# What is the force, in terms of Coulomb's constant, between two electrical charges of -4 C and 19 C that are 12 m  apart?

Aug 15, 2016

$F = - 6.156 \cdot {10}^{12} N$

#### Explanation:

$F = k \cdot \frac{{q}_{1} \cdot {q}_{2}}{d} ^ 2$

$k = {9.10}^{9} \text{ } N \cdot {m}^{2} \cdot {C}^{- 2}$

${q}_{1} = - 4 C$

${q}_{2} = 19 C$

$d = 12 \text{ } m$

${F}_{1} = - {F}_{2} = F$

$F = {9.10}^{9} \frac{\left(- 4\right) \cdot 19}{12} ^ 2$

$F = - \frac{9 \cdot {10}^{9} \cdot \cancel{4} \cdot 19}{\cancel{144}}$

$F = - 9 \cdot 36 \cdot 19 \cdot {10}^{9}$

$F = - 6156 \cdot {10}^{9}$

$F = - 6.156 \cdot {10}^{12} N$