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AP Physics 2 : Electricity and Magnetism

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

Example Questions

Example Question #1 : Gauss's Law

A sphere of radius 5cm contains a charge of 66nC , calculate the electric flux.

Possible Answers:

$3.55*10^3V\cdot m$

$6.21*10^3V\cdot m$

$7.45*10^3V\cdot m$

$9.55*10^3V\cdot m$

$2.45*10^3V\cdot m$

Correct answer:

$7.45*10^3V\cdot m$

Explanation:

Use the equation for electric flux:

$Flux_E=\frac{Q}{\epsilon_0}$

Plug in values:

$Flux_E=\frac{66*10^{-9}}{8.85*10^{-12}}$

$Flux_E=7.45*10^3V\cdot m$

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Example Question #151 : Electricity And Magnetism

A $15\textup{ nC}$ charge is placed inside of a metal sphere with radius $10\textup{ cm}$ . Determine the electric field at the surface of the sphere.

Possible Answers:

$1.35*10^4\ \frac{\textup{N}}{\textup{C}}$

$8.44*10^4\ \frac{\textup{N}}{\textup{C}}$

$3.95*10^4\ \frac{\textup{N}}{\textup{C}}$

$6.11*10^4\ \frac{\textup{N}}{\textup{C}}$

None of these

Correct answer:

$1.35*10^4\ \frac{\textup{N}}{\textup{C}}$

Explanation:

Using Electric Field Formula:

$E=k\frac{q}{r^2}$

Converting to , to and plugging in values:

$E=9*10^9 \frac{15*10^{-9}}{.10^2}$

$E=1.35*10^4 \frac{N}{C}$

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Example Question #4 : Gauss's Law

Determine the electric field on the surface of a sphere of radius $0.333\textup{ m}$ if $1969\textup{ nC}$ are contained within.

Possible Answers:

$3.2*10^5\ \frac{\textup{N}}{\textup{C}}$

$1.6*10^5\ \frac{\textup{N}}{\textup{C}}$

$2.4*10^5\ \frac{\textup{N}}{\textup{C}}$

$5.33*10^5\ \frac{\textup{N}}{\textup{C}}$

$4.8*10^5\ \frac{\textup{N}}{\textup{C}}$

Correct answer:

$1.6*10^5\ \frac{\textup{N}}{\textup{C}}$

Explanation:

Using Gauss's law:

$\Phi =\frac{Q}{\epsilon_0}$

$E_{surface}*A=\frac{Q}{\epsilon_0}$

Where $E_{surface}$ is the electric field at the surface of the enclosed shape

is the surface area of the shape

is the charge enclosed

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

Solving for $E_{surface}$

$E_{surface}=\frac{Q}{\epsilon_0 A}$

Surface area of a sphere:

$A=4\pi r^2$

Plugging in values:

$E_{surface}=\frac{1969*10^{-9}}{8.85*10^{-12} 4\pi *.333^2}$

$E_{surface}=1.6*10^5\frac{N}{C}$

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Example Question #10 : Gauss's Law

Determine the electric flux on the surface of a ball with radius 10cm with a helium nucleus inside.

Possible Answers:

$3.61*10^{-8}\frac{N\cdot m^2}{C}$

$7.77*10^{-8}\frac{N\cdot m^2}{C}$

None of these

$1.6*10^{-8}\frac{N\cdot m^2}{C}$

$9.16*10^{-8}\frac{N\cdot m^2}{C}$

Correct answer:

$3.61*10^{-8}\frac{N\cdot m^2}{C}$

Explanation:

$\oint{E*dA}=\frac{Q}{\epsilon_0}$

$\oint{E*dA}$ is considered the flux.

Plugging in values:

$\oint{E*dA}=\frac{2*1.6*10^{-19}}{8.85*10^{-12}}$

$\oint{E*dA}=3.61*10^{-8}\frac{N\cdot m^2}{C}$

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Example Question #91 : Electrostatics

Tom is given an uncharged, solid sphere of radius . He then decides to charge all the material within a distance $\frac{R}{2}$ from the center uniformly. He gives that part of the sphere a total charge of $\frac{-Q}{2}$ . He then charges the rest of the material uniformly, giving that part of the sphere a total charge of . Besides the very center point of the sphere, at what distance from the center will there be no electric field?

Possible Answers:

.75R

.606R

.833R

.825R

Correct answer:

.825R

Explanation:

We need the charge enclosed by a gaussian surface (in this case, we'll use a sphere because there's spherical symmetry) to be zero. When this happens, there cannot be any electric field as $\frac{q_{enc}}{\epsilon}= \int E\cdot da$

$q_{enc}=0$

$0=E*(4\pi r^2)$

E=0

All the charge is uniformly distributed, so the charge density of the inner sphere is $\frac{\frac{-Q}{2}}{\frac{4\pi (\frac{R}{2})^3}{3}}=\frac{-3Q}{\pi R^3}$ and on the outer sphere is $\frac{Q}{\frac{4\pi(R^3-(\frac{R}{2})^3)}{3}}=\frac{6Q}{7\pi R^3}$ . We only need to use the outer sphere's surface density since the zero electric field point cannot enclose zero charge in the inner sphere since it's only enclosing negative charge.

The distance where there is no electric field will be when the gaussian surface encloses $\frac{Q}{2}$ from the outer sphere in order to cancel the $\frac{-Q}{2}$ from the inner sphere.

Because everything is uniformly distributed:

$q=\rho*V$ . is charge, $\rho$ is charge density, and is volume. We need this to equal $\frac{Q}{2}$

$\frac{Q}{2}=\frac{6Q}{7\pi R}*V$

$V=\frac{7\pi R^3}{12}$

This is the volume of just the outer surface needed. To find the distance, we just use the formula for the volume of a thick spherical shell and algebra.

Let be the distance where there will be zero electric field.

$\frac{7\pi R^3}{12}=\frac{4\pi*(r^3-(\frac{R}{2})^3))}{3}$

This gives

$r\approx.825R$

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Example Question #91 : Electrostatics

You have 1.5kg of water. One mole of water has a mass of $18 \frac{g}{mol}$ , and a single molecule of water contains 10 electrons. What is the total amount of charge contributed by the electrons in the water?

$e=1.6\cdot 10^{-19}$

$N_A=6.022\cdot10^{23}$

Possible Answers:

$8.03\cdot10^7C$

$-8.03\cdot 10^7C$

$-7.45\cdot10^7C$

$-4.01\cdot10^7C$

$4.01\cdot10^7C$

Correct answer:

$-8.03\cdot 10^7C$

Explanation:

Because we're talking about electrons, the answer must be negative. The way to solve this is to find how many electrons are in 1.5 kg of water. First, we need to convert kilograms of water into grams of water:

$1.5 kg_{H_2O} \cdot \frac{1000g}{1 kg}=1500 g_{H_2O}$

Then, we can use the provided molar mass of water to calculate the number of moles of water in 1.5kg of water:

$1500 g_{H_2O}\cdot \frac{1 mol_{H_2O}}{18 g_{H_2O}}=83.33 mol_{H_2O}$

Once we know how many moles of water we have, we can use Avogadro's number ( $6.022\cdot10^{23}\frac{molecules}{mol}$ ) to calculate how many molecules of water are in 83.33mol.

$83.33 mol_{H_2O}\cdot\frac{6.022\cdot10^{23} molecules}{1 mol}=5.0181326\cdot10^{25} molecules_{H_2O}$

Once we know how many molecules of water we have, we can multiply by 10 to figure out how many electrons those molecules represent, since we are told that each water molecule has 10 electrons.

$5.0181326\cdot10^{25} molecules_{H_2O}\cdot\frac{10e^-}{1 molecule_{H_2O}}=5.0181326\cdot10^{26} e^-$

Finally, we can multiply by the provided charge of an electron to calculate the charge of those electrons.

$5.0181326\cdot10^{26} electrons\cdot\frac{-1.6\cdot10^{-19}C}{1e^-}=-80290121.6\approx-8.03\cdot10^{-7}C$

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Example Question #1 : Using Avogadro's Number

You have a neutral balloon. If you were to add 21,000 electrons to it, what would its net charge be?

