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AP Physics 2 : AP Physics 2

Study concepts, example questions & explanations for AP Physics 2

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All AP Physics 2 Resources

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

Example Questions

Example Question #61 : Electrostatics

Two electrons are deep in space and apart. Determine the force of one electron on the other.

Possible Answers:

7.5N

$7.5*10^{-18}N$

2.3N

$2.3*10^{-28}N$

None of these

Correct answer:

$2.3*10^{-28}N$

Explanation:

Using

$F_{elec}=k\frac{q_1q_2}{r^2}$

Plugging in values:

$F_{elec}=9.0*10^9\frac{({-1.6*10^{-19}})^2}{1^2}$

$F_{elec}=2.3*10^{-28}N$

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Example Question #122 : Electricity And Magnetism

A mobile 50g , 10nC charge is perfectly balanced on top of a stationary 50nC charge. What will be the equilibrium height of the mobile charge?

Possible Answers:

.038m

.007m

.003m

.075m

None of these

Correct answer:

.003m

Explanation:

Using

$F_{elec}=k\frac{q_1q_2}{r^2}$

$F_{net}=mg+F_{elec}$

Combining equations and plugging in values:

$.050*9.8+9.0*10^9\frac{10*10^{-9}*50*10^{-9}}{r^2}=0$

Solving for

r=.003m

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Example Question #21 : Electric Force Between Point Charges

There are two point charges suspended in space. Charge A has a charge of $-5\ \mu \textup{C}$ . Charge B has a charge of $4\ \mu \textup{C}$ . If they are 3 meters apart, what is the magnitude of the force between them? Is the force attractive or repulsive?

Possible Answers:

$0.02 \textup{ N}$ ; repulsive

None of these answers are correct.

$0.02 \textup{ N}$ ; attractive

$0.06 \textup{ N}$ ; repulsive

$0.06 \textup{ N}$ ; attractive

Correct answer:

$0.02 \textup{ N}$ ; attractive

Explanation:

To determine the force between two point charges, you use Coulomb's Law.

$F=\frac{k*q_1*q_2}{r^2}$

We have the values of q1, q2, and r. We already know the value of k: 9 x 10⁹. To find the force, we can plug in each of the values.

$\begin{align*} F&=\frac{(9E9)*(-5E{-6})*(4E{-6})}{3^2} \\ &= -0.02 \end{align*}$

Therefore, the magnitude of the force is 0.02. Now, to determine the direction of the force, we examine the sign. If the sign is negative, then the force is attractive. As our force value is negative, it therefore is an attractive force.

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Example Question #21 : Electric Force Between Point Charges

A -9nC charge is located at the origin. A -7nC charge is located at (5,0)mm . Where could an electron be placed where it would experience no net force?

Possible Answers:

None of these

(0,2.65)mm

(2.35,0)mm

(2.65,0)mm

(4.25,0)mm

Correct answer:

(2.65,0)mm

Explanation:

Since both charges are located on the x-axis, this is a one dimensional problem.

The electron, which has a negative charge, will be repelled by both charges.

Placing the electron somewhere "in the middle" will allow the forces to balance out.

$F_{net}=0=k\frac{q_1q_e}{r_1^2}-k\frac{q_2q_e}{r_2^2}$

Where

is the constant $9.0*10^{9}\frac{N\cdot m^2}{C^2}$

q_1 is the value of the first charge

q_2 is the value of the second charge

q_e is the charge of the electron

r_1 is the distance from the first charge to the electron

r_2 is the distance from the second charge to the electron

q_1 is 5mm away from q_2 , so:

r_1+r_2=5mm

Combining equations and plugging in values:

$F_{net}=0=9*10^9\frac{-9*10^{-9}*-1.6*10^{-19}}{r_1^2}-9*10^9\frac{-7*10^{-9}*-1.6*10^{-19}}{(.005-r_1)^2}$

$F_{net}=0=9*10^9\frac{-9*10^{-9}*-1.6*10^{-19}}{r_1^2}-9*10^9\frac{-7*10^{-9}*-1.6*10^{-19}}{r_2^2}$

$\frac{1.296*10^{-17}}{r_1^2}=\frac{1.008*10^{-17}}{r_2^2}$

$\frac{r_1^2}{1.296*10^{-17}}=\frac{r_2^2}{1.008*10^{-17}}$

$\frac{r_1^2}{1.296}=\frac{r_2^2}{1.008}$

$r_1=\sqrt{\frac{1.296r_2^2}{1.008}}$

r_1=1.13r_2

1.13r_2+r_2=5mm

r_2=2.35mm

r_1=2.65mm

Thus, the location of balanced forces is at:

(2.65,0)mm

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Example Question #71 : Electrostatics

A 30nC charge is located at the origin. A 50nC charge is located at (10,0)mm . Where could a proton be placed where it would experience no net force?

$k=9.0*10^{9}\frac{N\cdot m^2}{C^2}$

Possible Answers:

(8.74,0)mm

None of these

(2.19,0)mm

(4.37,0)mm

(0,4.37)mm

Correct answer:

(4.37,0)mm

Explanation:

Since both point charges are located on the x-axis, this is a one dimensional problem.

The proton, which has a positive charge, will be repelled by both charges.

Placing the proton somewhere "in the middle" will allow the forces to balance out.

$F_{net}=0=k\frac{q_1q_e}{r_1^2}-k\frac{q_2q_e}{r_2^2}$

Where

$k=9.0*10^{9}\frac{N\cdot m^2}{C^2}$

q_1 is the value of the first charge

q_2 is the value of the second charge

q_e is the charge of the proton

r_1 is the distance from the first charge to the proton

r_2 is the distance from the second charge to the proton

q_1 is 5mm away from q_2 , so:

r_1+r_2=10mm

Combining equations and plugging in values:

$F_{net}=0=9*10^9\frac{30*10^{-9}*1.6*10^{-19}}{r_1^2}-9*10^9\frac{50*10^{-9}*1.6*10^{-19}}{r_2^2}$

$\frac{4.32*10^{-17}}{r_1^2}=\frac{7.2*10^{-17}}{r_2^2}$

$\frac{r_1^2}{4.32*10^{-17}}=\frac{r_2^2}{7.2*10^{-17}}$

$\frac{r_1^2}{4.32}=\frac{r_2^2}{7.2}$

$r_1=\sqrt{\frac{4.32r_2^2}{7.2}}$

r_1=.775r_2

.775r_2+r_2=10mm

r_2=5.63mm

r_1=4.37mm

Thus, the location of balanced forces is at:

(4.37,0)mm

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Example Question #72 : Electrostatics

An electron is 500mm away from a point charge. It experiences a force of $2.6*10^{-16}N$ towards the point charge. Determine the value of the point charge.

Possible Answers:

$4.5*10^{-8}C$

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

$9*10^{-8}C$

$4.5*10^{-1}C$

None of these

Correct answer:

$4.5*10^{-8}C$

Explanation:

Using

$F=k\frac{q_1q_e}{r^2}$

Solving for q_1

$F\frac{r^2}{k*q_e}=q_1$

Converting to and plugging in values

$-2.6*10^{-16}\frac{.500^2}{(9*10^{9})*-1.6*10^{-19}}=q_1$

*Note: a negative sign is used for the force because it is an attractive force, if it was a repulsive force, the opposite sign would be used.

$q=4.5*10^{-8}C$

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Example Question #21 : Electric Force Between Point Charges

Which of the following will cut the magnitude of the electric force between two charged particles in half?

Possible Answers:

Reducing the charge of only one of the particles by half

Placing the two particles twice as far apart

Doubling the charge of each particle

Double the charge of only one of the particles

Reducing the charge of each particle by half

Correct answer:

Reducing the charge of only one of the particles by half

Explanation:

To answer this question, it's necessary to understand the factors that affect the electric force. To show this, we can write the electrical force expression.

$F_{e}=k\frac{q_{1}q_{2}}{r^{2}}$

Using the above expression, we can look at how each answer choice would change the electric force.

If we place the two particles twice as far apart, the magnitude of the electric force will be reduced by a factor of .

Doubling the charge of one particle would double the electric force. Doubling the charge on both particles would cause the electric force to become times as great.

