Electric Force Between Point Charges - AP Physics 1

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Question

Two point charges, each having a charge of +1C, are 2 meters apart. If the distance between them is doubled, by what factor does the force between them change?

$k=9.0\times10^9\frac{N\cdot m^2}{C^2}$

Answer

This is a question where knowing how to effectively sift through a problem statement and choose only the information you need will really help. We are given a bunch of values, but only need to know one thing, which is that the distance between the two charges is doubled.

Coulomb's law is as follows:

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

We can rewrite this for the initial and final scenarios:

$F_{ei}= k\frac{q_1q_2}{r_i^2}$

$F_{ef}= k\frac{q_1q_2}{r_f^2}$

We can divide one equation by the other to set up a ratio:

$\frac{F_{ei}}{F_{ef}} = \frac{r_f^2}{r_i^2}$

We know that the final radius is double the intial, which is written as:

Substituting this in we get:

$\frac{F_{ei}}{F_{ef}} = \frac{(2r_i)^2}{r_i^2} = 4$

Rerranging for the final force, we get:

$F_{ef} = \frac{1}{4}F_{ei}$

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Question

If $q_{1}=+3.1\mu C$ , $q_{2}=-1.6\mu C$ , and $q_{3}=-2.4\mu C$ , then what is the magnitude of the net force on charge 2?

$k=8.99\cdot10^{9}\frac{N\cdot m^2}{C^2}$

Answer

First lets set up two axes. Have be to the right of charge 3 and 2 in the diagram and be above charges 1 and 2 in the diagram with charge 2 at the origin.

Coloumb's law tells us the force between point charges is

$F_{21}=k\frac{q_{1}q_{2}}{r^{2}}$

The net force on charge 2 can be determined by combining the force on charge 2 due to charge 1 and the force on charge 2 due to charge 3.

Since charge 1 and charge 2 are of opposite polarities, they have an attractive force; therefore, charge 2 experiences a force towards charge 1 (in the direction). By using Coloumb's law, we can determine this force to be

$F_{21}=k\frac{q_{1}q_{2}}{{r_{12}}^{2}}$

$F_{21}=(8.99\cdot10^{9}\frac{N\cdot m^2}{C^2})\frac{(3.1\mu C)(1.6\mu C)}{(0.4m)^{2}}=0.279N$ in the direction

Since charge 2 and 3 have the same polarities, they have a repulsive force; therefore, charge 2 experiences a force away from charge 2 (in the direction). By using Coloumb's law, we can determine this force to be:

$F_{23}=k\frac{q_{2}q_{3}}{{r_{23}}^2}$

$(8.99\cdot10^{9}\frac{N\cdot m^2}{C^2})\frac{(1.6\mu C)(2.4\mu C)}{(0.3m)^{2}}=0.38357N$ in the -direction

If we draw out these two forces tip to tail, we can construct the net force:

From this, we can see that $F_{21}$ and $F_{23}$ create a right triangle with the net force on charge 2 as the hypotenuse. By using the Pythagorean theorem, we can calculate the magnitude of the net force:

$F_{2}=\sqrt{{F_{23}}^{2}+{F_{21}}^{2}}=\sqrt{(0.38357N)^{2}+(0.279N)^{2}}=0.474N$

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Question

What is the force exerted on a point charge of $6\mu C$ by a point charge of $9\mu C$ that is located away?

Answer

Use Coulomb's law.

$F=\frac{kq_{1}q_{2}}{r^{2}}$

Plug in known values and solve.

$F=\frac{(9\cdot 10^{9})(6\cdot 10^{-6})(9\cdot 10^{-6})}{0.75^{2}}$

Note that a positive value for electric force corresponds to a repulsive force. This should make sense since the charge on both particles are the same sign (positive).

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Question

If we have 2 charges, and , that are apart, what is the force exerted on by if we know that has a charge of $-6\mu C$ and has a charge of $-14\mu C$ ?

Answer

Use Coulomb's law.

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

$F=\frac{(9\cdot10^{9})(-6\cdot10^{-6})(-14\cdot10^{-6})}{0.1^{2}}$

Note that the electric force between two charges of the same sign (both positive or both negative) is a positive value. This indicates a repulsive force.

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Question

A point charge of magnitude $1.6 * 10^{-5} C$ is located 0.01m away from a point charge of magnitude $2 * 10^{-5} C$ . What is the electric force between the point charges?

