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?

Jun 3, 2017

$F = - \frac{1}{4} {k}_{e}$

Explanation:

The electric force $F$ between two point charges ${q}_{1}$ and ${q}_{2}$ a distance $r$ apart is given by the equation

$F = {k}_{e} \frac{{q}_{1} {q}_{2}}{{r}^{2}}$

where ${k}_{e}$ is Coulomb's constant, equal to $8.988 \times {10}^{9} \left({\text{N • m"^2)/("C}}^{2}\right)$

Plugging in known values, we have

$F = {k}_{e} \left({\left(6 \text{C")(-6"C"))/((12"m}\right)}^{2}\right)$

$= {k}_{e} \left(- 36 {\text{C"^2)/(144"m}}^{2}\right)$

$= - \left(1 {\text{C"^2)/(4"m}}^{2}\right) {k}_{e}$

So, in terms of Coulomb's constant (${k}_{e}$), the (attractive) electric force acting between the two point charges is

$\textcolor{red}{F = - \frac{1}{4} {k}_{e}}$

(excluding units)