Question

What is the force, in terms of Coulomb's constant, between two electrical charges of `6 C` and `2 C` that are `2 m` apart?

Answer 1

Electrostatic force of repulsion between two like charges of `q_1` and `q_2` separated by a distance of `r` is given as,

`F=(kq_1 q_2)/r^2` where, `k` is a constant

Now,here, given that `q_1=6C,q_2=2C,r=2m`

So putting the values we get,

`F=(12k)/2^2=3k` `N`

Answer 2

The force is `=(3k)N`

Explanation:

The net force between `2` charges is

`F=k(Q_1Q_2)/r^2`

Where `r` is the distance between the charges

The Coulomb Constant is `k=9*10^9Nm^2/C^2`

The charge `Q_1=6C`

The charge `Q_2=2C`

The distance is `r=2m`

The force is

`F=k(6*2)/(2^2)=(3k)N`