Call Now to Set Up Tutoring:

(888) 888-0446

AP Physics C: Mechanics : Capacitors

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

Create An Account Create Tests & Flashcards

All AP Physics C: Mechanics Resources

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

Example Questions

Example Question #1 :Capacitors

A parallel plate capacitor has a capacitance of. If the plates are 1cm apart, what is the area of the plates?

$\epsilon_0=8.85* 10^{-12} \frac{F}{m}$

Possible Answers:

$1.1*10^{11}m^2$

1.1*10^7m^2

1.1*10^9m^2

$1.1*10^{10}m^2$

Correct answer:

1.1*10^9m^2

Explanation:

The relationship between capacitance, distance, and area is $C=\epsilon_0 \frac{A}{d}$ . We can rearrange this equation to solve for area.

$A=\frac{Cd}{\epsilon_0}$

Now, we can use the values given in teh question to solve.

$A=\frac{1F(1*10^{-2}m)}{8.85* 10^{-12} \frac{F}{m}}=1.1*10^9m^2$

Report an Error

Example Question #1 :Electric Circuits

Charge is distributed uniformly over the area of the two plates of a parallel plate capacitor, resulting in a surface area charge density of $\pm \sigma$ on the plates (the top plate is positive and the bottom is negative, as shown below). Each plate has area和分离的距离. A material of dielectric constant $\kappa_{material}$ has been placed between the two plates.

Ps0_capacitor

Which of the following would not result in an increase in the measure of electric potential difference between the two plates?

Possible Answers:

Increase the area,, of the plates

Increase the distance,, between the plates

Replace the material between the two plates with one of a lower dielectric constant, $\kappa$

Increase the value of the surface charge density, $\sigma$ , on each plate

Correct answer:

Increase the area,, of the plates

Explanation:

The electric potential difference created between the plates of a parallel plate capactor is given by the equation:

$V = \frac{Q}{C}$

The charge can be calculated by using the equation:

$Q=\sigma A$

The value of the capacitance is related to the dimensions of the capacitor with the equation:

$C=\kappa \frac{\varepsilon _{o}A}{d}$

Combining these equations yields:

$V=\frac{\sigma A}{\frac{\kappa \epsilon_0A}{d}}=\frac{\sigma d}{\varepsilon_{o} \kappa }$

The area becomes inconsequential, while the potential is directly proportional to the surface charge density and the distance between the plates, and inversely proportional to the dielectric of the material between the plates. Changing the area does not cause any change in the potential difference measured between the plates, and changing any of the other variables would cause a resultant change in the potential difference.

Report an Error

Example Question #1 :Capacitors

哟u are hired to make a capacitor out of two parallel metal sheets. If someone wanted you to make a thin capacitor of $700\mu F$ out of those metal sheets, and the sheets needed to be 0.5mm apart, what area do the two metal plates need to be?

$\epsilon_0=8.85*10^{-12}\frac{C^2}{Nm^2}$

Possible Answers:

6.21*10^6m^2

2.77*10^3m^2

3.95*10^4m^2

5.47*10^8m^2

$3.95*10^{-4}m^2$

Correct answer:

3.95*10^4m^2

Explanation:

For parallel plate capacitors, the equation is $C=\frac{\epsilon_0A}{d}$ .

Solve for.

$A=\frac{dC}{\epsilon_0}$

Now we can plug in our given values.

$A=\frac{(0.5*10^{-3})(700*10^{-6})}{8.85*10^{-12}\frac{\text{C}^2}{\text{Nm}^2}}$

$A=3.95*10^4\ \text{m}^2$

Report an Error

Example Question #1 :Using Capacitor Equations

The plates of a parallel plate capacitor are 3mm apart and 5m^2 in area. A potential difference of 1000V is applied across the capacitor. Find the capacitance.

$\epsilon_0=8.85*10^{-12} \frac{F}{m}$

Possible Answers:

$1.5*10^{-9}F$

$1.5*10^{-7}F$

$1.5*10^{-8}F$

$1.5*10^{-10}F$

Correct answer:

$1.5*10^{-8}F$

Explanation:

Capacitance is related to plate area and distance by the equation $C=\epsilon_0\frac{A}{d}$ .

Given the area and distance, we can solve for capacitance. The voltage, in this case, is irrelevant.

