible Answers:

$\frac{\sqrt{3}Q}{4\pi\epsilon_0a}$

$\frac{3\sqrt{3}Q}{2\pi\epsilon_0a}$

$\frac{3\sqrt{3}Q}{4\pi\epsilon_0a}$

$\frac{3\sqrt{3}Q}{8\pi\epsilon_0a}$

$\frac{\sqrt{3}Q}{2\pi\epsilon_0a}$

Correct answer:

$\frac{3\sqrt{3}Q}{4\pi\epsilon_0a}$

Explanation:

Draw a line from the center perpendicular to any side of the triangle. This line divides the side into two equal pieces of length $\frac{a}{2}$ . From the center, draw another line to one of the vertices at the end of this side. This produces a 30-60-90 triangle with longer leg $\frac{a}{2}$ , so the hypotenuse (the distance from the vertex to the center) is $\frac{a}{\sqrt{3}}$ . The potential at the center is due to three of these charges, so it must be

$V=3\times \left ( \frac{Q}{4\pi\epsilon_0\left(\frac{a}{\sqrt{3}} \right )} \right)$

$=\frac{3\sqrt{3}Q}{4\pi\epsilon_0a}$

Report an Error

Example Question #171 : Ap Physics C

Ap physics c e m potential problems 2 6 16 1 1

A uniformly charged hollow disk has inner radius R_1 and outer radius R_2 , and carries a total charge . Calculate the potential a distance from the center, on the axis of the disk.

Possible Answers:

$\frac{1}{8\pi\epsilon_0}\frac{Q}{R_2^2-R_1^2}\left(\sqrt{R_2^2+a^2}-\sqrt{R_1^2+a^2} \right )$

$\frac{1}{4\pi\epsilon_0}\frac{Q}{R_2^2}\left(\sqrt{R_2^2+a^2}-\sqrt{R_1^2+a^2} \right )$

$\frac{1}{4\pi\epsilon_0}\frac{Q}{R_2^2-R_1^2}\left(\sqrt{R_2^2+a^2}-\sqrt{R_1^2+a^2} \right )$

$\frac{1}{2\pi\epsilon_0}\frac{Q}{R_2^2-R_1^2}\left(\sqrt{R_2^2+a^2}-\sqrt{R_1^2+a^2} \right )$

$\frac{1}{2\pi\epsilon_0}\frac{Q}{R_2^2}\left(\sqrt{R_2^2+a^2}-\sqrt{R_1^2+a^2} \right )$

Correct answer:

$\frac{1}{2\pi\epsilon_0}\frac{Q}{R_2^2-R_1^2}\left(\sqrt{R_2^2+a^2}-\sqrt{R_1^2+a^2} \right )$

Explanation:

Use a polar coordinate system with surface charge density $\frac{Q}{\pi(R_2^2-R_1^2)}$ and area element $Rdrd\theta$ . The distance from the point of interest to a point a distance from the center is $\sqrt{r^2+a^2}$ , so the potential is

$V=\frac{1}{4\pi\epsilon_0}\int_{0}^{2\pi}\int_{R_1}^{R_2}\left(\frac{Q}{\pi(R_2^2-R_1^2)}\right )\frac{rdrd\theta}{\sqrt{r^2+a^2}}$

$=\frac{1}{4\pi\epsilon_0}(2\pi)\left(\frac{Q}{\pi(R_2^2-R_1^2)}\right )\int_{R_1}^{R_2}\frac{r}{\sqrt{r^2+a^2}}dr$

$=\frac{1}{2\pi\epsilon_0}\frac{Q}{R_2^2-R_1^2}\left[\sqrt{r^2+a^2} \right ]_{R_1}^{R_2}$

$=\frac{1}{2\pi\epsilon_0}\frac{Q}{R_2^2-R_1^2}\left(\sqrt{R_2^2+a^2} -\sqrt{R_1^2+a^2}\right )$

Report an Error

Example Question #21 : Electricity

Ap physics c e m potential problems 2 6 16 1

Two point charges and are separated by a distance . Calculate the potential at point P, a distance from charge in the direction perpendicular to the line connecting the two charges.

Possible Answers:

$\frac{-Q}{48\pi\epsilon_0a}$

$\frac{Q}{24\pi\epsilon_0a}$

$\frac{-Q}{30\pi\epsilon_0a}$

$\frac{Q}{60\pi\epsilon_0a}$

$\frac{11Q}{60\pi\epsilon_0a}$

Correct answer:

$\frac{Q}{60\pi\epsilon_0a}$

Explanation:

By the Pythagorean theorem, the distance from point P to charge is . Because point P is also from charge , it follows that the potential is

$V=\frac{1}{4\pi\epsilon_0}\left( \frac{2Q}{5a}-\frac{Q}{3a}\right )$

$=\frac{Q}{60\pi\epsilon_0a}$

Report an Error

Example Question #181 : Ap Physics C

Ap physics c e m potential problems 2 6 16 5

A nonuniformly charged ring of radius carries a linear charge density of $\lambda(\theta) = b\sin^2\theta$ . Calculate the potential at the center of the ring.

