Electric Force Between Point Charges - AP Physics 2

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Question

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?

Answer

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

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

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:

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

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Question

What is the force experienced by a $20\mu C$ point charge away from a $30\mu C$ point charge?

Answer

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

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

In this equation, is force in Newtons, is the respective charge value in $\mu C$ , is radius in meters, and is the Coulomb constant, which has a value of $9 \cdot 10^9 \frac{Nm^2}{C^2}$ .

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

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

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

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

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Question

Two objects in space are placed apart. One has a charge of , the second has a charge of . What is the electric force between them?

Answer

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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Question

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

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

Answer

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

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

We have the values for , , , and , so we just need to rearrange the equation to solve for , 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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Question

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?

Answer

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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Question

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?

Answer

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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Question

If charge has a value of $-910^{-13}C$ , and charge has a value of $810^{-13}C$ , and the distance between their centers, , is what will be the magnitude of the force on charge ?

Answer

Using coulombs law to solve

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

Where:

it the first charge, in coulombs.

is the second charge, in coulombs.

is the distance between them, in meters

is the constant of $910^9\frac{Nm^2}{C^2}$

Converting into

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

Plugging values into coulombs law

$F=910^{9}\frac{-910^{-13}810^{-13}}{.23^2}$

$F=-1.2210^{-13}N$

The magnitude will be the absolute value.

$|F|=1.2210^{-13}N$

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Question

If charge has a value of $-910^{-13}C$ , and charge has a value of $810^{-13}C$ , and the distance between their centers, , has a value of , what will be the force on charge ?

Answer

Using coulombs law to solve

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

Where:

it the first charge, in coulombs.

is the second charge, in coulombs.

is the distance between them, in meters

is the constant of $910^9\frac{Nm^2}{C^2}$

Converting into

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

Plugging values into coulombs law

$F=910^9\frac{(-910^{-13})(810^{-13})}{{.66}^2}$

$F=-1.4910^{-14}N$

Magnitude will be the absolute value

$|F|=1.4910^{-14}N$

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Question

Charge has a charge of

Charge has a charge of

The distance between their centers, is .

What is the magnitude of the force on due to ?

Answer

Using the electric field equation:

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

Where is $9.010^9 N\frac{m^2}{C^2}$

is charge , in

is charge , in

is the distance, in .

Convert to and plug in values:

$\overrightarrow{F}=9.010^9\frac{910^{-9}-210^{-9}}{.025^2}$

$\overrightarrow{F}=-5.3510^{-3}{N}$

Magnitude is equivalent to absolute value:

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

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Question

Charge has a charge of

Charge has a charge of

The distance between their centers, is .

What is the magnitude of the force on due to ?

Answer

Use Coulomb's law:

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

Where is $9.010^9 N\frac{m^2}{C^2}$

is charge , in Coulombs

is charge , in Coulombs

is the distance, in meters.

Convert to and plug in values:

$\overrightarrow{F}=9.010^9\frac{1910^{-9}-2810^{-9}}{.0045^2}$

$\overrightarrow{F}=-2.410^{-1}{N}$

Magnitude is equivalent to absolute value:

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

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Question

Charge has a charge of

Charge has a charge of

The distance between their centers, is .

What is the magnitude of the force on due to ?

Answer

Using Coulomb's law:

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

Where is $9.010^9 N\frac{m^2}{C^2}$

is charge , in Coulombs

is charge , in Coulombs

is the distance, in meters.

Convert to and plug in values:

$\overrightarrow{F}=9.010^9\frac{410^{-9}-710^{-9}}{.0065^2}$

$\overrightarrow{F}=-6.010^{-3}{N}$

Magnitude is equivalent to absolute value:

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

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Question

Charge has a charge of

Charge has a charge of

The distance between their centers, is .

What is the magnitude of the force on due to ?

Answer

Using Coulomb's law:

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

Where is $9.010^9 N\frac{m^2}{C^2}$

is charge , in Coulombs

is charge , in Coulombs

is the distance, in meters.

Convert to and plug in values:

$\overrightarrow{F}=9.010^9\frac{410^{-9}5*10^{-9}}{.0015^2}$

$\overrightarrow{F}=8.010^{-2}{N}$

Magnitude is equivalent to absolute value:

$|\overrightarrow{F}|=8.010^{-2}{N}$

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Question

Charge has a charge of

Charge has a charge of

The distance between their centers, is .

Determine the magnitude of the electric force on due to .

Answer

Use Coulomb's law:

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

Where is $9.010^9 N\frac{m^2}{C^2}$

is charge , in

is charge , in

is the distance, in .

Convert to and plug in values:

$\overrightarrow{F}=9.010^9\frac{810^{-9}1010^{-9}}{.001^2}$

$\overrightarrow{F}=7.210^{-1}{N}$

Magnitude is equivalent to absolute value:

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

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Question

Charge has a charge of

Charge has a charge of

The distance between their centers, is .

What is the magnitude of the force on due to ?

Answer

Use Coulomb's law:

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

Where is $9.010^9 N\frac{m^2}{C^2}$

is charge , in

is charge , in

is the distance, in .

Convert to and plug in values:

$\overrightarrow{F}=9.010^9\frac{2010^{-9}1010^{-9}}{.15^2}$

$\overrightarrow{F}=810^{-5}{N}$

Magnitude is equivalent to absolute value:

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

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Question

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

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

Answer

Use Coulombs Law:

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

Where $k=910^9N\frac{m^2}{C^2}$

Convert to and plug in values:

$F=910^9\frac{3.210^{-19}3.210^{-19}}{(110^{-9})^2}$

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

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Question

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

Answer

k

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Question

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.

Answer

Using Coulomb's law:

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

Using

Combining equations:

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

Converting to and to and plugging in values:

$a=910^9\frac{210^{-9}410^{-9}}{2*10^{-3}.15^2}$

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

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Question

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.

Answer

Using Coulomb's law:

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

Using

Combining equations:

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

Converting to and to and plugging in values:

$a=910^9\frac{210^{-9}410^{-9}}{1010^{-3}.15^2}$

$a=3.210^{-4}\frac{m}{s^2}$

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Question

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

Answer

Using

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

Plugging in values:

$F_{elec}=9.010^9\frac{({-1.610^{-19}})^2}{1^2}$

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

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Question

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

Answer

Using

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

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

Combining equations and plugging in values:

$.0509.8+9.010^9\frac{1010^{-9}50*10^{-9}}{r^2}=0$

Solving for

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