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AP Physics 2 : Electric Force Between Point Charges

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Example Question #11 : Electric Force Between Point Charges

Coloumbs law for varsity

Charge has a charge of 4 nC

Charge has a charge of -7nC

The distance between their centers, is 6.5 mm .

What is the magnitude of the force on due to ?

Possible Answers:

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

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

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

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

None of these

Correct answer:

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

Explanation:

Using Coulomb's law:

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

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

q_A is charge , in Coulombs

q_B is charge , in Coulombs

is the distance, in meters.

Convert to and plug in values:

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

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

Magnitude is equivalent to absolute value:

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

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Example Question #11 : Electric Force Between Point Charges

Coloumbs law for varsity

Charge has a charge of 4 nC

Charge has a charge of 5nC

The distance between their centers, is 1.5 mm .

What is the magnitude of the force on due to ?

Possible Answers:

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

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

None of these

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

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

Correct answer:

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

Explanation:

Using Coulomb's law:

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

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

q_A is charge , in Coulombs

q_B is charge , in Coulombs

is the distance, in meters.

Convert to and plug in values:

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

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

Magnitude is equivalent to absolute value:

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

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Example Question #11 : Electric Force Between Point Charges

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 two times larger

The force on the Q charge is four 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 #761 : 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 forces are equal

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

The force on the Q charge is twice as large

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

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 #15 : Electric Force Between Point Charges

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

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

Possible Answers:

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

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

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

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

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

Correct answer:

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

Explanation:

Use Coulombs Law:

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

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

Convert to and plug in values:

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

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

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Example Question #63 : Electrostatics

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

Possible Answers:

The attractive forces will increase

The force will change from repulsive to attractive

The repulsive force will decrease

The repulsive forces will increase

It is impossible to determine

Correct answer:

The repulsive forces will increase

Explanation:

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Example Question #61 : Electrostatics

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

Possible Answers:

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

None of these

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

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

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

Correct answer:

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

Explanation:

Using Coulomb's law:

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

Using

F=ma

Combining equations:

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

Converting to and to and plugging in values:

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

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

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Example Question #11 : Electric Force Between Point Charges

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

Possible Answers:

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

None of these

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

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

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

Correct answer:

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

Explanation:

Using Coulomb's law:

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

Using

F=ma

Combining equations:

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

Converting to and to and plugging in values:

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

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

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Example Question #11 : Electric Force Between Point Charges

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

Possible Answers:

2.3N

$7.5*10^{-18}N$

7.5N

$2.3*10^{-28}N$

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 #20 : Electric Force Between Point Charges

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:

.007m

None of these

.003m

.075m

.038m

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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AP Physics 2 : Electric Force Between Point Charges

Study concepts, example questions & explanations for AP Physics 2

All AP Physics 2 Resources

Example Questions

Example Question #11 : Electric Force Between Point Charges

Example Question #11 : Electric Force Between Point Charges

Example Question #11 : Electric Force Between Point Charges

Example Question #761 : Ap Physics 2

Example Question #15 : Electric Force Between Point Charges

Example Question #63 : Electrostatics

Example Question #61 : Electrostatics

Example Question #11 : Electric Force Between Point Charges

Example Question #11 : Electric Force Between Point Charges

Example Question #20 : Electric Force Between Point Charges

All AP Physics 2 Resources

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