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

Answer

Use Coulomb's law to find the electric force between the charges:

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

$F=\frac{910^91.610^{-5}2*10^{-5}}{0.01^2}$

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Question

A point charge of magnitude $3 * 10^{-2} C$ is 2nm away from a point charge of identical charge. What is the electric force between the point charges?

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

Answer

The electric force between two point charges is given by Coulomb's law:

$F=\frac{kq_{1}q_{2}}{r^2}$

Now, plug in the given charges (both the same magnitude), the given constant, and the distance between the charges (in meters) to get our answer:

$F=\frac{910^9(3 10^{-2})^2}{(2* 10^{-9})^2}=2.025*10^{24} N$

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Question

Determine the magnitude of the electric force between 2 protons that are 3nm apart. Also determine if this force is attractive or repulsive.

$e=1.602*10^{-19}C$

Answer

Recall that Coulomb's law tells us that the magnitude of force between two point charges is given as:

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

Here, is force between two particles, are the charges of each of the two particles, and is the distance between the two charges. In our case, and are identical since each is the charge of a proton which is given as: $1.60210^{-19}C$ , and $r=310^{-9} m$

Thus, plug in known values and solve.

$F= \frac{(1.60210^{-19}C)(1.60210^{-19}C)}{(310^{-9}m)^2}=\frac{2.55610^{-38}C^2}{910^{-18}m^2}$

$F=2.840*10^{-21}N$

To determine if the force is attractive or repulsive, we only need to examine the sign of the charges. Since both protons have the same sign for their charge (positive charges) they will repel.

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Question

What is the magnitude of the electric force between two charged metals that are 3m apart, that have absolute value of the charges being 1C and 3C?

$k=9.0*10^9\frac{N\cdot m^2}{C^2}$

Answer

We are given all the necessary information to find the magnitude of the electric force by using Coulomb's law:

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

Where is Coulomb's constant given by $9.010^9\frac{N\cdot m^2}{C^2}$ , and are the respective charges, and is the distance between the charges. In our case:

$\mid\mid F\mid\mid= \mid\frac{9.010^913}{3^3}N\mid=10^9N$

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Question

Three charges are shown in the given figure. Find the net force on the "top" charge due to the other two (both magnitude and direction). Let and assume all charges are away from each other.

Let $q_{1}$ be the bottom left particle, $q_{2}$ be the top particle and $q_{3}$ be the bottom right particle. Note the axis.

$k=8.99\cdot10^{9} \frac{N\cdot m^{2}}{C^{2}}$

Answer

The method to solving Coulomb's law problems with electrostatic configurations is to find the magnitude of the force and then assign a direction based of what is known about the charges. Coulomb's law is given as:

$F=k \frac{q_{a} q_{b}}{r^{2}}$

Where $q_{a}$ and $q_{b}$ are the two particles we are finding the force between and is the electric constant and is:

$k=\frac{1}{4 \pi \epsilon _{o}}=8.99\cdot10^{9} \frac{N\cdot m^{2}}{C^{2}}$

Notice that the distances between $q_{1}$ and is the same as the $q_{2}$ and $q_{3}$ . Since the magnitudes of all charges are the same, that means that the magnitudes of the forces (not directions) are the same. So the force exerted on $q_{2}$ from $q_{1}$ is the same magnitude as the force exerted on $q_{2}$ from $q_{3}$ .

A sketch of the forces is shown below:

Remember that there are always equal and opposite force pairs. We only care about the forces acting on $q_{2}$ and the last picture shows the two forces that act on it from $q_{1}$ and $q_{3}$ . Notice that the vector arrows are of equal length (force magnitudes are equal) and in different directions. Coulomb forces obey the law of superposition and we can add them. Before we do that let's calculate the magnitude of the two forces pictured.

Remember to convert distances to meters and charge magnitudes to Coulombs so the units work out and you are not off by any factors of .

$F=k \frac{q_{a} q_{b}}{r^{2}}=89.9N$

Both the red vector arrow and the blue vector arrow have magnitudes of . Notice in the diagram below that if the charges are spaced equidistant, the will form an equilateral triangle.

The angle $\theta=60^{o}$ is the same angle that each vector on the right has relative to the line drawn. In order to add the vectors together we need to separate the components of the vectors into their x- and y-components and add the respective components. This is where symmetry can be handy to make the problem easier. Since the particles are equidistant and the charge magnitudes are all equal, this lead to the force magnitudes to be equal. By inspection it can be shown that the y-components must be equal and opposite and therefore cancel.

