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AP Physics 2 : Electrostatics

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 #1 : Electric Force Between Point Charges

You have two point charges, q_1 and q_2 , 3m away from each other. The value of q_1 is $3\mu C$ , and the force they both experience is 0.024 N . What is the value of q_2 ?

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

Possible Answers:

$6\mu C$

$8\mu C$

$16\mu C$

$5\mu C$

$12\mu C$

Correct answer:

$8\mu C$

Explanation:

The equation for the force between two point charges is as follows:

$F=\frac{kq_1q_2}{r^2}$

We have the values for , q_1 , , and , so we just need to rearrange the equation to solve for q_2 , then plug in the values we have.

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

$q_2= \frac{0.024 \cdot 3^2}{(9\cdot 10^9)\cdot (3\cdot 10^{-6})}$

$q_2= 8\cdot 10^{-6}$

Therefore, the value for the second charge is $8\mu C$ .

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Example Question #1 : Electric Force Between Point Charges

Two objects in space are placed 35m apart. One has a charge of 90e , the second has a charge of -95e . What is the electric force between them?

Possible Answers:

$1.6\cdot 10^{-27}N$ towards each other

$3.75\cdot 10^{-27}N$ towards each other

$1.6\cdot 10^{-27}N$ away from each other

$1.6\cdot 10^{27}N$ towards each other

$3.75\cdot 10^{-27}N$ away from each other

Correct answer:

$1.6\cdot 10^{-27}N$ towards each other

Explanation:

Use Coloumb's law:

$E_{force}=\frac{kq_1q_2}{r^2}$

First, convert the charges into coulombs.

$1 e = 1.6\cdot 10^{-19} C$

$q_1=1.44\cdot 10^{-17}C$

$q_2=-1.52\cdot 10^{-17}$

Plug in values into Coulomb's law.

$E_{force}=9\cdot10^{9}\frac{N\cdot m^{2}}{C^{2}}\cdot \frac{(1.44\cdot10^{-17}C)(-1.52\cdot10^{-17}C)}{(35m)^{2}}$

$E_{force}=-1.6\cdot 10^{-27}N$

Negative electric forces indicate attractive forces. Alternatively, we can reason that since one charge is positive, and the other is negative, the charges will attract.

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Example Question #756 : Ap Physics 2

Coloumbs law for varsity

If charge has a value of $-9*10^{-13}C$ , and charge has a value of $8*10^{-13}C$ , and the distance between their centers, , is 23 cm what will be the magnitude of the force on charge ?

Possible Answers:

$|F|=1.84*10^{-13}N$

None of these

$|F|=1.22*10^{-13}N$

$|F|=4.22*10^{-13}N$

$|F|=2.33*10^{-13}N$

Correct answer:

$|F|=1.22*10^{-13}N$

Explanation:

Using coulombs law to solve

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

Where:

q_1 it the first charge, in coulombs.

q_2 is the second charge, in coulombs.

is the distance between them, in meters

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

Converting into

$23cm*\frac{1m}{100cm}=.23m$

Plugging values into coulombs law

$F=9*10^{9}\frac{-9*10^{-13}8*10^{-13}}{.23^2}$

$F=-1.22*10^{-13}N$

The magnitude will be the absolute value.

$|F|=1.22*10^{-13}N$

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Example Question #1 : Electric Force Between Point Charges

Coloumbs law for varsity

If charge has a value of $-9*10^{-13}C$ , and charge has a value of $8*10^{-13}C$ , and the distance between their centers, , has a value of 66 cm , what will be the force on charge ?

Possible Answers:

None of these

$|F|=1.49*10^{-14}N$

$|F|=2.22*10^{-14}N$

$|F|=3.75*10^{-14}N$

|F|=1.49N

Correct answer:

$|F|=1.49*10^{-14}N$

Explanation:

Using coulombs law to solve

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

Where:

q_1 it the first charge, in coulombs.

q_2 is the second charge, in coulombs.

is the distance between them, in meters

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

Converting into

$66cm*\frac{1m}{100cm}=.66m$

Plugging values into coulombs law

$F=9*10^9\frac{(-9*10^{-13})(8*10^{-13})}{{.66}^2}$

$F=-1.49*10^{-14}N$

Magnitude will be the absolute value

$|F|=1.49*10^{-14}N$

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Example Question #757 : Ap Physics 2

Coloumbs law for varsity

Charge has a charge of 9 nC

Charge has a charge of -2nC

The distance between their centers, is 5.5 mm .

What is the magnitude of the force on due to ?

Possible Answers:

None of these

$|\overrightarrow{F}|=1.89*10^{-3}N$

$|\overrightarrow{F}|=5.35*10^{-3}N$

$|\overrightarrow{F}|=2.59*10^{-3}N$

$|\overrightarrow{F}|=9.87*10^{-3}N$

Correct answer:

$|\overrightarrow{F}|=5.35*10^{-3}N$

Explanation:

Using the electric field equation:

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

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:

$\overrightarrow{F}=9.0*10^9\frac{9*10^{-9}*-2*10^{-9}}{.025^2}$

$\overrightarrow{F}=-5.35*10^{-3}{N}$

Magnitude is equivalent to absolute value:

$|\overrightarrow{F}|=5.35*10^{-3}N$

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Example Question #3 : Electric Force Between Point Charges

Coloumbs law for varsity

Charge has a charge of 10 nC

Charge has a charge of 20nC

The distance between their centers, is 15 cm .

What is the magnitude of the force on due to ?

Possible Answers:

$|\overrightarrow{F}|=4*10^{-5}{N}$

$|\overrightarrow{F}|=6.9*10^{-5}{N}$

None of these

$|\overrightarrow{F}|=8*10^{-5}{N}$

$|\overrightarrow{F}|=3*10^{-5}{N}$

Correct answer:

$|\overrightarrow{F}|=8*10^{-5}{N}$

Explanation:

Use 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:

$\overrightarrow{F}=9.0*10^9\frac{20*10^{-9}*10*10^{-9}}{.15^2}$

$\overrightarrow{F}=8*10^{-5}{N}$

Magnitude is equivalent to absolute value:

$|\overrightarrow{F}|=8*10^{-5}{N}$

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Example Question #2 : Electric Force Between Point Charges

Coloumbs law for varsity

Charge has a charge of 8 nC

Charge has a charge of 10nC

The distance between their centers, is 1.0 mm .

Determine the magnitude of the electric force on due to .

Possible Answers:

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

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

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

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

None of these

Correct answer:

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

Explanation:

Use 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:

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

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

Magnitude is equivalent to absolute value:

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

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Example Question #3 : Electric Force Between Point Charges

Coloumbs law for varsity

Charge has a charge of 19 nC

Charge has a charge of -28nC

The distance between their centers, is 4.5 mm .

What is the magnitude of the force on due to ?

Possible Answers:

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

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

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

None of these

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

Correct answer:

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

Explanation:

Use 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:

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

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

Magnitude is equivalent to absolute value:

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

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Example Question #111 : Electricity And Magnetism

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:

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

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

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

None of these

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

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 #12 : 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:

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

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

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

None of these

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

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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AP Physics 2 : Electrostatics

Study concepts, example questions & explanations for AP Physics 2

All AP Physics 2 Resources

Example Questions

Example Question #1 : Electric Force Between Point Charges

Example Question #1 : Electric Force Between Point Charges

Example Question #756 : Ap Physics 2

Example Question #1 : Electric Force Between Point Charges

Example Question #757 : Ap Physics 2

Example Question #3 : Electric Force Between Point Charges

Example Question #2 : Electric Force Between Point Charges

Example Question #3 : Electric Force Between Point Charges

Example Question #111 : Electricity And Magnetism

Example Question #12 : Electric Force Between Point Charges

All AP Physics 2 Resources

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