# A charge of 2 C is at (-9,1 ) and a charge of -1 C is at ( 2,-5 ). If both coordinates are in meters, what is the force between the charges?

May 30, 2017

$1.15 \times {10}^{8} \text{N}$

#### Explanation:

The electric force $F$ (in $\text{N}$) between two point charges ${q}_{1}$ and ${q}_{2}$ (in $\text{C}$) is represented by the equation

$F = \frac{1}{4 \pi {\epsilon}_{0}} \frac{| {q}_{1} {q}_{2} |}{{r}^{2}}$

where

• ${\epsilon}_{0}$ is a physical constant called the permittivity of free space, equal to $8.8542 \times {10}^{-} 12 \frac{{C}^{2}}{N \cdot {m}^{2}}$, and

• $r$ is the distance (in $\text{m}$) between the two point charges.

To find the distance between the two charges, we can use the distance formula:

r = sqrt((1--5)^2 + (-9-2)^2) = color(red)(12.5"m"

Now that we have all the necessary variables, let's plug them into the equation to find the charge:

$F = \frac{1}{4 \pi \left(8.8542 \times {10}^{-} 12 \frac{\cancel{{C}^{2}}}{N \cdot \cancel{{m}^{2}}}\right)} \left(| {\left(2 \cancel{\text{C"))(-1cancel("C"))|)/((12.5cancel("m}}\right)}^{2}\right)$

= color(blue)(1.15 xx 10^8 "N"

Thus, the electric force between the two point charges is $1.15 \times {10}^{8} \text{N}$, and since the charges are opposite, the force is that of attraction.