This means that total magnitude of the force acting on $q_{2}$ is just the sum of the x-component forces. To get the x-components we can use the cosine of the angle. Since the angles are equal and the magnitudes are equal, the final answer will be:

$F=F_{1}_{x}+F_{2}_{x}=(89.9N)Sin(60^{o})+(89.9N)sin(60^{o})$

$\sin(60^{o})=\frac{1}{2}$

$F=2(89.9N)\sin(60^{o})=2(89.9N)\left ( \frac{1}{2}\right )=89.9N \hat{x}$

The final answer is in the positive x-direction, denoted by the positive answer and the $\hat{x}$ to indicate in the x-direction. The answer must have a magnitude and direction to describe the net force acting on the particle.

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Question

Two electric charges are placed $3\ cm$ apart, where and .

What is the magnitude of force between them? Is it replusive or attractive?

Answer

The force between the two charged particles is proportional to the product of their charges, according to Coulomb's Law. Whether the force is attractive or repulsive depends on the signs of the charges. Like signs will repel while opposite signs will attract.

Using Coulomb's Law to find the magnitude of the charge:

$\left |F \right |=\frac{\left |q_1 \right |\left |q_2 \right |}{r^2}=\frac{(8.9810^9)(4C)(10C)}{(310^{-2}\ m)^2}$

$\left |F \right |=3.9910^{14}N$

Therefore, the magnitude of the force has been discovered. Finally, since the signs are opposite ( and ), the force is attractive. Therefore the answer is:

$\left |F \right |=3.9910^{14}N, \ attractive$

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Question

A mole of electrons have a charge of , which is called Faraday's constant. Given that Faraday's constant is $F=96,000\frac{C}{mol}$ , determine the electric force per mole exerted by individual moles of electrons on one another separated by by . Assume charges are static. Use Coulomb's law, and assume that moles of electrons behave like a point charge.

Answer

From Coulomb's Law:

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

Where is the distance between point charges, $k=9.010^9 \frac{N}{m^2C^2}$ , and and are charges of the electrons. In our case, $Q=q=96,000\frac{C}{mol}$ .

$F=\frac{910^9\frac{N}{m^2C^2}(96,000\frac{C}{mol})^2}{10m^2}= 9.210^{17}\frac{N}{mol}$

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Question

Which of the following pairs of charges would exhibit the most electrostatic repulsive force?

Answer

The correct answer is the charge and charge apart.

This results in the largest repulsive force according to the following equation:

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

is the repulsive force, is the magnitude of the charge, is the distance between the charges and is Coulomb's constant.

and charges placed apart provides the greatest repulsive force.

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Question

Two point charges, each having a charge of +1C, are 2 meters apart. If the distance between them is doubled, by what factor does the force between them change?

$k=9.0\times10^9\frac{N\cdot m^2}{C^2}$

Answer

This is a question where knowing how to effectively sift through a problem statement and choose only the information you need will really help. We are given a bunch of values, but only need to know one thing, which is that the distance between the two charges is doubled.

Coulomb's law is as follows:

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

We can rewrite this for the initial and final scenarios:

$F_{ei}= k\frac{q_1q_2}{r_i^2}$

$F_{ef}= k\frac{q_1q_2}{r_f^2}$

We can divide one equation by the other to set up a ratio:

$\frac{F_{ei}}{F_{ef}} = \frac{r_f^2}{r_i^2}$

We know that the final radius is double the intial, which is written as:

Substituting this in we get:

$\frac{F_{ei}}{F_{ef}} = \frac{(2r_i)^2}{r_i^2} = 4$

Rerranging for the final force, we get:

$F_{ef} = \frac{1}{4}F_{ei}$

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Question

If $q_{1}=+3.1\mu C$ , $q_{2}=-1.6\mu C$ , and $q_{3}=-2.4\mu C$ , then what is the magnitude of the net force on charge 2?

$k=8.99\cdot10^{9}\frac{N\cdot m^2}{C^2}$

Answer

First lets set up two axes. Have be to the right of charge 3 and 2 in the diagram and be above charges 1 and 2 in the diagram with charge 2 at the origin.

Coloumb's law tells us the force between point charges is

$F_{21}=k\frac{q_{1}q_{2}}{r^{2}}$

The net force on charge 2 can be determined by combining the force on charge 2 due to charge 1 and the force on charge 2 due to charge 3.