$C=8.85* 10^{-12} \frac{F}{m}(\frac{5 m^2}{3*10^{-3}m})=8.85* 10^{-12} \frac{F}{m}(1.67*10^{3}m)=1.5*10^{-8}F$

Report an Error

Example Question #1 :Capacitors

The plates of a parallel plate capacitor are 3mm apart and 5m^2 in area. A potential difference of 1000V is applied across the capacitor. Compute the charge on each plate.

$\epsilon_0=8.85* 10^{-12} \frac{F}{m}$

Possible Answers:

$1.5*10^{-8}C$

$1.5*10^{-6}C$

$1.5*10^{-7}C$

$1.5*10^{-5}C$

Correct answer:

$1.5*10^{-5}C$

Explanation:

Charge on a capacitor is given by the equation Q=CV . We know that the voltage is 1000V , but we need to determine the capacitance based off of the area and distance between the plates.

$C=\epsilon_0\frac{A}{d}$

$C=(8.85*10^{-12}\frac{F}{m})(\frac{5m^2}{3*10^{-3}m})$

$C=(8.85*10^{-12}\frac{F}{m})(1.67*10^{3}m)=1.5*10^{-8}F$

We can plug this value into the equation for charge.

$Q=CV=(1.5*10^{-8}F)(1000V)$

$Q=1.5*10^{-5}C$

Report an Error

Example Question #1 :Capacitors

The plates of a parallel plate capacitor are 3mm apart. A potential difference of 1000V is applied across the capacitor. Compute the magnitude of the generated electric field.

$\epsilon_0=8.85*10^{-12}\frac{F}{m}$

Possible Answers:

$3.33*10^5\frac{N}{C}$

$3.33*10^3\frac{N}{C}$

$3.33*10^4\frac{N}{C}$

$3.33*10^6\frac{N}{C}$

Correct answer:

$3.33*10^5\frac{N}{C}$

Explanation:

The electric field given by a capacitor is given by the formula $E=\frac{Q}{\epsilon_0 A}$ . We do not have these variables, so we will have to adjust the equation.

$Q=CV\rightarrow E=\frac{CV}{\epsilon_0A}$

The capacitance can be determined by the area of the plates and the distance between them.

$C=\epsilon_0\frac{A}{d}\rightarrow E=\frac{\epsilon_0\frac{A}{d}V}{\epsilon_0A}$

This equation simplifies to $E=\frac{V}{d}$ , allowing us to solve using the values given in the question.

$E=\frac{1000V}{3*10^{-3}m}=3.33*10^{5}\frac{N}{C}$

Report an Error

Example Question #2 :Capacitors

If two identical parallel plate capacitors of capacitanceare connected in series, which is true of the equivalent capacitance, $C_{eq}$ ?

Possible Answers:

$C < C_{eq}< 2C$

$C_{eq}=C$

$C_{eq}= 2C$

$C_{eq}=\frac{C}{2}$

$\frac{C}{2} < C_{eq} < C$

Correct answer:

$C_{eq}=\frac{C}{2}$

Explanation:

Relevant equations:

$\frac{1}{C_{eq, series}}=\frac{1}{C_1}+\frac{1}{C_2}+ ...$

Use the series equation, replacing C1 and C2 with the given constant C:

$\frac{1}{C_{eq, series}}= \frac{1}{C}+ \frac{1}{C}$

$\frac{1}{C_{eq, series}}=\frac{2}{C}$

$C_{eq, series} = \frac{C}{2}$

This agrees with the general rule that the equivalent capacitance in series is less than the capacitance of any of the individual capacitors.

Report an Error

Example Question #1 :Electric Circuits

Two parallel conducting plates each have an area ofand are separated by a distance,. One plate has a chargeevenly distributed across it, and the other has a charge of. A proton (charge) is initially held near the positive plate, then released such that it accelerates towards the negative plate. How much kinetic energy has the proton gained in this process?

Possible Answers:

$\frac{eQ \epsilon _o A}{d^2}$

$\frac{eQ\epsilon _o A}{d}$

$\frac{eQ}{\epsilon _o A d}$

$\frac{e \epsilon _o A}{Qd}$

$\frac{eQd}{\epsilon _oA}$

Correct answer:

$\frac{eQd}{\epsilon _oA}$

Explanation:

Relevant equations:

$W = q\Delta V$

$W = \Delta KE$

$C = \frac{Q}{V}$

$C = \frac{\epsilon _o A}{d}$

According to the work-energy theorem, work done on the proton is equal to its change in kinetic energy.