Possible Answers:

$\frac{b}{4\pi\epsilon_0}$

$\frac{b}{4\epsilon_0R}$

$\frac{b}{2\pi\epsilon_0}$

$\frac{b}{2\epsilon_0}$

$\frac{b}{4\epsilon_0}$

Correct answer:

$\frac{b}{4\epsilon_0}$

Explanation:

Use a polar coordinate system, the given linear charge density $\lambda(\theta)=b\sin^2\theta$ , and length element $Rd\theta$ . Since every point on the ring is the same distance from the center, we calculate the potential as

$V = \frac{1}{4\pi\epsilon_0}\int_{0}^{2\pi}b\sin^2\theta\left (\frac{1}{R} \right )R d\theta$

$=\frac{b}{4\pi\epsilon_0}\int_{0}^{2\pi}\sin^2\theta d\theta$

$=\frac{b}{4\pi\epsilon_0}\int_{0}^{2\pi}\frac{1-\cos2\theta}{2} d\theta$

$=\frac{b}{4\pi\epsilon_0}\left [\frac{\theta}{2}-\frac{\sin2\theta}{4} \right ]_{0}^{2\pi}$

$=\frac{b}{4\epsilon_0}$

Report an Error

Example Question #11 : Calculating Electric Potential

Ap physics c e m potential problems 2 6 16 6

In this model of a dipole, two charges and are separated by a distance as shown in the figure, where the charges lie on the x-axis at $+\frac{a}{2}$ and $-\frac{a}{2}$ respectively. Calculate the exact potential a distance from the origin at angle $\theta$ from the axis of the dipole.

Possible Answers:

$\frac{q}{4\pi\epsilon_0} \left ( \frac{1}{\sqrt{R^2 + \left (\frac{a}{2} \right)^2 +2Ra \cos\theta}} - \frac{1}{\sqrt{R^2 + \left (\frac{a}{2} \right)^2 - 2Ra\cos\theta}} \right )$

$\frac{q}{4\pi\epsilon_0} \left ( \frac{1}{\sqrt{R^2 + \left (\frac{a}{2} \right)^2 +Ra \cos\theta}} - \frac{1}{\sqrt{R^2 + \left (\frac{a}{2} \right)^2 - Ra\cos\theta}} \right )$

$\frac{q}{4\pi\epsilon_0} \left ( \frac{1}{\sqrt{R^2 + \left (\frac{a}{2} \right)^2 -Ra \cos\theta}} - \frac{1}{\sqrt{R^2 + \left (\frac{a}{2} \right)^2 + Ra\cos\theta}} \right )$

$\frac{q}{4\pi\epsilon_0} \left ( \frac{1}{\sqrt{R^2 + \left (\frac{a}{2} \right)^2 -2Ra \cos\theta}} - \frac{1}{\sqrt{R^2 + \left (\frac{a}{2} \right)^2 + 2Ra\cos\theta}} \right )$

$\frac{q}{4\pi\epsilon_0} \left ( \frac{1}{\sqrt{R^2 + a^2 -2Ra \cos\theta}} - \frac{1}{\sqrt{R^2 + a^2 + 2Ra\cos\theta}} \right )$

Correct answer:

$\frac{q}{4\pi\epsilon_0} \left ( \frac{1}{\sqrt{R^2 + \left (\frac{a}{2} \right)^2 -Ra \cos\theta}} - \frac{1}{\sqrt{R^2 + \left (\frac{a}{2} \right)^2 + Ra\cos\theta}} \right )$

Explanation:

By the law of cosines, the distance from the point to charge is

$\sqrt{R^2+\left(\frac{a}{2} \right )^2 - 2R\left(\frac{a}{2} \right )\cos\theta}$ .

The distance to charge can be found by using the law of cosines using the supplementary angle $\pi-\theta$ , for which $\cos(\pi-\theta) = \cos(\pi)\cos\theta - \sin(\pi)\sin\theta = -\cos\theta$ . Therefore the distance to is

$\sqrt{R^2+\left(\frac{a}{2} \right )^2 - 2R\left(\frac{a}{2} \right )(-\cos\theta)}$ .

Lastly, the exact potential is given by

$\frac{q}{4\pi\epsilon_0} \left ( \frac{1}{\sqrt{R^2 + \left (\frac{a}{2} \right)^2 -Ra \cos\theta}} - \frac{1}{\sqrt{R^2 + \left (\frac{a}{2} \right)^2 + Ra\cos\theta}} \right )$ .

Remark: Far from the dipole (approximating $R\gg a$ ) gives the much simpler equation for the potential of an ideal electric dipole $\frac{qa\cos\theta}{4\pi\epsilon_0 R^2}$ .