Since charge 1 and charge 2 are of opposite polarities, they have an attractive force; therefore, charge 2 experiences a force towards charge 1 (in the direction). By using Coloumb's law, we can determine this force to be

$F_{21}=k\frac{q_{1}q_{2}}{{r_{12}}^{2}}$

$F_{21}=(8.99\cdot10^{9}\frac{N\cdot m^2}{C^2})\frac{(3.1\mu C)(1.6\mu C)}{(0.4m)^{2}}=0.279N$ in the direction

Since charge 2 and 3 have the same polarities, they have a repulsive force; therefore, charge 2 experiences a force away from charge 2 (in the direction). By using Coloumb's law, we can determine this force to be:

$F_{23}=k\frac{q_{2}q_{3}}{{r_{23}}^2}$

$(8.99\cdot10^{9}\frac{N\cdot m^2}{C^2})\frac{(1.6\mu C)(2.4\mu C)}{(0.3m)^{2}}=0.38357N$ in the -direction

If we draw out these two forces tip to tail, we can construct the net force:

From this, we can see that $F_{21}$ and $F_{23}$ create a right triangle with the net force on charge 2 as the hypotenuse. By using the Pythagorean theorem, we can calculate the magnitude of the net force:

$F_{2}=\sqrt{{F_{23}}^{2}+{F_{21}}^{2}}=\sqrt{(0.38357N)^{2}+(0.279N)^{2}}=0.474N$

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Question

What is the force exerted on a point charge of $6\mu C$ by a point charge of $9\mu C$ that is located away?

Answer

Use Coulomb's law.

$F=\frac{kq_{1}q_{2}}{r^{2}}$

Plug in known values and solve.

$F=\frac{(9\cdot 10^{9})(6\cdot 10^{-6})(9\cdot 10^{-6})}{0.75^{2}}$

Note that a positive value for electric force corresponds to a repulsive force. This should make sense since the charge on both particles are the same sign (positive).

Compare your answer with the correct one above

Question

If we have 2 charges, and , that are apart, what is the force exerted on by if we know that has a charge of $-6\mu C$ and has a charge of $-14\mu C$ ?

Answer

Use Coulomb's law.

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

$F=\frac{(9\cdot10^{9})(-6\cdot10^{-6})(-14\cdot10^{-6})}{0.1^{2}}$

Note that the electric force between two charges of the same sign (both positive or both negative) is a positive value. This indicates a repulsive force.

Compare your answer with the correct one above

Question

A point charge of magnitude $1.6 * 10^{-5} C$ is located 0.01m away from a point charge of magnitude $2 * 10^{-5} C$ . What is the electric force between the point charges?

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

Answer

Use Coulomb's law to find the electric force between the charges:

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

$F=\frac{910^91.610^{-5}2*10^{-5}}{0.01^2}$

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Question

A point charge of magnitude $3 * 10^{-2} C$ is 2nm away from a point charge of identical charge. What is the electric force between the point charges?

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

Answer

The electric force between two point charges is given by Coulomb's law:

$F=\frac{kq_{1}q_{2}}{r^2}$

Now, plug in the given charges (both the same magnitude), the given constant, and the distance between the charges (in meters) to get our answer:

$F=\frac{910^9(3 10^{-2})^2}{(2* 10^{-9})^2}=2.025*10^{24} N$

Compare your answer with the correct one above

Question

Determine the magnitude of the electric force between 2 protons that are 3nm apart. Also determine if this force is attractive or repulsive.

$e=1.602*10^{-19}C$

Answer

Recall that Coulomb's law tells us that the magnitude of force between two point charges is given as:

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

Here, is force between two particles, are the charges of each of the two particles, and is the distance between the two charges. In our case, and are identical since each is the charge of a proton which is given as: $1.60210^{-19}C$ , and $r=310^{-9} m$

Thus, plug in known values and solve.

$F= \frac{(1.60210^{-19}C)(1.60210^{-19}C)}{(310^{-9}m)^2}=\frac{2.55610^{-38}C^2}{910^{-18}m^2}$

$F=2.840*10^{-21}N$

To determine if the force is attractive or repulsive, we only need to examine the sign of the charges. Since both protons have the same sign for their charge (positive charges) they will repel.

Compare your answer with the correct one above

Question

What is the magnitude of the electric force between two charged metals that are 3m apart, that have absolute value of the charges being 1C and 3C?

$k=9.0*10^9\frac{N\cdot m^2}{C^2}$

Answer

We are given all the necessary information to find the magnitude of the electric force by using Coulomb's law:

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

Where is Coulomb's constant given by $9.010^9\frac{N\cdot m^2}{C^2}$ , and are the respective charges, and is the distance between the charges. In our case:

$\mid\mid F\mid\mid= \mid\frac{9.010^913}{3^3}N\mid=10^9N$

Compare your answer with the correct one above

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