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AP Physics 2 : AP Physics 2

Study concepts, example questions & explanations for AP Physics 2

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

Example Questions

Example Question #2 : Electric Force Between Point Charges

Coloumbs law for varsity

Charge $\displaystyle A$ has a charge of 8 nC $\displaystyle 8 nC$

Charge $\displaystyle B$ has a charge of 10nC $\displaystyle 10nC$

The distance between their centers, $\displaystyle r$ is 1.0 mm $\displaystyle 1.0 mm$.

Determine the magnitude of the electric force on $\displaystyle B$ due to $\displaystyle A$.

Possible Answers:

$|\overrightarrow{F}|=3.7*10^{-1}{N}$ $\displaystyle |\overrightarrow{F}|=3.7*10^{-1}{N}$

$|\overrightarrow{F}|=7.2*10^{-1}{N}$ $\displaystyle |\overrightarrow{F}|=7.2*10^{-1}{N}$

$|\overrightarrow{F}|=7.2*10^{-9}{N}$ $\displaystyle |\overrightarrow{F}|=7.2*10^{-9}{N}$

$|\overrightarrow{F}|=4.2*10^{-1}{N}$ $\displaystyle |\overrightarrow{F}|=4.2*10^{-1}{N}$

None of these

Correct answer:

$|\overrightarrow{F}|=7.2*10^{-1}{N}$ $\displaystyle |\overrightarrow{F}|=7.2*10^{-1}{N}$

Explanation:

Use Coulomb's law:

$\overrightarrow{E}=k\frac{q_A q_B}{r^2}$ $\displaystyle \overrightarrow{E}=k\frac{q_A q_B}{r^2}$

Where $\displaystyle k$ is $9.0*10^9 N\frac{m^2}{C^2}$ $\displaystyle 9.0*10^9 N\frac{m^2}{C^2}$

q_A $\displaystyle q_A$ is charge $\displaystyle A$, in $\displaystyle coulombs$

q_B $\displaystyle q_B$ is charge $\displaystyle B$, in $\displaystyle coulombs$

$\displaystyle r$ is the distance, in meters $\displaystyle meters$.

Convert $\displaystyle mm$ to $\displaystyle m$ and plug in values:

$\overrightarrow{F}=9.0*10^9\frac{8*10^{-9}*10*10^{-9}}{.001^2}$ $\displaystyle \overrightarrow{F}=9.0*10^9\frac{8*10^{-9}*10*10^{-9}}{.001^2}$

$\overrightarrow{F}=7.2*10^{-1}{N}$ $\displaystyle \overrightarrow{F}=7.2*10^{-1}{N}$

Magnitude is equivalent to absolute value:

$|\overrightarrow{F}|=7.2*10^{-1}{N}$ $\displaystyle |\overrightarrow{F}|=7.2*10^{-1}{N}$

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Example Question #3 : Electric Force Between Point Charges

Coloumbs law for varsity

Charge $\displaystyle A$ has a charge of 19 nC $\displaystyle 19 nC$

Charge $\displaystyle B$ has a charge of -28nC $\displaystyle -28nC$

The distance between their centers, $\displaystyle r$ is 4.5 mm $\displaystyle 4.5 mm$.

What is the magnitude of the force on $\displaystyle B$ due to $\displaystyle A$?

Possible Answers:

$|\overrightarrow{F}|=4.5*10^{-1}{N}$ $\displaystyle |\overrightarrow{F}|=4.5*10^{-1}{N}$

$|\overrightarrow{F}|=9.1*10^{-1}{N}$ $\displaystyle |\overrightarrow{F}|=9.1*10^{-1}{N}$

$|\overrightarrow{F}|=7.9*10^{-1}{N}$ $\displaystyle |\overrightarrow{F}|=7.9*10^{-1}{N}$

None of these

$|\overrightarrow{F}|=2.4*10^{-1}{N}$ $\displaystyle |\overrightarrow{F}|=2.4*10^{-1}{N}$

