Question

What is the force, in terms of Coulomb's constant, between two electrical charges of `-45 C` and `30 C` that are `15 m` apart?

Answer 1

-6k or `-5.4xx10^10N`

Explanation:

You would use Coulomb's Law for this question:

`F_e=(kq_1q_2)/(r^2)`
where q1 represents the quantity of charge on object 1 (in Coulombs), q2 represents the quantity of charge on object 2 (in Coulombs), and r represents the distance of separation between the two objects (in meters). "k" is a constant, and it's value is `9.0xx10^(9)(Nm^2)/C^2`

If we plug in the values we get:

`F_e=(k(-45C)(30C))/((15m)^2)`

In terms of "k", this would simplify to:

`F_e=k(-6)`

If we sub in the value for k, we get:

`F_e=(9.0xx10^(9)(Nm^2)/C^2)(-6)`

`F_e=-5.4xx10^10N`

A negative force indicates a force of ATTRACTION, whereas a positive force indicates a force of REPULSION.

Here's another video to help:

Khan Academy Video