Charge and Electric Force - MCAT Physical

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Question

Two charges of $Q$ coulombs are a distance $d$ apart from each other. Which of the following would reduce the force exerted between the charges by a factor of 4?

Answer

Given Coulomb's Law electrostatic forces: $F= \frac{kQq}{d^{2}}$

We can see that distance and force are inversly related. Also distance is squared, so if we increase the distance by 2, the force between the two charges will be reduced by a factor of four.

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Question

$e = 1.60 \times 10^{-19}$

$k = 8.99 \times 10^{9}$

How many electrons would it require to generate a 5 N attractive force on a +3 µC charge from 0.1 m away.

Answer

To find the answer we must first solve for q, using Coulomb's Law. Then after finding q = 1.85 x 10–6 C, we must divide by 1.6 x 10–19 C to get the amount of electrons required to make such a charge.

Compare your answer with the correct one above

Question

Two charges Q1 and Q2 are held stationary 1 m away from each other. The magnitude of charge of Q2 is twice that of Q1. A positive point charge A is placed directly in the middle of the two charges (Q1 and Q2). What is the ratio of the net force on charge A when Q1 and Q2 are both negative over the net force on charge A when Q1 is negative and Q2 is positive?

Answer

For this problem we must understand Coulomb's Law. The force on charge A will be dircetly affected by the charge of Q1 and Q2. Q1 and Q2 are both negative, they will both attract charge A, and the net force will be reduced as they pull in opposite directions. Switching Q2 to a positive charge will result in a repulsive force on charge A which will be in the same direction as the attractive force between Q1 and A.

Compare your answer with the correct one above

Question

Two point charges, Q1 and Q2, are placed on the x-axis. Q1 is placed at x = 0m and Q2 at x = 1m. If a positive charge, Q3, is placed on the x-axis at x = 0.25m, it experiences a net electric force of 0. Which of the following could be determined based on this observation?

Answer

In order for there to be 0 net force on Q3, the forces from Q1 and Q2 must be in opposite directions. So Q1 and Q2 must either be both attracting Q3 or both repelling Q3 (that is, Q1 and Q2 are either both negative or both positive).

Also, since these forces must have equal magnitude in order to cancel, such that $\small F_{13}=F_{23}$ , where $\small \small F_{13}$ is the magnitude of force between Q1 and Q3, and similarly for $\small F_{23}$ . Recalling Coulomb’s law for electric force, $\small F_{13}=\frac{kQ_{1}Q_{3}}{r_{13}^{2}}$ and $\small \small F_{23}=\frac{kQ_{2}Q_{3}}{r_{23}^{2}}$ (where Q1, Q2, and Q3 are the magnitudes of the respective charges). Since $\small r_{23}>r_{13}$ , it must also be true that $\small kQ_{2}Q_{3}>kQ_{1}Q_{3}$ for the two fractions to have the same value. So, Q2 must have a greater magnitude than Q1.

Compare your answer with the correct one above

Question

Electronegativity is an important concept in physical chemistry, and often used to help quantify the dipole moment of polar compounds. Polar compounds are different from those compounds that are purely nonpolar or purely ionic. An example can be seen by contrasting sodium chloride, NaCl, with an organic molecule, R-C-OH. The former is purely ionic, and the latter is polar covalent.

When comparing more than one polar covalent molecule, we use the dipole moment value to help us determine relative strength of polarity. Dipole moment, however, is dependent on the electronegativity of the atoms making up the bond. Electronegativity is a property inherent to the atom in question, whereas dipole moment is a property of the bond between them.

For example, oxygen has an electronegativity of 3.44, and hydrogen of 2.20. In other words, oxygen more strongly attracts electrons when in a bond with hydrogen. This leads to the O-H bond having a dipole moment.

When all the dipole moments of polar bonds in a molecule are summed, the molecular dipole moment results, as per the following equation.

Dipole moment = charge * separation distance

Electronegativity is based on the principle that the closer an electron is to the atomic nucleus, where postive charge is concentrated, the more attractive force the positive charge can exert on the electron.

As a result, as atomic radius decreases by one half, what happens to the force pulling the electron toward that atom?

