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AP Physics 2 : Electrostatics

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Example Questions

Example Question #61 : Electrostatics

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 forces are equal

The force on the 3Q charge is two times larger

The force on the Q charge is four times larger

The force on the Q charge is two times larger

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 #62 : Electrostatics

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 3Q charge is one-fourth as large

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

The force on the Q charge is twice 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 #63 : 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 A due to B.

Possible Answers:

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

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

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

None of these

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

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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Example Question #64 : Electrostatics

Two electrons are deep in space and $\displaystyle 1m$ apart. Determine the force of one electron on the other.

Possible Answers:

7.5N $\displaystyle 7.5N$

$7.5*10^{-18}N$ $\displaystyle 7.5*10^{-18}N$

$2.3*10^{-28}N$ $\displaystyle 2.3*10^{-28}N$

None of these

2.3N $\displaystyle 2.3N$

Correct answer:

$2.3*10^{-28}N$ $\displaystyle 2.3*10^{-28}N$

Explanation:

Using

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

Plugging in values:

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

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

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Example Question #122 : Electricity And Magnetism

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

Possible Answers:

.038m $\displaystyle .038m$

.007m $\displaystyle .007m$

.003m $\displaystyle .003m$

.075m $\displaystyle .075m$

None of these

Correct answer:

.003m $\displaystyle .003m$

Explanation:

Using

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

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

Combining equations and plugging in values:

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

Solving for $\displaystyle r$

r=.003m $\displaystyle r=.003m$

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Example Question #122 : Electricity And Magnetism

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

Possible Answers:

$0.02 \textup{ N}$ $\displaystyle 0.02 \textup{ N}$; attractive

$0.02 \textup{ N}$ $\displaystyle 0.02 \textup{ N}$; repulsive

None of these answers are correct.

$0.06 \textup{ N}$ $\displaystyle 0.06 \textup{ N}$; repulsive

$0.06 \textup{ N}$ $\displaystyle 0.06 \textup{ N}$; attractive

Correct answer:

$0.02 \textup{ N}$ $\displaystyle 0.02 \textup{ N}$; attractive

Explanation:

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

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

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.

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

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

Possible Answers:

None of these

$\displaystyle (0,2.65)mm$

$\displaystyle (2.35,0)mm$

$\displaystyle (2.65,0)mm$

$\displaystyle (4.25,0)mm$

Correct answer:

$\displaystyle (2.65,0)mm$

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.

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

Where

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

q_1 $\displaystyle q_1$ is the value of the first charge

q_2 $\displaystyle q_2$ is the value of the second charge

q_e $\displaystyle q_e$ is the charge of the electron

r_1 $\displaystyle r_1$ is the distance from the first charge to the electron

r_2 $\displaystyle r_2$ is the distance from the second charge to the electron

q_1 $\displaystyle q_1$ is 5mm $\displaystyle 5mm$ away from q_2 $\displaystyle q_2$, so:

$\displaystyle r_1+r_2=5mm$

Combining equations and plugging in values:

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

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

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

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

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

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

$\displaystyle r_1=1.13r_2$

$\displaystyle 1.13r_2+r_2=5mm$

$\displaystyle r_2=2.35mm$

$\displaystyle r_1=2.65mm$

Thus, the location of balanced forces is at:

$\displaystyle (2.65,0)mm$

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AP Physics 2 : Electrostatics

Study concepts, example questions & explanations for AP Physics 2

All AP Physics 2 Resources

Example Questions

Example Question #61 : Electrostatics

Example Question #62 : Electrostatics

Example Question #63 : Electrostatics

Example Question #111 : Electricity And Magnetism

Example Question #63 : Electrostatics

Example Question #66 : Electrostatics

Example Question #64 : Electrostatics

Example Question #122 : Electricity And Magnetism

Example Question #122 : Electricity And Magnetism

Example Question #21 : Electric Force Between Point Charges

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