What is the force, in terms of Coulomb's constant, between two electrical charges of -6 C and -16 C that are 9 m apart?

2 Answers
Jun 12, 2018

I get approximately 10.7k.

Explanation:

Coulomb's law states that:

F=k(q_1q_2)/r^2

where:

  • k is Coulomb's constant

  • q_1,q_2 are charges of the two charges in coulombs

  • r is the distance between the charges in meters

So, we get:

F=k(-6*-16)/9

~~10.7k

Jun 12, 2018

F~~1.0652*10^10 N

Explanation:

Coulomb's Law with Coulomb's constant:

F=(kQ_1Q_2)/r^2

We will assume that the charges occur in a vacuum.

k= 8.988*10^9 (Nm^2)/C^2
Q_1=-6 C
Q_2=-16 C
r=9 m

F=(( 8.988*10^9)(-6)(-16))/9^2

F~~1.0652*10^10 N