# What is the force, in terms of Coulomb's constant, between two electrical charges of 22 C and -32 C that are 6 m  apart?

Sep 13, 2017

It is an attraction force of $1.76 \cdot {10}^{11} N$.

#### Explanation:

Coulomb's Law, in equation form, is

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

where F is the magnitude of the force of attraction, or repulsion, between 2 charges,
k is known as Coulomb's constant and has a value of 8.99×10^9 (N m^2)/C^2,
${q}_{1} \mathmr{and} {q}_{2}$ are the 2 charges, in Coulombs,
and r is the distance between the charges, in meters.

The force is a force of attraction if F has a negative value and repulsion if the value of F has a positive value.

$\textcolor{g r e e n}{E \mathrm{di} t \text{ The green lines between here and the red text below replace those red lines.}}$

$\textcolor{g r e e n}{\text{Plugging the values into the formula}}$
$\textcolor{g r e e n}{F = k \cdot \frac{22 C \cdot \left(- 32 C\right)}{6 m} ^ 2}$
$\textcolor{g r e e n}{F = - k \cdot 19.6 N}$

$\textcolor{g r e e n}{\text{So it is an attraction force of } k \cdot 19.6 N}$.

$\textcolor{red}{\text{Plugging the values into the formula}}$
color(red)(F = (8.99×10^9 (N m^2)/C^2) * (22 C*(-32 C))/(6 m)^2)
$\textcolor{red}{F = - 175.8 \cdot {10}^{9} N = - 1.76 \cdot {10}^{11} N}$

$\textcolor{red}{\text{So it is an attraction force of } 1.76 \cdot {10}^{11} N}$.

My apologies for the need to edit, I originally overlooked the request that the answer be in terms of k."

I hope this helps,
Steve