Question

Two charges of `-2 C` and `4 C` are at points `(-2 , 4, 8)` and `( 2 ,4, 8 )`, respectively. Assuming that both coordinates are in meters, what is the force between the two points?

Answer 1

The distance formula for Cartesian coordinates is

`d=sqrt((x_2-x_1)^2+(y_2-y_1)^2+(z_2-z_1)^2`
Where `x_1, y_1, z_1`, and`x_2, y_2, z_2` are the Cartesian coordinates of two points respectively.
Let `(x_1,y_1,z_1)` represent `(-2,4,8)` and `(x_2,y_2,z_2)` represent `(2,4,8)`.

`d=sqrt((2-(-2))^2+(4-4)^2+(8-8)^2`

`d=sqrt((2+2)^2+(0)^2+(0)^2`

`d=sqrt((4)^2+0+0)=sqrt(16)=4`

Hence the distance between the two charges is `4` meters.

The electrostatic force between two charges is given by
`F=(kq_1q_2)/r^2`

Where `F` is the force between the charges, `k` is the constant

and its value is `9*10^9 Nm^2/C^2`,`q_1` and `q_2` are the

magnitudes of the charges and `r` is the distance between the two charges.

Here `F=??`, `k=9*10^9Nm^2/C^2`, `q_1=-2C`, `q_2=4C` and `r=d=4`.

`implies F=(9*10^9*(-2)*4)/(4)^2=(9*10^9*(-8))/16=(-72*10^9)/16=-4.5*10^9N`

`implies F=-4.5*10^9N`

Negative sign shows that the force between the two charges is attractive.