Answer

This is a function of the equation $F = k\frac{q_1q_2}{d^2}$ .

In other words, as the distance between two charges doubles, the force between them goes increases, as the denominator decreases from to $(\frac{1}{2}d)^2$ .

Here, we are decreasing the distance by 1/2, decreasing the denominator by 1/4, so our force goes up by 4 times.

Compare your answer with the correct one above

Question

Two charges of $Q$ coulombs are a distance $d$ apart from each other. Which of the following would reduce the force exerted between the charges by a factor of 4?

Answer

Given Coulomb's Law electrostatic forces: $F= \frac{kQq}{d^{2}}$

We can see that distance and force are inversly related. Also distance is squared, so if we increase the distance by 2, the force between the two charges will be reduced by a factor of four.

Compare your answer with the correct one above

Question

$e = 1.60 \times 10^{-19}$

$k = 8.99 \times 10^{9}$

How many electrons would it require to generate a 5 N attractive force on a +3 µC charge from 0.1 m away.

Answer

To find the answer we must first solve for q, using Coulomb's Law. Then after finding q = 1.85 x 10–6 C, we must divide by 1.6 x 10–19 C to get the amount of electrons required to make such a charge.

Compare your answer with the correct one above

Question

Two charges Q1 and Q2 are held stationary 1 m away from each other. The magnitude of charge of Q2 is twice that of Q1. A positive point charge A is placed directly in the middle of the two charges (Q1 and Q2). What is the ratio of the net force on charge A when Q1 and Q2 are both negative over the net force on charge A when Q1 is negative and Q2 is positive?

Answer

For this problem we must understand Coulomb's Law. The force on charge A will be dircetly affected by the charge of Q1 and Q2. Q1 and Q2 are both negative, they will both attract charge A, and the net force will be reduced as they pull in opposite directions. Switching Q2 to a positive charge will result in a repulsive force on charge A which will be in the same direction as the attractive force between Q1 and A.

Compare your answer with the correct one above

Question

Two point charges, Q1 and Q2, are placed on the x-axis. Q1 is placed at x = 0m and Q2 at x = 1m. If a positive charge, Q3, is placed on the x-axis at x = 0.25m, it experiences a net electric force of 0. Which of the following could be determined based on this observation?

Answer

In order for there to be 0 net force on Q3, the forces from Q1 and Q2 must be in opposite directions. So Q1 and Q2 must either be both attracting Q3 or both repelling Q3 (that is, Q1 and Q2 are either both negative or both positive).

Also, since these forces must have equal magnitude in order to cancel, such that $\small F_{13}=F_{23}$ , where $\small \small F_{13}$ is the magnitude of force between Q1 and Q3, and similarly for $\small F_{23}$ . Recalling Coulomb’s law for electric force, $\small F_{13}=\frac{kQ_{1}Q_{3}}{r_{13}^{2}}$ and $\small \small F_{23}=\frac{kQ_{2}Q_{3}}{r_{23}^{2}}$ (where Q1, Q2, and Q3 are the magnitudes of the respective charges). Since $\small r_{23}>r_{13}$ , it must also be true that $\small kQ_{2}Q_{3}>kQ_{1}Q_{3}$ for the two fractions to have the same value. So, Q2 must have a greater magnitude than Q1.

Compare your answer with the correct one above

Question

Electronegativity is an important concept in physical chemistry, and often used to help quantify the dipole moment of polar compounds. Polar compounds are different from those compounds that are purely nonpolar or purely ionic. An example can be seen by contrasting sodium chloride, NaCl, with an organic molecule, R-C-OH. The former is purely ionic, and the latter is polar covalent.

When comparing more than one polar covalent molecule, we use the dipole moment value to help us determine relative strength of polarity. Dipole moment, however, is dependent on the electronegativity of the atoms making up the bond. Electronegativity is a property inherent to the atom in question, whereas dipole moment is a property of the bond between them.

For example, oxygen has an electronegativity of 3.44, and hydrogen of 2.20. In other words, oxygen more strongly attracts electrons when in a bond with hydrogen. This leads to the O-H bond having a dipole moment.

When all the dipole moments of polar bonds in a molecule are summed, the molecular dipole moment results, as per the following equation.

Dipole moment = charge * separation distance

Electronegativity is based on the principle that the closer an electron is to the atomic nucleus, where postive charge is concentrated, the more attractive force the positive charge can exert on the electron.

As a result, as atomic radius decreases by one half, what happens to the force pulling the electron toward that atom?

Answer

This is a function of the equation $F = k\frac{q_1q_2}{d^2}$ .

In other words, as the distance between two charges doubles, the force between them goes increases, as the denominator decreases from to $(\frac{1}{2}d)^2$ .

Here, we are decreasing the distance by 1/2, decreasing the denominator by 1/4, so our force goes up by 4 times.

Compare your answer with the correct one above

Question

Two charges of $Q$ coulombs are a distance $d$ apart from each other. Which of the following would reduce the force exerted between the charges by a factor of 4?

Answer

Given Coulomb's Law electrostatic forces: $F= \frac{kQq}{d^{2}}$

We can see that distance and force are inversly related. Also distance is squared, so if we increase the distance by 2, the force between the two charges will be reduced by a factor of four.

Compare your answer with the correct one above

Question

$e = 1.60 \times 10^{-19}$

$k = 8.99 \times 10^{9}$

How many electrons would it require to generate a 5 N attractive force on a +3 µC charge from 0.1 m away.

Answer

To find the answer we must first solve for q, using Coulomb's Law. Then after finding q = 1.85 x 10–6 C, we must divide by 1.6 x 10–19 C to get the amount of electrons required to make such a charge.

Compare your answer with the correct one above

Question

Two charges Q1 and Q2 are held stationary 1 m away from each other. The magnitude of charge of Q2 is twice that of Q1. A positive point charge A is placed directly in the middle of the two charges (Q1 and Q2). What is the ratio of the net force on charge A when Q1 and Q2 are both negative over the net force on charge A when Q1 is negative and Q2 is positive?

Answer

For this problem we must understand Coulomb's Law. The force on charge A will be dircetly affected by the charge of Q1 and Q2. Q1 and Q2 are both negative, they will both attract charge A, and the net force will be reduced as they pull in opposite directions. Switching Q2 to a positive charge will result in a repulsive force on charge A which will be in the same direction as the attractive force between Q1 and A.

Compare your answer with the correct one above

Question

Two point charges, Q1 and Q2, are placed on the x-axis. Q1 is placed at x = 0m and Q2 at x = 1m. If a positive charge, Q3, is placed on the x-axis at x = 0.25m, it experiences a net electric force of 0. Which of the following could be determined based on this observation?

Answer

In order for there to be 0 net force on Q3, the forces from Q1 and Q2 must be in opposite directions. So Q1 and Q2 must either be both attracting Q3 or both repelling Q3 (that is, Q1 and Q2 are either both negative or both positive).

Also, since these forces must have equal magnitude in order to cancel, such that $\small F_{13}=F_{23}$ , where $\small \small F_{13}$ is the magnitude of force between Q1 and Q3, and similarly for $\small F_{23}$ . Recalling Coulomb’s law for electric force, $\small F_{13}=\frac{kQ_{1}Q_{3}}{r_{13}^{2}}$ and $\small \small F_{23}=\frac{kQ_{2}Q_{3}}{r_{23}^{2}}$ (where Q1, Q2, and Q3 are the magnitudes of the respective charges). Since $\small r_{23}>r_{13}$ , it must also be true that $\small kQ_{2}Q_{3}>kQ_{1}Q_{3}$ for the two fractions to have the same value. So, Q2 must have a greater magnitude than Q1.

Compare your answer with the correct one above

Question

Electronegativity is an important concept in physical chemistry, and often used to help quantify the dipole moment of polar compounds. Polar compounds are different from those compounds that are purely nonpolar or purely ionic. An example can be seen by contrasting sodium chloride, NaCl, with an organic molecule, R-C-OH. The former is purely ionic, and the latter is polar covalent.

