Question

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

Answer 1

`F=4.95*10^8N`

Explanation:

The distance formula for Cartesian coordinates is

`d=sqrt((x_2-x_1)^2+(y_2-y_1)^2`
Where `x_1, y_1`, and`x_2, y_2` are the Cartesian coordinates of two points respectively.
Let `(x_1,y_1)` represent `(6,-5)` and `(x_2,y_2)` represent `(3,5)`.

`implies d=sqrt((3-6)^2+(5-(-5))^2)`
`implies d=sqrt((-3)^2+(10)^2`
`implies d=sqrt(9+100)`
`implies d=sqrt(109)`

Hence the distance between the two charges is `sqrt(109)`.

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=-3C` and `r=d=sqrt109`.

`implies F=(9*10^9(-2)(-3))/(sqrt109)^2=(9*10^9*6)/109=(54*10^9)/109=0.495*10^9N=4.95*10^8N`
`implies F=4.95*10^8N`

Since the force is positive therefore the force is repullsive.