Call Now to Set Up Tutoring:

(888) 888-0446

AP Physics 2 : Electricity and Magnetism

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 #2 : Capacitors And Capacitance

Imagine a capacitor with a magnitude of charge Q on either plate. This capacitor has area A, separation distance D, and is connected to a battery of voltage V. If some external agent pulls the capacitor apart such that D doubles, did the charge on each plate increase, decrease or stay the same?

Possible Answers:

The charge is zero

Decrease

Stays constant

Increase

Correct answer:

Decrease

Explanation:

Relevant equations:

Q=CV

$C=\frac{\varepsilon_{o}A}{D}$

Plug the second equation into the first:

$Q=\frac{\varepsilon _{o}AV}{D}$

Considering all the variables in the numerator are held fixed for this problem, we see that increasing D will decrease the charge stored on each plate.

Report an Error

Example Question #1 : Capacitors And Capacitance

Imagine a capacitor with a magnitude of charge Q on either plate. This capacitor has area A, separation distance D, and is not connected to a battery of voltage V. If some external agent pulls the capacitor apart such that D doubles, did the charge on each plate increase, decrease or stay the same?

Possible Answers:

Decrease

We need to know the capacitance of the capacitor

Increase

Stays constant

Correct answer:

Stays constant

Explanation:

The charge has no where to go. Without the battery connected, the charge has no physical avenue on or off the plates.

Report an Error

Example Question #931 : Ap Physics 2

Imagine a capacitor with a magnitude of charge Q on either plate. This capacitor has area A, separation distance D, and is connected to a battery of voltage V. If some external agent pulls the capacitor apart such that D doubles, how did the capacitance change? Also, do we need to use the information provided about the battery?

Possible Answers:

Increase, yes

Decrease, yes

Decrease, no

Increase, no

Correct answer:

Decrease, no

Explanation:

$C=\frac{\varepsilon_{o}A}{D}$

As D increases, C will decrease

Does the battery matter? No. Capacitance is a "geometric" quantity. This means that it can be determined solely by physical parameters like the area and separation distance. The charge on the plates, voltage difference, electric field and any other quantity you could think of does not influence the capacitance other than A or D.

Report an Error

Example Question #932 : Ap Physics 2

Lazy capacitor

Consider the given diagram. If V_1=4 V , each plate of the capacitor has surface area 100 cm^2 , and the plates are 0.1mm apart, determine the number of excess electrons on the negative plate.

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

Possible Answers:

$n=2.21*10^{9}$

None of these

$n=3.33*10^{9}$

$n=5.15*10^{9}$

$n=5.85*10^{9}$

Correct answer:

$n=2.21*10^{9}$

Explanation:

The voltage rise through the source must be the same as the drop through the capacitor.

$V_1=V_{C1}$

The voltage drop across the capacitor is the equal to the electric field multiplied by the distance.

$V_{C1}=E*d$

Combine equations and find the electric field:

V_1=E*d

$\frac{V_1}{d}=E$

Convert mm to m and plugg in values:

$\frac{V_1}{d}=E$

$4000\frac{N}{C}=E$

Use the electric field in a capacitor equation:

$E=\frac{q}{A*\epsilon_0}$

$E*\epsilon_0*A=q$

Convert cm^2 to m^2 and plug in values:

$(4000)*(8.86*10^{-12})*(100*10^{-4})=q$

$3.54*10^{-10}=q$

The magnitude of total charge on the positive plate is equal to the total charge on the negative plate, so to find the number of excess elections:

$\frac{3.54*10^{-10}}{1.6*10^{-19}}=n$

$n=2.21*10^{9}$

Report an Error

Example Question #281 : Electricity And Magnetism

Lazy capacitor

Consider the given diagram. If V_1=4 V , each plate of the capacitor has surface area 100 cm^2 , and the plates at 1mm apart, determine the excess charge on the positive plate.

Possible Answers:

$6.11*10^{-10}C$

$9.26*10^{-10}C$

$4.44*10^{-10}C$

$1.11*10^{-10}C$

$3.54*10^{-10}C$

Correct answer:

$3.54*10^{-10}C$

Explanation:

The voltage rise through the source must be the same as the drop through the capacitor.