$e=1.6\cdot 10^{-19}$ = charge of one electron

Possible Answers:

$-3.36\cdot 10^{-9}\mu C$

None of the other answers is correct

$3.08\cdot 10^{-9}\mu C$

$-3.08\cdot 10^{-9}\mu C$

$3.36\cdot 10^{-9}\mu C$

Correct answer:

$-3.36\cdot 10^{-9}\mu C$

Explanation:

The elemental charge is the magnitude of charge, in Coulombs, that each electron or proton has. Because electrons have a negative charge, don't forget to add a negative sign into the equation.

$q=-(\#\ of\ electrons)(e)$

$q= -(21000)(1.6\cdot 10^{-19})$

$q= -3.36\cdot 10^{-15}$

When you convert the answer to microcoulombs, the answer is $-3.36\cdot 10^{-9}\mu C$ :

$-3.36\cdot 10^{-15} C\cdot\frac{1.0\cdot10^6 \mu C}{1 C}= -3.36\cdot 10^{-9}\mu C$

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Example Question #101 : Electrostatics

Which of the following best represents the charge of an electron?

$1C=6.24*10^{18}e^-$

Possible Answers:

$-6.24 * 10^{-19}\textup{C}$

$1.6 *10^{-18}\textup C$

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

$-1\textup C$

$6.24 * 10^{-18}{\textup{C}$

Correct answer:

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

Explanation:

The equation for the quantity of charge is:

q=ne

where is the charge quantity, represents the number of electrons, and is the charge of an electron, also known as the elementary charge.

Rewrite the equation.

$e=\frac{q}{n}$

One coulomb, , consists of $6.24 * 10^{18}$ electrons,

Substitute these two values into the formula.

$e=\frac{q}{n}=\frac{1 \textup{ C}}{6.24 * 10^{18} \textup{electrons}}$

$e= 1.6 * 10^{-19} \frac{\textup{C}}{\textup{1electron}}$

This number represents the electron's fundamental charge.

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

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Example Question #1 : Charge Distribution

Imagine you have a neutral balloon. If you remove 16,000 electrons from it, what is the net charge on the balloon?

$e=1.6\cdot 10^{-19}C$

Possible Answers:

$5.32\cdot10^{-15}C$

$2.56\cdot10^{-15}C$

$-2.56\cdot10^{-15}C$

$-6.55\cdot10^{-15}C$

$6.55\cdot10^{-14}C$

Correct answer:

$-2.56\cdot10^{-15}C$

Explanation:

Because this is a neutral balloon, the net charge is equal to the charge the was removed, but opposite in sign. There were 16,000 electrons removed, each of which has a charge of $1.6\cdot 10^{-19}C$ . Therefore, the total charge that was removed is:

$1.6\cdot 10^{-19}C \cdot 16000$

$=2.56\cdot10^{-15}C$

To answer the question, we must remember that if that much charge was removed from the balloon, the balloon will now be negative.

$=-2.56\cdot10^{-15}C$

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Example Question #1 : Charge Distribution

A hollow metal sphere of radius 3.4cm has a charge of 19.9nC distributed evenly on the entirety of the surface. Find the surface charge density.

Possible Answers:

$.500\frac{nC}{cm^2}$

$.137\frac{nC}{cm^2}$

$.459\frac{nC}{cm^2}$

$.334\frac{nC}{cm^2}$

$.887\frac{nC}{cm^2}$

Correct answer:

$.137\frac{nC}{cm^2}$

Explanation:

Surface area of sphere:

$4\pi r^2$

Plug in values:

$145cm^2=4\pi*r^2$

$\frac{19.9}{145}=.137\frac{nC}{cm^2}$

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AP Physics 2 : Electricity and Magnetism

Study concepts, example questions & explanations for AP Physics 2

All AP Physics 2 Resources

Example Questions

Example Question #1 : Gauss's Law

Example Question #151 : Electricity And Magnetism

Example Question #4 : Gauss's Law

Example Question #10 : Gauss's Law

Example Question #91 : Electrostatics

Example Question #91 : Electrostatics

Example Question #1 : Using Avogadro's Number

Example Question #101 : Electrostatics

Example Question #1 : Charge Distribution

Example Question #1 : Charge Distribution

All AP Physics 2 Resources

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