Question

Two charges, qA and qB, are separated by a distance, r, and exert a force, F, on each other. What new force will exist for the following scenario? Your answer should be given in multiples of F (i.e. 2F, ½F, etc.)

qA is doubled and r is doubled

Answer 1

`(1)/(2)F`

Step-by-step explanation:

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

`F=k(q_1 q_2)/(r^2)`

where:

`k=9\cdot 10^9 Nm^(-2)C^(-2)` is the Coulomb's constant

`q_1, q_2` are the two charges

r is the separation between the two charges

In this problem, at the beginning we have:

`q_1=q_A` is the first charge

`q_2=q_B` is the second charge

r is their initial separation

So the initial force is

`F=k(q_A q_B)/(r^2)`

Later, we have:

qA is doubled and r is doubled

This means that:

`q_A'=2q_A` is the new charge

`r'=2r` is the new separation

So the new force is:

`F'=k(q_A' q_B)/(r'^2)=k((2q_A)(q_B))/((2r)^2)=(1)/(2)(k(q_A q_B)/(r^2))=(1)/(2)F`