Question

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

Answer 1

`F=-1.08*10^9 N`

Explanation:

Charge `q_1=3C` is placed at `(3,-1)`
Charge `q_2=-4C` is placed at `(-3,7)`

Distance between both charges is
`r=sqrt((x_2-x_1)^2+(y_2-y_1)^2`
`r=sqrt((-3-(3))^2+(7-(-1))^2`
`r=sqrt((-6)^2+(8)^2`
`r=sqrt(36+64)`
`r=sqrt(100) m = 10 m` and `r^2=100`

Force between two charges `q_1` and `q_2` is
`F=(kq_1q_2)/r^2` where `k=9*10^9 Nm^2/C^2`
`=>F=((9*10^9)(3)(-4))/(100)`
`=>F=((9*10^9)(-12))/(100)`
`=>F=((-108*10^9))/(100) N`
`=>F=-1.08*10^9 N`