$V_1=V_{C1}$

The voltage drop across the capacitor is the equal to the electric field multiplied by the distance.

$V_{C1}=E*d$

Combine equations and solve for the electric field:

V_1=E*d

$\frac{V_1}{d}=E$

Convert mm to m and plug in values:

$\frac{V_1}{d}=E$

$4000\frac{N}{C}=E$

Using the electric field in a capacitor equation:

$E=\frac{q}{A*\epsilon_0}$

Rearrange to solve for the charge:

$E*\epsilon_0*A=q$

Convert cm^2 to m^2 and plug in values:

$(4000)*(8.86*10^{-12})*(100*10^{-4})=q$

$3.54*10^{-10}C=q$

Report an Error

Example Question #12 : Capacitors And Capacitance

Suppose I have a uniform electric field within a parallel plate capacitor with field strength of $E=41\frac{N}{C}$ .

Suppose the capacitor's plates are .5m in length and .001m in width, and the space between the plates is 1.5cm .

What is the voltage difference across the capacitor?

Possible Answers:

.475V

5.5V

61.5V

.615V

Correct answer:

.615V

Explanation:

Voltage difference $\Delta V$ is given by

$\Delta V=E*d$ , where is the electric field strength and is the distance between the two plates.

For this problem,

$\Delta V=(41*.015)V= .615V$

Report an Error

Example Question #15 : Circuit Components

Suppose I have a uniform electric field within a parallel plate capacitor with field strength of $E=41\frac{N}{C}$ .

Suppose the capacitor's plates are .5m in length and .001m in width, and the space between the plates is 1.5cm .

Determine the capacitance of this system given that the space in between is a vacuum, and that the permittivity of empty space $\epsilon_0$ is $8.85*10^{-12}\frac{F}{m}$ .

Possible Answers:

$C=2.95*10^{-12}F$

$C=2.95*10^{-13}F$

$C=3.38*10^{-6}F$

$C=3.37*10^{6}F$

Correct answer:

$C=2.95*10^{-13}F$

Explanation:

The formula for capacitance is given by:

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

Where $\kappa$ is dielectric strength, is distance between plates, $\epsilon _0$ is permittivity of empty space, and is cross sectional area.

To determine , we do

$A=(.5*.001)m^2=5*10^{-4}m^2$

$\kappa=1$ because between the plates is a vacuum.

d=.015m

Putting it all together,

$C=2.95*10^{-13}F$

Report an Error

Example Question #1 : Capacitor Energy

Consider the circuit:

Circuit_3__capacitors_

If the voltage drop across C2 is 5V, what is the total energy stored in C2 and C3?

$C1 = 5F,\ C2 = 7F,\ C3 = 1F$

Possible Answers:

25J

50J

75J

100J

Correct answer:

100J

Explanation:

In parallel branches of a circuit, the voltage drops are all the same. Therefore, we know that the voltage drop across C3 is also 5V.

We can then use the following equation to calculate the total stored energy:

$U=\frac{1}{2}CV^2$

Since the voltage is the same for both capacitors, we can simply add the two capacitances to do one calculation for energy:

$U = \frac{1}{2}(8F)(5V)^2 = 100J$

Report an Error

Example Question #282 : Electricity And Magnetism

A $15\mu F$ capacitor is connected to a battery. Once the capacitor is fully charged, how much energy is stored?

Possible Answers:

$6.75 \cdot 10^{-5} J$

$1.35 \cdot 10^{-5} J$

None of the other answers is correct

$6.075 \cdot 10^{-4} J$

$1.215 \cdot 10^{-3} J$

Correct answer:

$6.075 \cdot 10^{-4} J$

Explanation:

To find the amount of energy stored in a capactior, we use the equation

$U=\frac{1}{2}\frac{Q^2}{C}=\frac{1}{2}QV=\frac{1}{2}CV^2$ .

We're given the capacitance (), and the voltage (), so we'll use the third equation.

$U=\frac{1}{2}(15\mu F)(9V)$

$U= 6.075\cdot 10^{-4}J$

Report an Error

Example Question #3 : Capacitor Energy

You have 4 capacitors, , , , and , arranged as shown in the diagram below.

