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.