Halving the charge of both particles would cause the electric force to decrease by a factor of .

If the charge on only one of the particles is cut in half, then the electric force would be cut in half as well. Thus, this is the correct answer.

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Example Question #22 : Electric Force Between Point Charges

Two charges are a fixed distance apart. Both charges have charge . If another charge of charge and mass is placed a distance $\frac{d}{4}$ from one of the charges and $\frac{3d}{4}$ from the other, what will be the magnitude of its acceleration the moment it's released?

Possible Answers:

$\frac{16kqQ}{md^2}$

$\frac{8kqQ}{3md}$

$\frac{8kqQ}{9md}$

$\frac{128kqQ}{9md^2}$

$\frac{16kqQ}{9md^2}$

Correct answer:

$\frac{128kqQ}{9md^2}$

Explanation:

All we have to do is find the sum of the forces on the charge and divide by its mass. To find the force from each charge, we can use Coulomb's law:

$F=\frac{kq_{1}q_{2}}{d^2}$

Let's let the force from the charge a distance $\frac{d}{4}$ away be positive. That force is $\frac{kqQ}{(\frac{d}{4})^2}=\frac{16kqQ}{d^2}$ . The other force will be negative because it's acting in the opposite direction. This force is $\frac{-kqQ}{(\frac{3d}{4})^2}=\frac{-16kqQ}{9d^2}$ . Adding these two together we get

$\frac{128kqQ}{9d^2}$

This is the magnitude of the net force. To find acceleration, we divide by the mass to get

$\frac{128kqQ}{9md^2}$

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Example Question #1 : Electric Potential Energy

$6.1 \cdot 10^{20}$ electrons pass through a $20\Omega$ resistor in 11 minutes. What is the potential difference across that resistor?

$e=1.602\cdot 10^{-19}$

Possible Answers:

6.12V

8.11V

4.45V

1.34V

2.96V

Correct answer:

2.96V

Explanation:

We need to find the voltage, and we have the resistance, so if we can find the current, then we can use Ohm's Law to find the voltage.

The definition of current is amount of charge that flows through a point in time, so current can be calculated using this equation:

$I=\frac{\Delta Q}{\Delta t}$

We're told how many electrons have passed through a resistor, and we know how much charge a single electron has. If we convert the number of electrons into total amount of charge, we can divide that number by the amount of time in seconds to find the current.

$6.1\cdot 10^{20} e^-\cdot 1.602\cdot 10^{-19}\frac{C}{electron}=97.722 C$

$11min\cdot60\frac{s}{min}=660s$

Now we have the amount of charge in and the amount of time in seconds, so we divide the two.

$I=\frac{\Delta Q}{\Delta t}$

$I=\frac{97.722}{660}$

I= 0.1481

Now that we have the current, we can use Ohm's Law to find the voltage.

V=IR

V=(0.1481)(20)

V= 2.961

Therefore, the potential difference is 2.96V.

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Example Question #1 : Electric Potential Energy

What is the electric potential between two terminals of a cell if it requires $\textup {7J}$ of work to transfer $\textup {3C}$ between the terminals?

Possible Answers:

$\frac{7}{3}\textup{A}$

$\frac{3}{7}\textup{V}$

$\frac{3}{49}\textup{A}$

$\frac{7}{3}\textup{V}$

$\frac{3}{7}\textup{eV}$

Correct answer:

$\frac{7}{3}\textup{V}$

Explanation:

Write the formula for the potential difference.

$V=\frac{W}{Q}$

Substitute the work and charge. The unit is in volts.

$V=\frac{W}{Q}= \frac{7\textup{J}}{3\textup{C}}=\frac{7}{3}\textup{V}$

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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 #61 : Electrostatics

Example Question #122 : Electricity And Magnetism

Example Question #21 : Electric Force Between Point Charges

Example Question #21 : Electric Force Between Point Charges

Example Question #71 : Electrostatics

Example Question #72 : Electrostatics

Example Question #21 : Electric Force Between Point Charges

Example Question #22 : Electric Force Between Point Charges

Example Question #1 : Electric Potential Energy

Example Question #1 : Electric Potential Energy

All AP Physics 2 Resources

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