# A charge of -2 C is at (-5 , 1 ) and a charge of -1 C is at (1 ,-3) . If both coordinates are in meters, what is the force between the charges?

Aug 7, 2017

${F}_{\text{e}} = 3.46 \times {10}^{8}$ $\text{N}$

#### Explanation:

We're asked to find the magnitude of the electric force between two point charges.

To do this, we can use Coulomb's law:

$\underline{\overline{| \stackrel{\text{ ")(" "F_"e" = k(|q_1q_2|)/(r^2)" }}{|}}}$

where

• ${F}_{\text{e}}$ is the magnitude of the electric force

• $k$ is Coulomb's constant, equal to $8.988 \times {10}^{9} \left({\text{N"*"m"^2)/("C}}^{2}\right)$

• ${q}_{1}$ and ${q}_{2}$ are the point charges, in no particular order

• $r$ is the distance, in meters, between the two point charges

We have:

• $k = 8.988 \times {10}^{9} \left({\text{N"*"m"^2)/("C}}^{2}\right)$

• ${q}_{1} = - 2$ $\text{C}$

• ${q}_{2} = - 1$ $\text{C}$

• to calculate $r$, we use the distance formula:

r = sqrt((-5-1)^2 + (1-(-3))^2) = color(red)(ul(7.21

Plugging these in:

F_"e" = 8.988xx10^9("N"*cancel("m"^2))/(cancel("C"^2))((|(-2cancel("C"))(-1cancel("C"))|)/((color(red)(7.21)cancel(color(red)("m")))^2)) = color(blue)(ulbar(|stackrel(" ")(" "3.46xx10^8color(white)(l)"N"" ")|)