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AP Physics 2 : Charge Distribution

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Example Questions

Example Question #1 : Charge Distribution

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:

$8.03\cdot10^7C$

$-7.45\cdot10^7C$

$-8.03\cdot 10^7C$

$4.01\cdot10^7C$

$-4.01\cdot10^7C$

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 #2 : Charge Distribution

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:

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

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

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

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

None of the other answers is correct

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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All AP Physics 2 Resources

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AP Physics 2 : Charge Distribution

Study concepts, example questions & explanations for AP Physics 2

All AP Physics 2 Resources

Example Questions

Example Question #1 : Charge Distribution

Example Question #2 : Charge Distribution

Example Question #101 : Electrostatics

Example Question #1 : Charge Distribution

Example Question #1 : Charge Distribution

All AP Physics 2 Resources

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