Question

4. In general, the electric force is directly related to the magnitude (strength) of the charges. This means that as the magnitude of the charges increases, the electric force will also:
A. Increase
B. Decrease
C. Not change at all
D. There is no way to know

5. Therefore, when the magnitude of one charge is increased by a factor, the magnitude of the force will…
A. Increase by the same factor
B. Decrease by the same factor
C. Increase by the square of the factor
D. Decrease by the square of the factor
E. Not change at all

Based on your previous answers, which statement best describes how the electric force would change if the magnitude of Charge 1 were DECREASED?
A. The electric force would decrease.
B. The electric force would increase.
C. The electric force would not change.
D. There is no way to know.

My teacher didn't explain this well

Answer 1

4.  A. Increase

The magnitude of the electric force between two charges is given by Coulomb's law:

$F=k\frac{q_1 q_2}{r^2}$

where

k is the Coulomb's constant

q_1 and q_2 are the two charges

r is the separation between the two charges

As we see from the formula, the strength of the electric force is:

- directly proportional to the two charges q1 and q2

- inversely proportional to the square of the distance r

Therefore, we can conclude that:

- if the magnitude of the charges increases, then the magnitude of the force will increases as well

- if the separation between the charges increases, then the magnitude of the force will decrease

5. A. Increase by the same factor

Again, the magnitude of the electric force between the two charges is given by Coulomb's law:

$F=k\frac{q_1 q_2}{r^2}$

we have said that the force F is directly proportional to the two charges q_1 and q_2.

Let's assume now that charge q_1 is increases by a certain factor A. Then the electric force will change as follows:

$F'=k\frac{(A q_1) q_2}{r^2}=A(k\frac{q_1 q_2}{r^2})=A\cdot F$

so, the magnitude of the electric force has increased by the same factor.

6. A. The electric force would decrease.

This part is similar to part 5, however this time the charge q_1 is decreased.

This means that we can rewrite the new charge q1 as

$q_1 ' = \frac{q_1}{A}$

where A > 1. Let's see how the electric force changes:

$F'=k\frac{(\frac{q_1}{A}) q_2}{r^2}=\frac{1}{A}(k\frac{q_1 q_2}{r^2})=\frac{F}{A}$

This means that the magnitude of the electric force has decreased by the same factor.