Report an Error

Example Question #21 : Electricity And Magnetism Exam

Ap physics c e m potential problems 2 6 16 4

A uniformly charged hollow spherical shell of radius carries a total charge . Calculate the potential a distance (where a>R ) from the center of the sphere.

Possible Answers:

$\frac{Q}{8\pi\epsilon_0a}$

$\frac{Q}{4\pi\epsilon_0R}$

$\frac{Q}{4\pi\epsilon_0a^2}$

$\frac{Q}{4\pi\epsilon_0a}$

$\frac{Q}{8\pi\epsilon_0R}$

Correct answer:

$\frac{Q}{4\pi\epsilon_0a}$

Explanation:

Use a spherical coordinate system and place the point of interest a distance from the center on the z-axis. By the law of cosines, the distance from this point to any point on the sphere is $\sqrt{R^2 + a^2 -2Ra\cos\theta}$ . Using surface charge density $\frac{Q}{4\pi R^2}$ and area element $R^2\sin\theta d\theta d\phi$ , we evaluate the potential as:

$V =\frac{1}{4\pi\epsilon_0} \int_{0}^{2\pi}\int_{0}^{\pi}\left (\frac{Q}{4\pi R^2} \right )\frac{R^2\sin\theta d\theta d\phi}{\sqrt{R^2 + a^2-2Ra\cos\theta}}$

. $=\frac{1}{4\pi\epsilon_0} \left (2\pi \right )\left (\frac{Q}{4\pi} \right )\int_{0}^{\pi}\frac{\sin\theta d\theta d\phi}{\sqrt{R^2 + a^2-2Ra\cos\theta}}$

$=\frac{1}{4\pi\epsilon_0} \left (\frac{Q}{2} \right ) \left [\frac{1}{Ra}\sqrt{R^2 + a^2-2Ra\cos\theta} \right ]_{0}^{\pi}$

$=\frac{1}{4\pi\epsilon_0} \left (\frac{Q}{2} \right ) \frac{1}{Ra}\left (|R+a|-|R-a| \right )$

$=\frac{1}{4\pi\epsilon_0} \left (\frac{Q}{2} \right ) \frac{1}{Ra}(2R)$

$=\frac{Q}{4\pi\epsilon_0a}$

Remarkably, this is the same potential that would exist a distance from a point charge.

Report an Error

View Tutors

Kaylee
Certified Tutor

Brigham Young University, Bachelor, MFA - Animation.

View Tutors

Angela
Certified Tutor

North Georgia College State University, Bachelor of Science, Special Education. North Georgia College State University, Mas...

View Tutors

Dereika
Certified Tutor

Southern University and A & M College, Bachelor of Science, Accounting. Grand Canyon University, Master of Science, Accounting.

All AP Physics C: Mechanics Resources

2 Diagnostic Tests 92 Practice Tests Question of the Day Flashcards Learn by Concept

Popular Subjects

SSAT Tutors in San Diego, Algebra Tutors in Denver, Spanish Tutors in Seattle, Chemistry Tutors in Philadelphia, GMAT Tutors in Los Angeles, Calculus Tutors in Seattle, Algebra Tutors in New York City, Algebra Tutors in San Francisco-Bay Area, Spanish Tutors in Atlanta, SAT Tutors in Washington DC

Popular Courses & Classes

ACT Courses & Classes in Los Angeles, ISEE Courses & Classes in Washington DC, SSAT Courses & Classes in Seattle, ACT Courses & Classes in Houston, GRE Courses & Classes in San Diego, SAT Courses & Classes in Phoenix, Spanish Courses & Classes in San Francisco-Bay Area, Spanish Courses & Classes in New York City, MCAT Courses & Classes in Washington DC, Spanish Courses & Classes in Atlanta

Popular Test Prep

ISEE Test Prep in Washington DC, LSAT Test Prep in Miami, SSAT Test Prep in Seattle, ACT Test Prep in Los Angeles, GRE Test Prep in Los Angeles, ISEE Test Prep in Philadelphia, GMAT Test Prep in Philadelphia, MCAT Test Prep in Chicago, GMAT Test Prep in Washington DC, GRE Test Prep in Atlanta

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 C: Mechanics : Electricity

Study concepts, example questions & explanations for AP Physics C: Mechanics

All AP Physics C: Mechanics Resources

Example Questions

Example Question #21 : Electricity And Magnetism Exam

Example Question #21 : Electricity

Example Question #1 : Calculating Electric Potential

Example Question #2 : Calculating Electric Potential

Example Question #11 : Calculating Electric Potential

Example Question #171 : Ap Physics C

Example Question #21 : Electricity

Example Question #181 : Ap Physics C

Example Question #11 : Calculating Electric Potential

Example Question #21 : Electricity And Magnetism Exam

All AP Physics C: Mechanics Resources

Report an issue with this question

DMCA Complaint

Contact Information

Address

Complaint Details

Email address:
Your name:
Feedback: