AP Physics 2 : Electric Force Between Point Charges

Study concepts, example questions & explanations for AP Physics 2

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Example Questions

Example Question #1 : Electric Force Between Point Charges

Two charges are placed a certain distance apart such that the force that each charge experiences is 20 N. If the distance between the charges is doubled, what is the new force that each charge experiences?

Possible Answers:

There is no way to determine the new force

\(\displaystyle 10N\)

\(\displaystyle 80N\)

\(\displaystyle 5N\)

\(\displaystyle 40N\)

Correct answer:

\(\displaystyle 5N\)

Explanation:

To solve this problem, we'll need to utilize the equation for the electric force:

\(\displaystyle F_{e}=k\frac{Q_{1}Q_{2}}{r^{2}}\)

We're told that the force each charge experiences is 20 N at a certain distance, but then that distance is doubled. Thus, the new electric force will be:

\(\displaystyle F_{e}^{'}=k\frac{Q_{1}Q_{2}}{(2r)^{2}}=k\frac{Q_{1}Q_{2}}{4r^{2}}=\frac{F_{e}}{4}=\frac{20N}{4}=5N\)

Example Question #1 : Electric Force Between Point Charges

What is the force experienced by a \(\displaystyle 20\mu C\) point charge \(\displaystyle 5m\) away from a \(\displaystyle 30\mu C\) point charge?

Possible Answers:

\(\displaystyle 0.216N\)

\(\displaystyle 0.108N\)

\(\displaystyle 1.08N\)

The point charge experiences no force

\(\displaystyle 5.4N\)

Correct answer:

\(\displaystyle 0.216N\)

Explanation:

The equation to find the force from two point charges is called Coulomb's Law.

\(\displaystyle F=\frac{kq_1q_2}{r^2}\)

In this equation, \(\displaystyle F\) is force in Newtons, \(\displaystyle q\) is the respective charge value in \(\displaystyle \mu C\), \(\displaystyle r\) is radius in meters, and \(\displaystyle k\) is the Coulomb constant, which has a value of \(\displaystyle 9 \cdot 10^9 \frac{N*m^2}{C^2}\).

Now, we just plug in the numbers. Note: the charge values are in microcoulombs (the Greek letter \(\displaystyle \mu\), called "mu," stands for micro), which is equal to \(\displaystyle 10^{-6}\).

\(\displaystyle F&=\frac{kq_1q_2}{r^2}\)

\(\displaystyle F=\frac{(9\cdot 10^9\frac{N*m^2}{C^2}) \cdot (20\cdot 10^{-6}\mu C) \cdot (30 \cdot 10^{-6}\mu C)}{(5m)^2}\)

\(\displaystyle F= 0.216\)

Therefore, the force experienced on either charge is 0.216N of force.

Example Question #103 : Electricity And Magnetism

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

 \(\displaystyle k=9\cdot 10^9\frac{N\cdot m^2}{C^2}\)

Possible Answers:

\(\displaystyle 5\mu C\)

\(\displaystyle 16\mu C\)

\(\displaystyle 6\mu C\)

\(\displaystyle 12\mu C\)

\(\displaystyle 8\mu C\)

Correct answer:

\(\displaystyle 8\mu C\)

Explanation:

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

\(\displaystyle F=\frac{kq_1q_2}{r^2}\)

We have the values for \(\displaystyle F\), \(\displaystyle q_1\), \(\displaystyle r\), and \(\displaystyle k\), so we just need to rearrange the equation to solve for \(\displaystyle q_2\), then plug in the values we have.

\(\displaystyle q_2=\frac{F\cdot r^2}{k\cdot q_1}\)

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

\(\displaystyle q_2= 8\cdot 10^{-6}\)

Therefore, the value for the second charge is \(\displaystyle 8\mu C\).

Example Question #4 : Electric Force Between Point Charges

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

Possible Answers:

\(\displaystyle 3.75\cdot 10^{-27}N\) away from each other

\(\displaystyle 1.6\cdot 10^{-27}N\) towards each other

\(\displaystyle 1.6\cdot 10^{27}N\) towards each other

\(\displaystyle 3.75\cdot 10^{-27}N\) towards each other

\(\displaystyle 1.6\cdot 10^{-27}N\) away from each other

Correct answer:

\(\displaystyle 1.6\cdot 10^{-27}N\) towards each other

Explanation:

Use Coloumb's law:

\(\displaystyle E_{force}=\frac{kq_1q_2}{r^2}\) 

First, convert the charges into coulombs.

\(\displaystyle 1 e = 1.6\cdot 10^{-19} C\)

\(\displaystyle q_1=1.44\cdot 10^{-17}C\)

\(\displaystyle q_2=-1.52\cdot 10^{-17}\)

Plug in values into Coulomb's law.

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

\(\displaystyle E_{force}=-1.6\cdot 10^{-27}N\)

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.

Example Question #5 : Electric Force Between Point Charges

Coloumbs law for varsity

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

Possible Answers:

None of these

\(\displaystyle |F|=1.22*10^{-13}N\)

\(\displaystyle |F|=2.33*10^{-13}N\)

\(\displaystyle |F|=1.84*10^{-13}N\)

\(\displaystyle |F|=4.22*10^{-13}N\)

Correct answer:

\(\displaystyle |F|=1.22*10^{-13}N\)

Explanation:

Using coulombs law to solve

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

 Where:

\(\displaystyle q_1\) it the first charge, in coulombs.

\(\displaystyle q_2\) is the second charge, in coulombs.

\(\displaystyle r\) is the distance between them, in meters

\(\displaystyle k\) is the constant of \(\displaystyle 9*10^9\frac{N*m^2}{C^2}\)

Converting \(\displaystyle cm\) into \(\displaystyle m\)

\(\displaystyle 23cm*\frac{1m}{100cm}=.23m\)

Plugging values into coulombs law

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

\(\displaystyle F=-1.22*10^{-13}N\)

The magnitude will be the absolute value.

\(\displaystyle |F|=1.22*10^{-13}N\)

Example Question #6 : Electric Force Between Point Charges

Coloumbs law for varsity

If charge \(\displaystyle A\) has a value of \(\displaystyle -9*10^{-13}C\), and charge \(\displaystyle B\) has a value of \(\displaystyle 8*10^{-13}C\), and the distance between their centers, \(\displaystyle r\), has a value of \(\displaystyle 66 cm\), what will be the force on charge \(\displaystyle B\)?

Possible Answers:

\(\displaystyle |F|=1.49*10^{-14}N\)

None of these

\(\displaystyle |F|=1.49N\)

\(\displaystyle |F|=2.22*10^{-14}N\)

\(\displaystyle |F|=3.75*10^{-14}N\)

Correct answer:

\(\displaystyle |F|=1.49*10^{-14}N\)

Explanation:

Using coulombs law to solve

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

 Where:

\(\displaystyle q_1\) it the first charge, in coulombs.

\(\displaystyle q_2\) is the second charge, in coulombs.

\(\displaystyle r\) is the distance between them, in meters

\(\displaystyle k\) is the constant of \(\displaystyle 9*10^9\frac{N*m^2}{C^2}\)

Converting \(\displaystyle cm\) into \(\displaystyle m\)

\(\displaystyle 66cm*\frac{1m}{100cm}=.66m\)

Plugging values into coulombs law

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

\(\displaystyle F=-1.49*10^{-14}N\)

Magnitude will be the absolute value

 \(\displaystyle |F|=1.49*10^{-14}N\)

Example Question #7 : Electric Force Between Point Charges

Coloumbs law for varsity

Charge \(\displaystyle A\) has a charge of \(\displaystyle 9 nC\)

Charge \(\displaystyle B\) has a charge of \(\displaystyle -2nC\)

The distance between their centers, \(\displaystyle r\) is \(\displaystyle 5.5 mm\).

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

Possible Answers:

\(\displaystyle |\overrightarrow{F}|=5.35*10^{-3}N\)

\(\displaystyle |\overrightarrow{F}|=9.87*10^{-3}N\)

\(\displaystyle |\overrightarrow{F}|=2.59*10^{-3}N\)

\(\displaystyle |\overrightarrow{F}|=1.89*10^{-3}N\)

None of these

Correct answer:

\(\displaystyle |\overrightarrow{F}|=5.35*10^{-3}N\)

Explanation:

Using the electric field equation:

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

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

\(\displaystyle q_A\) is charge \(\displaystyle A\), in \(\displaystyle coulombs\)

\(\displaystyle q_B\) is charge \(\displaystyle B\), in \(\displaystyle coulombs\)

\(\displaystyle r\) is the distance, in \(\displaystyle meters\).

Convert \(\displaystyle mm\) to \(\displaystyle m\) and plug in values:

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

\(\displaystyle \overrightarrow{F}=-5.35*10^{-3}{N}\)

Magnitude is equivalent to absolute value:

\(\displaystyle |\overrightarrow{F}|=5.35*10^{-3}N\)

Example Question #8 : Electric Force Between Point Charges

Coloumbs law for varsity

Charge \(\displaystyle A\) has a charge of \(\displaystyle 10 nC\)

Charge \(\displaystyle B\) has a charge of \(\displaystyle 20nC\)

The distance between their centers, \(\displaystyle r\) is \(\displaystyle 15 cm\).

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

Possible Answers:

\(\displaystyle |\overrightarrow{F}|=6.9*10^{-5}{N}\)

\(\displaystyle |\overrightarrow{F}|=4*10^{-5}{N}\)

\(\displaystyle |\overrightarrow{F}|=8*10^{-5}{N}\)

\(\displaystyle |\overrightarrow{F}|=3*10^{-5}{N}\)

None of these

Correct answer:

\(\displaystyle |\overrightarrow{F}|=8*10^{-5}{N}\)

Explanation:

Use Coulomb's law:

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

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

\(\displaystyle q_A\) is charge \(\displaystyle A\), in \(\displaystyle coulombs\)

\(\displaystyle q_B\) is charge \(\displaystyle B\), in \(\displaystyle coulombs\)

\(\displaystyle r\) is the distance, in \(\displaystyle meters\).

Convert \(\displaystyle cm\) to \(\displaystyle m\) and plug in values:

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

\(\displaystyle \overrightarrow{F}=8*10^{-5}{N}\)

Magnitude is equivalent to absolute value:

\(\displaystyle |\overrightarrow{F}|=8*10^{-5}{N}\)

Example Question #9 : Electric Force Between Point Charges

Coloumbs law for varsity

Charge \(\displaystyle A\) has a charge of \(\displaystyle 8 nC\)

Charge \(\displaystyle B\) has a charge of \(\displaystyle 10nC\)

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

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

Possible Answers:

None of these

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

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

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

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

Correct answer:

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

Explanation:

Use Coulomb's law:

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

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

\(\displaystyle q_A\) is charge \(\displaystyle A\), in \(\displaystyle coulombs\)

\(\displaystyle q_B\) is charge \(\displaystyle B\), in \(\displaystyle coulombs\)

\(\displaystyle r\) is the distance, in \(\displaystyle meters\).

Convert \(\displaystyle mm\) to \(\displaystyle m\) and plug in values:

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

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

Magnitude is equivalent to absolute value:

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

Example Question #10 : Electric Force Between Point Charges

Coloumbs law for varsity

Charge \(\displaystyle A\) has a charge of \(\displaystyle 19 nC\)

Charge \(\displaystyle B\) has a charge of \(\displaystyle -28nC\)

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

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

Possible Answers:

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

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

None of these

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

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

Correct answer:

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

Explanation:

Use Coulomb's law:

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

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

\(\displaystyle q_A\) is charge \(\displaystyle A\), in Coulombs

\(\displaystyle q_B\) is charge \(\displaystyle B\), in Coulombs

\(\displaystyle r\) is the distance, in meters.

Convert \(\displaystyle mm\) to \(\displaystyle m\) and plug in values:

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

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

Magnitude is equivalent to absolute value:

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

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