When comparing more than one polar covalent molecule, we use the dipole moment value to help us determine relative strength of polarity. Dipole moment, however, is dependent on the electronegativity of the atoms making up the bond. Electronegativity is a property inherent to the atom in question, whereas dipole moment is a property of the bond between them.

For example, oxygen has an electronegativity of 3.44, and hydrogen of 2.20. In other words, oxygen more strongly attracts electrons when in a bond with hydrogen. This leads to the O-H bond having a dipole moment.

When all the dipole moments of polar bonds in a molecule are summed, the molecular dipole moment results, as per the following equation.

Dipole moment = charge * separation distance

Electronegativity is based on the principle that the closer an electron is to the atomic nucleus, where postive charge is concentrated, the more attractive force the positive charge can exert on the electron.

As a result, as atomic radius decreases by one half, what happens to the force pulling the electron toward that atom?

Answer

This is a function of the equation $F = k\frac{q_1q_2}{d^2}$ .

In other words, as the distance between two charges doubles, the force between them goes increases, as the denominator decreases from to $(\frac{1}{2}d)^2$ .

Here, we are decreasing the distance by 1/2, decreasing the denominator by 1/4, so our force goes up by 4 times.

Compare your answer with the correct one above

Question

Two charges of $Q$ coulombs are a distance $d$ apart from each other. Which of the following would reduce the force exerted between the charges by a factor of 4?

Answer

Given Coulomb's Law electrostatic forces: $F= \frac{kQq}{d^{2}}$

We can see that distance and force are inversly related. Also distance is squared, so if we increase the distance by 2, the force between the two charges will be reduced by a factor of four.

Compare your answer with the correct one above

Question

$e = 1.60 \times 10^{-19}$

$k = 8.99 \times 10^{9}$

How many electrons would it require to generate a 5 N attractive force on a +3 µC charge from 0.1 m away.

Answer

To find the answer we must first solve for q, using Coulomb's Law. Then after finding q = 1.85 x 10–6 C, we must divide by 1.6 x 10–19 C to get the amount of electrons required to make such a charge.

Compare your answer with the correct one above

Question

Two charges Q1 and Q2 are held stationary 1 m away from each other. The magnitude of charge of Q2 is twice that of Q1. A positive point charge A is placed directly in the middle of the two charges (Q1 and Q2). What is the ratio of the net force on charge A when Q1 and Q2 are both negative over the net force on charge A when Q1 is negative and Q2 is positive?

Answer

For this problem we must understand Coulomb's Law. The force on charge A will be dircetly affected by the charge of Q1 and Q2. Q1 and Q2 are both negative, they will both attract charge A, and the net force will be reduced as they pull in opposite directions. Switching Q2 to a positive charge will result in a repulsive force on charge A which will be in the same direction as the attractive force between Q1 and A.

Compare your answer with the correct one above

Question

Two point charges, Q1 and Q2, are placed on the x-axis. Q1 is placed at x = 0m and Q2 at x = 1m. If a positive charge, Q3, is placed on the x-axis at x = 0.25m, it experiences a net electric force of 0. Which of the following could be determined based on this observation?

Answer

In order for there to be 0 net force on Q3, the forces from Q1 and Q2 must be in opposite directions. So Q1 and Q2 must either be both attracting Q3 or both repelling Q3 (that is, Q1 and Q2 are either both negative or both positive).

Also, since these forces must have equal magnitude in order to cancel, such that $\small F_{13}=F_{23}$ , where $\small \small F_{13}$ is the magnitude of force between Q1 and Q3, and similarly for $\small F_{23}$ . Recalling Coulomb’s law for electric force, $\small F_{13}=\frac{kQ_{1}Q_{3}}{r_{13}^{2}}$ and $\small \small F_{23}=\frac{kQ_{2}Q_{3}}{r_{23}^{2}}$ (where Q1, Q2, and Q3 are the magnitudes of the respective charges). Since $\small r_{23}>r_{13}$ , it must also be true that $\small kQ_{2}Q_{3}>kQ_{1}Q_{3}$ for the two fractions to have the same value. So, Q2 must have a greater magnitude than Q1.

Compare your answer with the correct one above

Question

Electronegativity is an important concept in physical chemistry, and often used to help quantify the dipole moment of polar compounds. Polar compounds are different from those compounds that are purely nonpolar or purely ionic. An example can be seen by contrasting sodium chloride, NaCl, with an organic molecule, R-C-OH. The former is purely ionic, and the latter is polar covalent.

When comparing more than one polar covalent molecule, we use the dipole moment value to help us determine relative strength of polarity. Dipole moment, however, is dependent on the electronegativity of the atoms making up the bond. Electronegativity is a property inherent to the atom in question, whereas dipole moment is a property of the bond between them.

For example, oxygen has an electronegativity of 3.44, and hydrogen of 2.20. In other words, oxygen more strongly attracts electrons when in a bond with hydrogen. This leads to the O-H bond having a dipole moment.

When all the dipole moments of polar bonds in a molecule are summed, the molecular dipole moment results, as per the following equation.

Dipole moment = charge * separation distance

Electronegativity is based on the principle that the closer an electron is to the atomic nucleus, where postive charge is concentrated, the more attractive force the positive charge can exert on the electron.

As a result, as atomic radius decreases by one half, what happens to the force pulling the electron toward that atom?

Answer

This is a function of the equation $F = k\frac{q_1q_2}{d^2}$ .

In other words, as the distance between two charges doubles, the force between them goes increases, as the denominator decreases from to $(\frac{1}{2}d)^2$ .

Here, we are decreasing the distance by 1/2, decreasing the denominator by 1/4, so our force goes up by 4 times.

Compare your answer with the correct one above

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Charge and Electric Force - MCAT Physical

Two charges of coulombs are a distance apart from each other. Which of the following would reduce the force exerted between the charges by a factor of 4?

Given Coulomb's Law electrostatic forces: We can see that distance and force are inversly related. Also distance is squared, so if we increase the distance by 2, the force between the two charges will be reduced by a factor of four.

How many electrons would it require to generate a 5 N attractive force on a +3 µC charge from 0.1 m away.

To find the answer we must first solve for q, using Coulomb's Law. Then after finding q = 1.85 x 10–6 C, we must divide by 1.6 x 10–19 C to get the amount of electrons required to make such a charge.

Two point charges, Q1 and Q2, are placed on the x-axis. Q1 is placed at x = 0m and Q2 at x = 1m. If a positive charge, Q3, is placed on the x-axis at x = 0.25m, it experiences a net electric force of 0. Which of the following could be determined based on this observation?

This is a function of the equation . In other words, as the distance between two charges doubles, the force between them goes increases, as the denominator decreases from to . Here, we are decreasing the distance by 1/2, decreasing the denominator by 1/4, so our force goes up by 4 times.

Two charges of coulombs are a distance apart from each other. Which of the following would reduce the force exerted between the charges by a factor of 4?

Given Coulomb's Law electrostatic forces: We can see that distance and force are inversly related. Also distance is squared, so if we increase the distance by 2, the force between the two charges will be reduced by a factor of four.

How many electrons would it require to generate a 5 N attractive force on a +3 µC charge from 0.1 m away.

To find the answer we must first solve for q, using Coulomb's Law. Then after finding q = 1.85 x 10–6 C, we must divide by 1.6 x 10–19 C to get the amount of electrons required to make such a charge.

Two point charges, Q1 and Q2, are placed on the x-axis. Q1 is placed at x = 0m and Q2 at x = 1m. If a positive charge, Q3, is placed on the x-axis at x = 0.25m, it experiences a net electric force of 0. Which of the following could be determined based on this observation?

This is a function of the equation . In other words, as the distance between two charges doubles, the force between them goes increases, as the denominator decreases from to . Here, we are decreasing the distance by 1/2, decreasing the denominator by 1/4, so our force goes up by 4 times.

Two charges of coulombs are a distance apart from each other. Which of the following would reduce the force exerted between the charges by a factor of 4?

Given Coulomb's Law electrostatic forces: We can see that distance and force are inversly related. Also distance is squared, so if we increase the distance by 2, the force between the two charges will be reduced by a factor of four.