Correct answer:

$|\overrightarrow{F}|=2.4*10^{-1}{N}$ $\displaystyle |\overrightarrow{F}|=2.4*10^{-1}{N}$

Explanation:

Use Coulomb's law:

$\overrightarrow{E}=k\frac{q_A q_B}{r^2}$ $\displaystyle \overrightarrow{E}=k\frac{q_A q_B}{r^2}$

Where $\displaystyle k$ is $9.0*10^9 N\frac{m^2}{C^2}$ $\displaystyle 9.0*10^9 N\frac{m^2}{C^2}$

q_A $\displaystyle q_A$ is charge $\displaystyle A$, in Coulombs

q_B $\displaystyle q_B$ is charge $\displaystyle B$, in Coulombs

$\displaystyle r$ is the distance, in meters.

Convert $\displaystyle mm$ to $\displaystyle m$ and plug in values:

$\overrightarrow{F}=9.0*10^9\frac{19*10^{-9}*-28*10^{-9}}{.0045^2}$ $\displaystyle \overrightarrow{F}=9.0*10^9\frac{19*10^{-9}*-28*10^{-9}}{.0045^2}$

$\overrightarrow{F}=-2.4*10^{-1}{N}$ $\displaystyle \overrightarrow{F}=-2.4*10^{-1}{N}$

Magnitude is equivalent to absolute value:

$|\overrightarrow{F}|=2.4*10^{-1}{N}$ $\displaystyle |\overrightarrow{F}|=2.4*10^{-1}{N}$

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Example Question #11 : Electric Force Between Point Charges

Coloumbs law for varsity

Charge $\displaystyle A$ has a charge of 4 nC $\displaystyle 4 nC$

Charge $\displaystyle B$ has a charge of -7nC $\displaystyle -7nC$

The distance between their centers, $\displaystyle r$ is 6.5 mm $\displaystyle 6.5 mm$.

What is the magnitude of the force on $\displaystyle B$ due to $\displaystyle A$?

Possible Answers:

$|\overrightarrow{F}|=3.8*10^{-3}{N}$ $\displaystyle |\overrightarrow{F}|=3.8*10^{-3}{N}$

$|\overrightarrow{F}|=9.1*10^{-3}{N}$ $\displaystyle |\overrightarrow{F}|=9.1*10^{-3}{N}$

$|\overrightarrow{F}|=6.0*10^{-3}{N}$ $\displaystyle |\overrightarrow{F}|=6.0*10^{-3}{N}$

$|\overrightarrow{F}|=7.5*10^{-3}{N}$ $\displaystyle |\overrightarrow{F}|=7.5*10^{-3}{N}$

None of these

Correct answer:

$|\overrightarrow{F}|=6.0*10^{-3}{N}$ $\displaystyle |\overrightarrow{F}|=6.0*10^{-3}{N}$

Explanation:

Using Coulomb's law:

$\overrightarrow{E}=k\frac{q_A q_B}{r^2}$ $\displaystyle \overrightarrow{E}=k\frac{q_A q_B}{r^2}$

Where $\displaystyle k$ is $9.0*10^9 N\frac{m^2}{C^2}$ $\displaystyle 9.0*10^9 N\frac{m^2}{C^2}$

q_A $\displaystyle q_A$ is charge $\displaystyle A$, in Coulombs

q_B $\displaystyle q_B$ is charge $\displaystyle B$, in Coulombs

$\displaystyle r$ is the distance, in meters.

