Call Now to Set Up Tutoring:

(888) 888-0446

AP Physics 2 : Capacitors and Electric Fields

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 : Capacitors And Electric Fields

A $3.0\: \mu F$ $\displaystyle 3.0\: \mu F$ parallel-plate capacitor is connected to a constant voltage source. If the distance between the plates of this capacitor is $4\:mm$ $\displaystyle 4\:mm$ and the capacitor holds a charge of $12.0\: \mu C$ $\displaystyle 12.0\: \mu C$, what is the value of the electric field between the plates of this capacitor?

Possible Answers:

$1\frac{V}{m}$ $\displaystyle 1\frac{V}{m}$

$10\frac{V}{m}$ $\displaystyle 10\frac{V}{m}$

$1000\frac{V}{m}$ $\displaystyle 1000\frac{V}{m}$

An electric field does not exist between the plates of a parallel-plate capacitor

$100\frac{V}{m}$ $\displaystyle 100\frac{V}{m}$

Correct answer:

$1000\frac{V}{m}$ $\displaystyle 1000\frac{V}{m}$

Explanation:

To solve this problem, we first need to solve for the voltage across the capacitor. For this, we'll need the formula for capacitance:

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

Solving for the voltage:

$V=\frac{Q}{C}=\frac{12.0\cdot 10^{-6}C}{3.0\cdot 10^{-6}F}=4.0V$ $\displaystyle V=\frac{Q}{C}=\frac{12.0\cdot 10^{-6}C}{3.0\cdot 10^{-6}F}=4.0V$

Now that we have the voltage, we can make use of the following equation to solve for the electric field:

V=Ed $\displaystyle V=Ed$

$E=\frac{V}{d}=\frac{4.0V}{4\cdot 10^{-3}m}=1000\frac{V}{m}$ $\displaystyle E=\frac{V}{d}=\frac{4.0V}{4\cdot 10^{-3}m}=1000\frac{V}{m}$

Report an Error

Example Question #2 : Capacitors And Electric Fields

A set of parallel plate capacitors with a surface area of 12m^2 $\displaystyle 12m^2$ has a total amount of charge equal to $\displaystyle 4C$. What is the electric field between the plates?

$\epsilon=8.85\cdot 10^{-12}\frac{F}{m}$ $\displaystyle \epsilon=8.85\cdot 10^{-12}\frac{F}{m}$

Possible Answers:

$5.34\cdot 10^{10}\frac{N}{C}$ $\displaystyle 5.34\cdot 10^{10}\frac{N}{C}$

$8.54\cdot 10^{9}\frac{N}{C}$ $\displaystyle 8.54\cdot 10^{9}\frac{N}{C}$

$6.67\cdot 10^{9}\frac{N}{C}$ $\displaystyle 6.67\cdot 10^{9}\frac{N}{C}$

$8.12\cdot 10^{10}\frac{N}{C}$ $\displaystyle 8.12\cdot 10^{10}\frac{N}{C}$

$3.77\cdot 10^{10}\frac{N}{C}$ $\displaystyle 3.77\cdot 10^{10}\frac{N}{C}$

Correct answer:

$3.77\cdot 10^{10}\frac{N}{C}$ $\displaystyle 3.77\cdot 10^{10}\frac{N}{C}$

Explanation:

The equation for the electric field between two parallel plate capacitors is:

$E=\frac{\sigma}{\epsilon}$ $\displaystyle E=\frac{\sigma}{\epsilon}$

Sigma is the charge density of the plates, which is equal to:

$\sigma = \frac{Q}{A}$ $\displaystyle \sigma = \frac{Q}{A}$

We are given the area and total charge, so we use them to find the charge density.

$\sigma&=\frac{4}{12}$ $\displaystyle \sigma&=\frac{4}{12}$

$\sigma= 0.3334\frac{C}{m^2}$ $\displaystyle \sigma= 0.3334\frac{C}{m^2}$

Now that we have the charge density, divide it by the vacuum permittivity to find the electric field.

$E=\frac{0.3334}{8.85\times10^{-12}}=3.77\times10^{10}\frac{N}{C}$ $\displaystyle E=\frac{0.3334}{8.85\times10^{-12}}=3.77\times10^{10}\frac{N}{C}$

Report an Error

Example Question #3 : Capacitors And Electric Fields

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 electric field, E, stored in the capacitor increase, decrease or stay the same?

