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?

1 Answer
Jan 12, 2016

Answer:

-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