Convert $\displaystyle mm$ to $\displaystyle m$ and plug in values:

$\overrightarrow{F}=9.0*10^9\frac{4*10^{-9}*-7*10^{-9}}{.0065^2}$ $\displaystyle \overrightarrow{F}=9.0*10^9\frac{4*10^{-9}*-7*10^{-9}}{.0065^2}$

$\overrightarrow{F}=-6.0*10^{-3}{N}$ $\displaystyle \overrightarrow{F}=-6.0*10^{-3}{N}$

Magnitude is equivalent to absolute value:

$|\overrightarrow{F}|=6.0*10^{-3}{N}$ $\displaystyle |\overrightarrow{F}|=6.0*10^{-3}{N}$

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Example Question #12 : Electric Force Between Point Charges

Coloumbs law for varsity

Charge $\displaystyle A$ has a charge of 4 nC $\displaystyle 4 nC$

Charge $\displaystyle B$ has a charge of 5nC $\displaystyle 5nC$

The distance between their centers, $\displaystyle r$ is 1.5 mm $\displaystyle 1.5 mm$.

What is the magnitude of the force on $\displaystyle B$ due to $\displaystyle A$?

Possible Answers:

$|\overrightarrow{F}|=8.0*10^{-2}{N}$ $\displaystyle |\overrightarrow{F}|=8.0*10^{-2}{N}$

$|\overrightarrow{F}|=8.0*10^{-1}{N}$ $\displaystyle |\overrightarrow{F}|=8.0*10^{-1}{N}$

$|\overrightarrow{F}|=1.3*10^{-2}{N}$ $\displaystyle |\overrightarrow{F}|=1.3*10^{-2}{N}$

None of these

$|\overrightarrow{F}|=4.3*10^{-2}{N}$ $\displaystyle |\overrightarrow{F}|=4.3*10^{-2}{N}$

Correct answer:

$|\overrightarrow{F}|=8.0*10^{-2}{N}$ $\displaystyle |\overrightarrow{F}|=8.0*10^{-2}{N}$

Explanation:

Using Coulomb's law:

$\overrightarrow{E}=k\frac{q_A q_B}{r^2}$ $\displaystyle \overrightarrow{E}=k\frac{q_A q_B}{r^2}$

Where $\displaystyle k$ is $9.0*10^9 N\frac{m^2}{C^2}$ $\displaystyle 9.0*10^9 N\frac{m^2}{C^2}$

q_A $\displaystyle q_A$ is charge $\displaystyle A$, in Coulombs

q_B $\displaystyle q_B$ is charge $\displaystyle B$, in Coulombs

$\displaystyle r$ is the distance, in meters.

Convert $\displaystyle mm$ to $\displaystyle m$ and plug in values:

$\overrightarrow{F}=9.0*10^9\frac{4*10^{-9}*5*10^{-9}}{.0015^2}$ $\displaystyle \overrightarrow{F}=9.0*10^9\frac{4*10^{-9}*5*10^{-9}}{.0015^2}$

$\overrightarrow{F}=8.0*10^{-2}{N}$ $\displaystyle \overrightarrow{F}=8.0*10^{-2}{N}$

Magnitude is equivalent to absolute value:

$|\overrightarrow{F}|=8.0*10^{-2}{N}$ $\displaystyle |\overrightarrow{F}|=8.0*10^{-2}{N}$

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Example Question #761 : Ap Physics 2

A point charge of Q sits at rest next to another charge of 3Q at a distance of R away. How does the force on Q from 3Q compare with the force on 3Q from Q?

Possible Answers:

The force on the Q charge is four times larger

The force on the Q charge is two times larger

The force on the 3Q charge is two times larger

The forces are equal

Correct answer:

The forces are equal

Explanation:

Newton's third law states that any force interactions between two bodies must be equal. This is not so simple when another body is introduced. However, with two bodies, any force that acts on the other will have an equal and opposite reaction force.

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Example Question #762 : Ap Physics 2

A point charge of Q and speed v heads towards another charge of 3Q. Assume the charges are relatively close to one another How does the force on Q from 3Q compare with the force on 3Q from Q?

Possible Answers:

The force on the Q charge is twice as large

The force on the 3Q charge is four times as large

The force on the 3Q charge is one-fourth as large

The forces are equal

Correct answer:

The forces are equal

Explanation:

Newton's third law states that any force interactions between two bodies must be equal. This is not so simple when another body is introduced. However, with two bodies, any force on acts on the other will have an equal and opposite reaction force. The fact that one of the bodies is moving is irrelevant when considering the relative magnitudes of the Newton's Third law pair.

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Example Question #63 : Electrostatics

Two $\displaystyle He$ nuclei are 1nm $\displaystyle 1nm$ apart. Determine the magnitude of the electrical force of one on the other.

$k=9*10^9N\frac{m^2}{C^2}$ $\displaystyle k=9*10^9N\frac{m^2}{C^2}$

Possible Answers:

$F=7.7*10^{-10}N$ $\displaystyle F=7.7*10^{-10}N$

$F=9.2*10^{-10}N$ $\displaystyle F=9.2*10^{-10}N$

$F=6.9*10^{-10}N$ $\displaystyle F=6.9*10^{-10}N$

$F=5.5*10^{-10}N$ $\displaystyle F=5.5*10^{-10}N$

$F=3.1*10^{-10}N$ $\displaystyle F=3.1*10^{-10}N$

Correct answer:

$F=9.2*10^{-10}N$ $\displaystyle F=9.2*10^{-10}N$

Explanation:

Use Coulombs Law:

$F=k\frac{q_1q_2}{r^2}$ $\displaystyle F=k\frac{q_1q_2}{r^2}$

Where $k=9*10^9N\frac{m^2}{C^2}$ $\displaystyle k=9*10^9N\frac{m^2}{C^2}$

Convert $\displaystyle nm$ to $\displaystyle m$ and plug in values:

$F=9*10^9\frac{3.2*10^{-19}*3.2*10^{-19}}{(1*10^{-9})^2}$ $\displaystyle F=9*10^9\frac{3.2*10^{-19}*3.2*10^{-19}}{(1*10^{-9})^2}$

$F=9.2*10^{-10}N$ $\displaystyle F=9.2*10^{-10}N$

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Example Question #111 : Electricity And Magnetism

How will the force between two positively charged objects change as they are brought closer together?

Possible Answers:

The repulsive forces will increase

It is impossible to determine

The repulsive force will decrease

The attractive forces will increase

The force will change from repulsive to attractive

Correct answer:

The repulsive forces will increase

Explanation:

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Example Question #13 : Electric Force Between Point Charges

Charge A and B are $15\textup{ cm}$ $\displaystyle 15\textup{ cm}$ apart. If charge A has a charge of $2 \textup{ nC}$ $\displaystyle 2 \textup{ nC}$ and a mass of $2\textup{ mg}$ $\displaystyle 2\textup{ mg}$, charge B has a charge of $4\textup{ nC}$ $\displaystyle 4\textup{ nC}$ and a mass of $10\textup{ mg}$ $\displaystyle 10\textup{ mg}$, determine the acceleration of A due to B.

Possible Answers:

$.0011\ \frac{\textup{m}}{\textup{ s}^2}$ $\displaystyle .0011\ \frac{\textup{m}}{\textup{ s}^2}$

$.0032\ \frac{\textup{m}}{\textup{ s}^2}$ $\displaystyle .0032\ \frac{\textup{m}}{\textup{ s}^2}$

None of these

$.0044\ \frac{\textup{m}}{\textup{ s}^2}$ $\displaystyle .0044\ \frac{\textup{m}}{\textup{ s}^2}$

$.0016\ \frac{\textup{m}}{\textup{ s}^2}$ $\displaystyle .0016\ \frac{\textup{m}}{\textup{ s}^2}$

Correct answer:

$.0016\ \frac{\textup{m}}{\textup{ s}^2}$ $\displaystyle .0016\ \frac{\textup{m}}{\textup{ s}^2}$

Explanation:

Using Coulomb's law:

$F=k\frac{q_1q_2}{r^2}$ $\displaystyle F=k\frac{q_1q_2}{r^2}$

Using

F=ma $\displaystyle F=ma$

Combining equations:

$a=k\frac{q_1q_2}{mr^2}$ $\displaystyle a=k\frac{q_1q_2}{mr^2}$

Converting $\displaystyle cm$ to $\displaystyle m$ and $\displaystyle mg$ to $\displaystyle kg$ and plugging in values:

$a=9*10^9\frac{2*10^{-9}*4*10^{-9}}{2*10^{-3}.15^2}$ $\displaystyle a=9*10^9\frac{2*10^{-9}*4*10^{-9}}{2*10^{-3}.15^2}$

$a=.0016\frac{m}{s^2}$ $\displaystyle a=.0016\frac{m}{s^2}$

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Example Question #66 : Electrostatics

Charge A and B are $15\textup{ cm}$ $\displaystyle 15\textup{ cm}$ apart. If charge A has a charge of $2\textup{ nC}$ $\displaystyle 2\textup{ nC}$ and a mass of $2\textup{ mg}$ $\displaystyle 2\textup{ mg}$, charge B has a charge of $4\textup{ nC}$ $\displaystyle 4\textup{ nC}$ and a mass of $10\textup{ mg}$ $\displaystyle 10\textup{ mg}$, determine the acceleration of B due to A.

Possible Answers:

$1.1*10^{-4}\ \frac{\textup{m}}{\textup{ s}^2}$ $\displaystyle 1.1*10^{-4}\ \frac{\textup{m}}{\textup{ s}^2}$

$6.6*10^{-4}\ \frac{\textup{m}}{\textup{ s}^2}$ $\displaystyle 6.6*10^{-4}\ \frac{\textup{m}}{\textup{ s}^2}$

$3.2*10^{-4}\ \frac{\textup{m}}{\textup{ s}^2}$ $\displaystyle 3.2*10^{-4}\ \frac{\textup{m}}{\textup{ s}^2}$

None of these

$8.4*10^{-4}\ \frac{\textup{m}}{\textup{ s}^2}$ $\displaystyle 8.4*10^{-4}\ \frac{\textup{m}}{\textup{ s}^2}$

Correct answer:

$3.2*10^{-4}\ \frac{\textup{m}}{\textup{ s}^2}$ $\displaystyle 3.2*10^{-4}\ \frac{\textup{m}}{\textup{ s}^2}$

Explanation:

Using Coulomb's law:

$F=k\frac{q_1q_2}{r^2}$ $\displaystyle F=k\frac{q_1q_2}{r^2}$

Using

F=ma $\displaystyle F=ma$

Combining equations:

$a=k\frac{q_1q_2}{mr^2}$ $\displaystyle a=k\frac{q_1q_2}{mr^2}$

Converting $\displaystyle cm$ to $\displaystyle m$ and $\displaystyle mg$ to $\displaystyle kg$ and plugging in values:

$a=9*10^9\frac{2*10^{-9}*4*10^{-9}}{10*10^{-3}.15^2}$ $\displaystyle a=9*10^9\frac{2*10^{-9}*4*10^{-9}}{10*10^{-3}.15^2}$

$a=3.2*10^{-4}\frac{m}{s^2}$ $\displaystyle a=3.2*10^{-4}\frac{m}{s^2}$

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AP Physics 2 : AP Physics 2

Study concepts, example questions & explanations for AP Physics 2

All AP Physics 2 Resources

Example Questions

Example Question #2 : Electric Force Between Point Charges

Example Question #3 : Electric Force Between Point Charges

Example Question #11 : Electric Force Between Point Charges

Example Question #12 : Electric Force Between Point Charges

Example Question #761 : Ap Physics 2

Example Question #762 : Ap Physics 2

Example Question #63 : Electrostatics

Example Question #111 : Electricity And Magnetism

Example Question #13 : Electric Force Between Point Charges

Example Question #66 : Electrostatics

All AP Physics 2 Resources

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