Possible Answers:

Decreases

Stays constant

We need to know

Increases

Correct answer:

Stays constant

Explanation:

Relevant equations:

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

Q=CV $\displaystyle Q=CV$

$U=\frac{1}{2}\frac{Q^{2}}{C}$ $\displaystyle U=\frac{1}{2}\frac{Q^{2}}{C}$

Considering we are dealing with regions of charge instead of a point charge (or a charge that is at least spherical symmetric), it would be wise to consider using the definition of the electric field here:

$E=\frac{V}{D}$ $\displaystyle E=\frac{V}{D}$

We are only considering magnitude so the direction of the electric field is not of concern. Considering a battery is not connected, we can't claim the V is constant. So we use the second equation to substitute for V in the above equation:

$E=\frac{Q}{CD}$ $\displaystyle E=\frac{Q}{CD}$

Substitute C from the first equation:

$E=\frac{Q\varepsilon _{o}D}{AD}$ $\displaystyle E=\frac{Q\varepsilon _{o}D}{AD}$

We see here that D actually cancels. This would not have been obvious without the previous substitution. Final result:

$E=\frac{\varepsilon _{o}Q}{A}$ $\displaystyle E=\frac{\varepsilon _{o}Q}{A}$

These 3 quantities are static in this situation so E does not change.

Report an Error

Example Question #2 : Capacitors And Electric Fields

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 electric field, E, stored in the capacitor increase, decrease or stay the same?

Possible Answers:

Stays constant

Decreases by exactly $\frac{1}{4}$ $\displaystyle \frac{1}{4}$

Increases by exactly $\frac{1}{2}$ $\displaystyle \frac{1}{2}$

Decreases by exactly $\frac{1}{2}$ $\displaystyle \frac{1}{2}$

Correct answer:

Decreases by exactly $\frac{1}{2}$ $\displaystyle \frac{1}{2}$

Explanation:

Relevant equations:

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

Q=CV $\displaystyle Q=CV$

$U=\frac{1}{2}\frac{Q^{2}}{C}$ $\displaystyle U=\frac{1}{2}\frac{Q^{2}}{C}$

$E=\frac{V}{D}$ $\displaystyle E=\frac{V}{D}$

The battery maintains a constant V so we can directly relate E and D. Since D is doubled, E will be halved.

Report an Error

Example Question #3 : Capacitors And Electric Fields

Lazy capacitor

If V_1=24V $\displaystyle V_1=24V$, each plate of the capacitor has surface area $\displaystyle 240 cm^2$, and the plates are .5mm $\displaystyle .5mm$ apart, determine the electric field between the plates.

Possible Answers:

$1.6*10^4 \frac{N}{C}$ $\displaystyle 1.6*10^4 \frac{N}{C}$

$4.8*10^4 \frac{N}{C}$ $\displaystyle 4.8*10^4 \frac{N}{C}$

$4.0*10^4 \frac{N}{C}$ $\displaystyle 4.0*10^4 \frac{N}{C}$

$5.5*10^4 \frac{N}{C}$ $\displaystyle 5.5*10^4 \frac{N}{C}$

$6.6*10^4 \frac{N}{C}$ $\displaystyle 6.6*10^4 \frac{N}{C}$

Correct answer:

$4.8*10^4 \frac{N}{C}$ $\displaystyle 4.8*10^4 \frac{N}{C}$

Explanation:

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

$V_1=V_{C1}$ $\displaystyle V_1=V_{C1}$

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

$V_{C1}=E*d$ $\displaystyle V_{C1}=E*d$

Combine these equations and solve for the electric field:

V_1=E*d $\displaystyle V_1=E*d$

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

Convert mm to m and plug in values:

$\frac{24}{.0005}=E$ $\displaystyle \frac{24}{.0005}=E$

$E=4.8*10^4 \frac{N}{C}$ $\displaystyle E=4.8*10^4 \frac{N}{C}$

Report an Error

Example Question #4 : Capacitors And Electric Fields

Lazy capacitor

If V_1=60V $\displaystyle V_1=60V$, each plate of the capacitor has surface area $\displaystyle 600 cm^2$, and the plates are 1mm $\displaystyle 1mm$ apart, determine the excess charge on the positive plate.