How many electrons would it require to generate a 5 N attractive force on a +3 µC charge from 0.1 m away.

To find the answer we must first solve for q, using Coulomb's Law. Then after finding q = 1.85 x 10–6 C, we must divide by 1.6 x 10–19 C to get the amount of electrons required to make such a charge.

Two point charges, Q1 and Q2, are placed on the x-axis. Q1 is placed at x = 0m and Q2 at x = 1m. If a positive charge, Q3, is placed on the x-axis at x = 0.25m, it experiences a net electric force of 0. Which of the following could be determined based on this observation?

This is a function of the equation . In other words, as the distance between two charges doubles, the force between them goes increases, as the denominator decreases from to . Here, we are decreasing the distance by 1/2, decreasing the denominator by 1/4, so our force goes up by 4 times.

Two charges of coulombs are a distance apart from each other. Which of the following would reduce the force exerted between the charges by a factor of 4?

Given Coulomb's Law electrostatic forces: We can see that distance and force are inversly related. Also distance is squared, so if we increase the distance by 2, the force between the two charges will be reduced by a factor of four.

How many electrons would it require to generate a 5 N attractive force on a +3 µC charge from 0.1 m away.

To find the answer we must first solve for q, using Coulomb's Law. Then after finding q = 1.85 x 10–6 C, we must divide by 1.6 x 10–19 C to get the amount of electrons required to make such a charge.

Two point charges, Q1 and Q2, are placed on the x-axis. Q1 is placed at x = 0m and Q2 at x = 1m. If a positive charge, Q3, is placed on the x-axis at x = 0.25m, it experiences a net electric force of 0. Which of the following could be determined based on this observation?

This is a function of the equation . In other words, as the distance between two charges doubles, the force between them goes increases, as the denominator decreases from to . Here, we are decreasing the distance by 1/2, decreasing the denominator by 1/4, so our force goes up by 4 times.

Two charges of $Q$ coulombs are a distance $d$ apart from each other. Which of the following would reduce the force exerted between the charges by a factor of 4?

Given Coulomb's Law electrostatic forces: $F= \frac{kQq}{d^{2}}$

We can see that distance and force are inversly related. Also distance is squared, so if we increase the distance by 2, the force between the two charges will be reduced by a factor of four.

$e = 1.60 \times 10^{-19}$

$k = 8.99 \times 10^{9}$

How many electrons would it require to generate a 5 N attractive force on a +3 µC charge from 0.1 m away.

Two charges of $Q$ coulombs are a distance $d$ apart from each other. Which of the following would reduce the force exerted between the charges by a factor of 4?

Given Coulomb's Law electrostatic forces: $F= \frac{kQq}{d^{2}}$

We can see that distance and force are inversly related. Also distance is squared, so if we increase the distance by 2, the force between the two charges will be reduced by a factor of four.

$e = 1.60 \times 10^{-19}$

$k = 8.99 \times 10^{9}$

How many electrons would it require to generate a 5 N attractive force on a +3 µC charge from 0.1 m away.

Two charges of $Q$ coulombs are a distance $d$ apart from each other. Which of the following would reduce the force exerted between the charges by a factor of 4?

Given Coulomb's Law electrostatic forces: $F= \frac{kQq}{d^{2}}$

We can see that distance and force are inversly related. Also distance is squared, so if we increase the distance by 2, the force between the two charges will be reduced by a factor of four.

$e = 1.60 \times 10^{-19}$

$k = 8.99 \times 10^{9}$

How many electrons would it require to generate a 5 N attractive force on a +3 µC charge from 0.1 m away.

Two charges of $Q$ coulombs are a distance $d$ apart from each other. Which of the following would reduce the force exerted between the charges by a factor of 4?

Given Coulomb's Law electrostatic forces: $F= \frac{kQq}{d^{2}}$

We can see that distance and force are inversly related. Also distance is squared, so if we increase the distance by 2, the force between the two charges will be reduced by a factor of four.

$e = 1.60 \times 10^{-19}$

$k = 8.99 \times 10^{9}$

How many electrons would it require to generate a 5 N attractive force on a +3 µC charge from 0.1 m away.