# A charge of 3 C is at (4 ,-7 ) and a charge of -1 C is at (-2 , 4 ) . If both coordinates are in meters, what is the force between the charges?

Jun 30, 2017

${F}_{e} = 1.72 \times {10}^{8}$ $\text{N}$

#### Explanation:

We're asked to calculate the magnitude of the electric force between two point charges, for which we can use the equation

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

where

• $k$ 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 two point charges (no particular order), in coulombs, $\text{C}$

• $r$ is the distance between the two charges, in meters, $\text{m}$

The distance between the two point charges is found using the distance formula:

r = sqrt((4-(-2))^2 + (-7-4)^2) = color(red)(12.5 color(red)("m"

Plugging in our known values, we have

F_e = (8.988xx10^9("N"•"m"^2)/("C"^2))(|(3color(white)(l)"C")(-1color(white)(l)"C")|)/((color(red)(12.5)color(white)(l)color(red)("m"))^2)

= color(blue)(1.72 xx 10^8 color(blue)("N"