### All MCAT Physical Resources

## Example Questions

### Example Question #1 : Electrostatics And Electrical Fields

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?

**Possible Answers:**

Decrease distance D by a factor of two.

Increase distance D by a factor of two.

Decrease the each charge Q by a factor of four.

Increase distance d by a factor of 4.

**Correct answer:**

Increase distance D by a factor of two.

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.

### Example Question #2 : Electrostatics And Electrical Fields

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?

**Possible Answers:**

**Correct 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*.

### Example Question #1 : Electrostatics And Electrical Fields

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?

**Possible Answers:**

Q1 and Q2 are both positive and have the same magnitude

Q1 and Q2 have opposite signs and the same magnitude

Q1 and Q2 have the same sign, but Q1 has a smaller magnitude than Q2

Q1 and Q2 are both negative and have the same magnitude

Q1 and Q2 have the same sign, but Q1 has a greater magnitude than Q2

**Correct answer:**

Q1 and Q2 have the same sign, but Q1 has a smaller magnitude than Q2

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 , where is the magnitude of force between Q1 and Q3, and similarly for . Recalling Coulomb’s law for electric force, and (where Q1, Q2, and Q3 are the magnitudes of the respective charges). Since , it must also be true that for the two fractions to have the same value. So, Q2 must have a greater magnitude than Q1.

### Example Question #1 : Charge And Electric Force

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?

**Possible Answers:**

It doubles

It is reduced by half

It goes up by a factor of eight

It quadruples

It stays constant

**Correct answer:**

It quadruples

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.

### Example Question #1 : Electrostatics And Electrical Fields

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

**Possible Answers:**

1.15 x 10^{–14} electrons

2.3 x 10^{–13} electrons

1.15 x 10^{13} electrons

0.68 x 10^{14} electrons

**Correct answer:**

1.15 x 10^{13} electrons

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.

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