MCAT Physical : Charge and Electric Force
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Example Questions
Example Question #12 : Ap Physics 2
Two charges of 
Decrease the each charge Q by a factor of four.
Decrease distance D by a factor of two.
Increase distance d by a factor of 4.
Increase distance D by a factor of two.
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 #16 : Ap Physics 2
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?
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?
Q1 and Q2 have the same sign, but Q1 has a greater 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 smaller magnitude than Q2
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
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 
Example Question #15 : Electricity And Magnetism
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?
It stays constant
It quadruples
It doubles
It goes up by a factor of eight
It is reduced by half
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 
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 #201 : Ap Physics 2
How many electrons would it require to generate a 5 N attractive force on a +3 µC charge from 0.1 m away.
2.3 x 10–13 electrons
1.15 x 1013 electrons
1.15 x 10–14 electrons
0.68 x 1014 electrons
1.15 x 1013 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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