1. Find an expression for potential difference,, in terms ofand, by setting the two capacitance equations equal to each other:

$\frac{Q}{V}=\frac{\epsilon _o A}{d}$

$V = \frac{Qd}{\epsilon _o A}$

2. Multiply the potential difference times the proton charge to find work (and thus kinetic energy):

$W = \frac {eQd}{\epsilon _o A} = \Delta K$

Report an Error

Example Question #1 :Using Capacitor Equations

A $5 \mu F$ and $10 \mu F$ capacitor are connected in series with a 12V battery. What is the magnitude of the charge on one plate of the $5 \mu F$ capacitor?

Possible Answers:

$2.4 \mu C$

$60 \mu C$

$10 \mu C$

$180 \mu C$

$40 \mu C$

Correct answer:

$40 \mu C$

Explanation:

Relevant equations:

Q = CV

$\frac{1}{C_{eq, series}} = \frac{1}{C_1} + \frac{1}{C_2}$

First, find the equivalent capacitance of the whole circuit:

$\frac{1}{C_{eq,series}}= \frac{1}{5} + \frac{1}{10} = \frac{3}{10}$

$C_{eq, series} = \frac{10}{3} \mu F$

Use this equivalent capacitance to find the total charge:

$Q = CV = (\frac{10}{3} \mu F) * (12 V) = 40 \mu C$

Capacitors in series always have the same charge on each unit, so the charge on the $5 \mu F$ capacitor is also $40 \mu C$ .

Report an Error

View Tutors

Jiban
Certified Tutor

Tribhuvan University, Bachelor of Science, Mathematics.

View Tutors

Michael
Certified Tutor

Indiana University of Pennsylvania-Main Campus, Bachelor of Science, Finance. Virginia Polytechnic Institute and State Univer...

View Tutors

Haoyi
Certified Tutor

Southern Methodist University, Bachelor of Accountancy, Accounting. Southern Methodist University, Master of Science, Finance.

All AP Physics C: Mechanics Resources

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

Popular Subjects

SAT Tutors in Denver,Calculus Tutors in Dallas Fort Worth,French Tutors in Philadelphia,Statistics Tutors in Denver,ACT Tutors in San Diego,GRE Tutors in Philadelphia,LSAT Tutors in New York City,Computer Science Tutors in Houston,Computer Science Tutors in Dallas Fort Worth,MCAT Tutors in Seattle

Popular Courses & Classes

LSAT Courses & Classes in Denver,ACT Courses & Classes in Philadelphia,SAT Courses & Classes in Atlanta,Spanish Courses & Classes in Phoenix,Spanish Courses & Classes in Boston,GRE Courses & Classes in New York City,LSAT Courses & Classes in Dallas Fort Worth,GMAT Courses & Classes in Dallas Fort Worth,MCAT Courses & Classes in Seattle,GMAT Courses & Classes in Miami

Popular Test Prep

ISEE Test Prep in San Francisco-Bay Area,SAT Test Prep in San Diego,SSAT Test Prep in San Francisco-Bay Area,LSAT Test Prep in Dallas Fort Worth,GMAT Test Prep in Dallas Fort Worth,LSAT Test Prep in Houston,ACT Test Prep in Philadelphia,GMAT Test Prep in Miami,MCAT Test Prep in Seattle,SAT Test Prep in Phoenix

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.

哟ur 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:

哟u 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.

把你的投诉我们指定的代理:

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.

我有一个好信仰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 : Capacitors

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

All AP Physics C: Mechanics Resources

Example Questions

Example Question #1 :Capacitors

Example Question #1 :Electric Circuits

Example Question #1 :Capacitors

Example Question #1 :Using Capacitor Equations

Example Question #1 :Capacitors

Example Question #1 :Capacitors

Example Question #2 :Capacitors

Example Question #1 :Electric Circuits

Example Question #1 :Using Capacitor Equations

All AP Physics C: Mechanics Resources

Report an issue with this question

DMCA Complaint

Contact Information

Address

Complaint Details

Find the Best Tutors

Email address:
哟ur name:
Feedback: