Question

What is the force, in terms of Coulomb's constant, between two electrical charges of `18 C` and `25 C` that are `15 m` apart?

Answer 1

The force is `=+(0.72k)N`

Explanation:

The force between `2` charges `q_1` and `q_2` is

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

Where `r` is the distance between the charges.

And the constant `k=9*10^9Nm^2C^-2`

The charge `q_1=18C`

The charge `q_2=25C`

The distance is

`r=15m`

Therefore,

The force is `=k*((18)*(25))/(15^2)=+(0.72k)N`

The positive sign indicates that the force is repulsive