Question

What is the force, in terms of Coulomb's constant, between two electrical charges of `6 C` and `-6 C` that are `12 m` apart?

Answer 1

`F = -1/4k_e`

Explanation:

The electric force `F` between two point charges `q_1` and `q_2` a distance `r` apart is given by the equation

`F =k_e(q_1q_2)/(r^2)`

where `k_e` is Coulomb's constant, equal to `8.988 xx 10^9 ("N • m"^2)/("C"^2)`

Plugging in known values, we have

`F =k_e((6"C")(-6"C"))/((12"m")^2)`

`= k_e(-36"C"^2)/(144"m"^2)`

`= -(1"C"^2)/(4"m"^2)k_e`

So, in terms of Coulomb's constant (`k_e`), the (attractive) electric force acting between the two point charges is

`color(red)(F = -1/4k_e)`

(excluding units)