Possible Answers:

50.6nC $\displaystyle 50.6nC$

31.9nC $\displaystyle 31.9nC$

25.4nC $\displaystyle 25.4nC$

48.8nC $\displaystyle 48.8nC$

27.7nC $\displaystyle 27.7nC$

Correct answer:

31.9nC $\displaystyle 31.9nC$

Explanation:

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

$V_1=V_{C1}$ $\displaystyle V_1=V_{C1}$

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

$V_{C1}=E*d$ $\displaystyle V_{C1}=E*d$

Combine equations and solve for the electric field:

V_1=E*d $\displaystyle V_1=E*d$

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

Convert mm to m and plugging in values:

$\frac{60V}{.001}=E$ $\displaystyle \frac{60V}{.001}=E$

$60000\frac{N}{C}=E$ $\displaystyle 60000\frac{N}{C}=E$

Use the electric field in a capacitor equation:

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

Combine equations:

$E*\epsilon_0*A=q$ $\displaystyle E*\epsilon_0*A=q$

Converting cm^2 $\displaystyle cm^2$ to m^2 $\displaystyle m^2$ and plug in values:

$60000*8.85*10^{-12}*.06=q$ $\displaystyle 60000*8.85*10^{-12}*.06=q$

$\displaystyle 31.9nC=q$

Report an Error

Example Question #5 : Capacitors And Electric Fields

Lazy capacitor

Consider the given diagram. If V_1=4V $\displaystyle V_1=4V$, each plate of the capacitor has surface area $\displaystyle 100 cm^2$, and the plates at 1mm $\displaystyle 1mm$ apart, determine the electric field between the plates.

Possible Answers:

$2000\frac{N}{C}=E$ $\displaystyle 2000\frac{N}{C}=E$

$7000\frac{N}{C}=E$ $\displaystyle 7000\frac{N}{C}=E$

$9000\frac{N}{C}=E$ $\displaystyle 9000\frac{N}{C}=E$

$3333\frac{N}{C}=E$ $\displaystyle 3333\frac{N}{C}=E$

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

Correct answer:

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

Explanation:

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

$V_1=V_{C1}$ $\displaystyle V_1=V_{C1}$

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

$V_{C1}=E*d$ $\displaystyle V_{C1}=E*d$

Combine equations and solve for the electric field:

V_1=E*d $\displaystyle V_1=E*d$

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

Convert mm to m and plug in values:

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

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

Report an Error

Example Question #1 : Capacitors And Electric Fields

Lazy capacitor

In the given circuit, the capacitor is made of two parallel circular plates of radius .3cm $\displaystyle .3cm$ that are 2.5mm $\displaystyle 2.5mm$ apart. If V_1 $\displaystyle V_1$ is equal to $\displaystyle 5V$, determine the electric field between the plates.

Possible Answers:

None of these

$1500\frac{N}{C}$ $\displaystyle 1500\frac{N}{C}$

$2000\frac{N}{C}$ $\displaystyle 2000\frac{N}{C}$

$2500\frac{N}{C}$ $\displaystyle 2500\frac{N}{C}$

$200\frac{N}{C}$ $\displaystyle 200\frac{N}{C}$

Correct answer:

$2000\frac{N}{C}$ $\displaystyle 2000\frac{N}{C}$

Explanation:

Since it is the only element in the circuit besides the source, the voltage drop across the capacitor must be equal to the voltage gain in the capacitor.

$V_1=V_{C1}$ $\displaystyle V_1=V_{C1}$

Definition of voltage:

V=E*d $\displaystyle V=E*d$

Combine equations:

$V_1=E_{C1}*d$ $\displaystyle V_1=E_{C1}*d$

Solve for $E_{C1}$ $\displaystyle E_{C1}$

$\frac{V}{d}=E_{C1}$ $\displaystyle \frac{V}{d}=E_{C1}$

Convert to meters and plug in values:

$\frac{5}{.0025}=E_{C1}$ $\displaystyle \frac{5}{.0025}=E_{C1}$

$2000\frac{N}{C}=E_{C1}$ $\displaystyle 2000\frac{N}{C}=E_{C1}$

Report an Error

Example Question #7 : Capacitors And Electric Fields

A parallel plate capacitor with a separation of $.5\textup{ mm}$ $\displaystyle .5\textup{ mm}$ and surface area $150\textup{ cm}^2$ $\displaystyle 150\textup{ cm}^2$ is in series with a $9\textup{ V}$ $\displaystyle 9\textup{ V}$ battery. How will the electric field in the capacitor change if the separation is doubled?