4capacitorcircuit

Their capacitances are as follows:

$\begin{align*} A&=1F \\ B&=3F \\ C&=2F \\ D&=4F \end{align*}$

If you have a 6V battery connected to the circuit, what's the total energy stored in the capacitors?

Possible Answers:

90.5J

144.0J

35.0J

100.5J

121.5J

Correct answer:

121.5J

Explanation:

The equation for energy stored in a capacitor is

$W=\frac{Q^2}{2C}=\frac{CV^2}{2}=\frac{QV}{2}$

We can find the capacitance by adding the capacitors together, and we have the voltage, so we'll use the second equation, $W=\frac{1}{2}CV^2$ .

When adding capacitors, remember how to add in series and parallel.

$(Series)\ \frac{1}{C_{tot}}&=\frac{1}{C_1}+\frac{1}{C_2}+...+\frac{1}{C_n}$

$(Parallel)\ C_{tot}&=C_1+C_2+...+C_n$

Capacitors and are in series, and are in parallel, and ABC and are in parallel.

$C_{AB}&=(\frac{1}{C_A}+\frac{1}{C_B})^{-1}$

$C_{AB}=(\frac{1}{1}+\frac{1}{3})^{-1}$

$C_{AB}= \frac{3}{4}$

$C_{ABC}&=C_{AB}+C_C$

$C_{ABC}=\frac{3}{4}+2$

$C_{ABC}= \frac{11}{4}$

$C_{tot}&=C_{ABC}+C_{D}$

$C_{total}=\frac{11}{4}+4$

$C_{total}=\frac{27}{4}$

Now that we have the total capacitance, we can use the earlier equation to find the energy.

$W=\frac{1}{2}CV^2$

$W= \frac{1}{2}(\frac{27}{4})(6)^2$

W= 121.5

The total energy stored is 121.5J.

Report an Error

View Tutors

Salman
Certified Tutor

Payam Noor University,, Bachelor of Science, Physics. University of Kashan, Doctor of Philosophy, Physics.

View Tutors

Loren
Certified Tutor

Sonoma State University, Bachelor in Arts, Philosophy. Saint Marys College of California, Certificate, Special Education.

View Tutors

Lauren
Certified Tutor

Ohio University-Eastern Campus, Bachelor in Arts, Early Childhood Education.

All AP Physics 2 Resources

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

Popular Subjects

LSAT Tutors in Houston, Spanish Tutors in Miami, Statistics Tutors in Washington DC, Physics Tutors in Los Angeles, Algebra Tutors in Washington DC, Calculus Tutors in Atlanta, Statistics Tutors in Seattle, Computer Science Tutors in Philadelphia, Spanish Tutors in Atlanta, Math Tutors in Miami

Popular Courses & Classes

SAT Courses & Classes in Atlanta, GMAT Courses & Classes in Philadelphia, GMAT Courses & Classes in Washington DC, GMAT Courses & Classes in Chicago, MCAT Courses & Classes in Denver, LSAT Courses & Classes in Houston, SSAT Courses & Classes in Atlanta, ACT Courses & Classes in Seattle, GMAT Courses & Classes in Dallas Fort Worth, SSAT Courses & Classes in Houston

Popular Test Prep

LSAT Test Prep in New York City, ISEE Test Prep in Atlanta, ISEE Test Prep in Denver, SAT Test Prep in Houston, SSAT Test Prep in San Diego, LSAT Test Prep in Boston, MCAT Test Prep in New York City, SAT Test Prep in San Francisco-Bay Area, GMAT Test Prep in Dallas Fort Worth, ACT Test Prep in Los Angeles

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 : Electricity and Magnetism

Study concepts, example questions & explanations for AP Physics 2

All AP Physics 2 Resources

Example Questions

Example Question #2 : Capacitors And Capacitance

Example Question #1 : Capacitors And Capacitance

Example Question #931 : Ap Physics 2

Example Question #932 : Ap Physics 2

Example Question #281 : Electricity And Magnetism

Example Question #12 : Capacitors And Capacitance

Example Question #15 : Circuit Components

Example Question #1 : Capacitor Energy

Example Question #282 : Electricity And Magnetism

Example Question #3 : Capacitor Energy

All AP Physics 2 Resources

Report an issue with this question

DMCA Complaint

Contact Information

Address

Complaint Details

Email address:
Your name:
Feedback: