# What is the force, in terms of Coulomb's constant, between two electrical charges of 44 C and 39 C that are 12 m  apart?

Nov 29, 2016

The force '$F ' = 11.91 k {C}^{2} / {m}^{2}$.
The coulomb's law gives $F = \frac{k {q}_{1} {q}_{2}}{r} ^ 2$, where 'k' is the proportionality constant. It is also called as coulomb's constant as it as derived from coulomb's law.
Proportionality constant 'k' is equal to '$\frac{1}{4 \pi {\epsilon}_{0}}$'
.$F = \frac{k \left(44 C\right) \left(39 C\right)}{12 m} ^ 2$
$\therefore$ The force in terms of coulomb's constant 'k' is $11.9 k \left({C}^{2} / {m}^{2}\right) \approx 12 k \left({C}^{2} / {m}^{2}\right)$