Possible Answers:

Halved

It will stay the same

Tripled

Quadrupled

Doubled

Correct answer:

Halved

Explanation:

The voltage drop through the capacitor needs to be equal to the voltage of the battery.

V_b=V_c $\displaystyle V_b=V_c$

The voltage drop of a parallel plate capacitor is equal to the internal electric field times the distance between them.

V_c=E*d $\displaystyle V_c=E*d$

Combing equations and solving for $\displaystyle E$

$E=\frac{V_b}{d}$ $\displaystyle E=\frac{V_b}{d}$

From this, it can be seen that doubling the separation will halve the electric field.

Report an Error

Example Question #1 : Capacitors And Electric Fields

Possible Answers:

Impossible to determine

No change

Tripled

Doubled

Halved

Correct answer:

Doubled

Explanation:

The voltage drop through the capacitor needs to be equal to the voltage of the battery.

V_b=V_c $\displaystyle V_b=V_c$

The voltage drop of a parallel plate capacitor is equal to the internal electric field times the distance between them.

V_c=E*d $\displaystyle V_c=E*d$

Combing equations and solving for $\displaystyle E$

$E=\frac{V_b}{d}$ $\displaystyle E=\frac{V_b}{d}$

From this, it can be seen that doubling the voltage of the battery will doubled the electric field inside the capacitor.

Report an Error

View Tutors

Leslie
Certified Tutor

McNeese State University, Bachelor in Arts, Elementary School Teaching.

View Tutors

Mariam
Certified Tutor

Yarmouk University, Bachelor of Science, Mathematics. Touro College, Master of Science, Teacher Education, Multiple Levels.

View Tutors

Angie
Certified Tutor

Boise State University, Bachelor of Science, Nursing (RN). Walden University, Master of Science, Nursing (RN).

All AP Physics 2 Resources

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

Popular Subjects

GRE Tutors in Philadelphia, GMAT Tutors in Philadelphia, Statistics Tutors in Boston, GRE Tutors in Houston, Math Tutors in Atlanta, Statistics Tutors in New York City, English Tutors in Atlanta, SAT Tutors in Miami, Biology Tutors in Washington DC, English Tutors in Washington DC

Popular Courses & Classes

Spanish Courses & Classes in San Francisco-Bay Area, Spanish Courses & Classes in Washington DC, GRE Courses & Classes in Seattle, MCAT Courses & Classes in Philadelphia, SAT Courses & Classes in Houston, GMAT Courses & Classes in San Diego, MCAT Courses & Classes in Los Angeles, LSAT Courses & Classes in Washington DC, ACT Courses & Classes in San Diego, ACT Courses & Classes in Atlanta

Popular Test Prep

GRE Test Prep in Atlanta, SSAT Test Prep in San Francisco-Bay Area, GMAT Test Prep in Phoenix, SSAT Test Prep in Seattle, ACT Test Prep in Seattle, ACT Test Prep in San Diego, GMAT Test Prep in New York City, MCAT Test Prep in Atlanta, LSAT Test Prep in Denver, GRE 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 : Capacitors and Electric Fields

Study concepts, example questions & explanations for AP Physics 2

All AP Physics 2 Resources

Example Questions

Example Question #1 : Capacitors And Electric Fields

Example Question #2 : Capacitors And Electric Fields

Example Question #3 : Capacitors And Electric Fields

Example Question #2 : Capacitors And Electric Fields

Example Question #3 : Capacitors And Electric Fields

Example Question #4 : Capacitors And Electric Fields

Example Question #5 : Capacitors And Electric Fields

Example Question #1 : Capacitors And Electric Fields

Example Question #7 : Capacitors And Electric Fields

Example Question #1 : Capacitors And 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: