# A charge of #5 C# is at #(-6, 3 )# and a charge of #-2 C# is at #(-5 , 7 ) #. If both coordinates are in meters, what is the force between the charges?

##### 1 Answer

Recall that **Coulomb's law** tells you the electric force of attraction or repulsion between two point charges:

#bb(F_E = (kq_1q_2)/(r^2) = (q_1q_2)/(4piepsilon_0r^2))# where

#k = 1/(4piepsilon_0)# is a constant,#epsilon_0 = 8.854xx10^(-12) "C"^2"/N"cdot"m"^2# is the vacuum permittivity, and#r# is the distance between the two point charges.

Since you were given coordinates, recall the distance formula to calculate

#r = sqrt((x_2 - x_1)^2 + (y_2 - y_1)^2)#

#= sqrt((-5 - -6)^2 + (7 - 3)^2)#

#= sqrt((1)^2 + (4)^2) = sqrt17# ,or,

#color(green)(r^2 = 17)# .

The constant

#color(green)(k) = 1/(4piepsilon_0) = 1/((4)(3.1415926535cdots)*(8.854xx10^(-12) "C"^2"/N"cdot"m"^2))#

#= color(green)(8.987xx10^9)# #color(green)("N"cdot"m"^2"/C"^2)#

So the **electric force**

#color(blue)(F_E) = (kq_1q_2)/(r^2)#

#= ((8.987xx10^9"N"cdotcancel"m"^2"/"cancel("C"^2))(5 cancel"C")(-2 cancel"C"))/(17 cancel("m"^2))#

#= color(blue)(-5.287xx10^9)# #color(blue)("N")#

This *negative* value means that the two point charges are highly *attracted* to each other. This makes sense because they're *oppositely*-signed charges.