Question

A charge of `5 C` is at `(-6, 1 )` and a charge of `-3 C` is at `(-2, 1)`. If both coordinates are in meters, what is the force between the charges?

Answer 1

The force between the charges is `8\times10^9` N.

Explanation:

Use Coulomb's law:
`F=\frac{k\abs{q_1q_2}}{r^2}`

Calculate `r`, the distance between the charges, using the Pythagorean theorem
`r^2 = \Delta x^2 + \Delta y^2`
`r^2 = (-6-(-2))^2 + (1-1)^2`
`r^2 = (-6+2)^2 + (1-1)^2`
`r^2 = 4^2 + 0^2`
`r^2 = 16`
`r=4`

The distance between the charges is `4`m. Substitute this into Coulomb's law. Substitute in the charge strengths as well.

`F=\frac{k\abs{q_1q_2}}{r^2}`
`F=k\frac{\abs{(5)(-3)}}{4^2}`
`F=k\frac{15}{16}`
`F = 8.99×10^9(\frac{15}{16})` (Substitute in the value of Coulomb's constant)
`F=8.4281\times 10^9` N
`F=8 \times 10^9` N (As you're working with one significant figure)