# What is the force, in terms of Coulomb's constant, between two electrical charges of -6 C and -16 C that are 9 m  apart?

Jun 12, 2018

I get approximately $10.7 k$.

#### Explanation:

Coulomb's law states that:

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

where:

• $k$ is Coulomb's constant

• ${q}_{1} , {q}_{2}$ are charges of the two charges in coulombs

• $r$ is the distance between the charges in meters

So, we get:

$F = k \frac{- 6 \cdot - 16}{9}$

$\approx 10.7 k$

Jun 12, 2018

$F \approx 1.0652 \cdot {10}^{10} N$

#### Explanation:

Coulomb's Law with Coulomb's constant:

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

We will assume that the charges occur in a vacuum.

$k = 8.988 \cdot {10}^{9} \frac{N {m}^{2}}{C} ^ 2$
${Q}_{1} = - 6 C$
${Q}_{2} = - 16 C$
$r = 9 m$

$F = \frac{\left(8.988 \cdot {10}^{9}\right) \left(- 6\right) \left(- 16\right)}{9} ^ 2$

$F \approx 1.0652 \cdot {10}^{10} N$