Question

Two charges of `-2 C` and `3 C` are positioned on a line at points `3` and `4`, respectively. What is the net force on a charge of `-9 C` at `0`?

Answer 1

`2.81*10^9N`

Explanation:

Coulomb's law: `F=(k_eQ_1Q_2)/r^2`, where:
`F` = the force (`N`)
`k_e` = the coulomb constant(`~~8.99*10^9N` `m^2` `C^(-2)`)
`Q_1 &Q_2` = the charge of each particle (`C`)
`r` = the distance (`m`)

Assuming `1` `"unit" = 1m`

Also, assuming a positive resultant force = repulsion, and a negative resultant force = attraction.

For `-9C` and `-2C`:
`F=((8.99*10^9)(-9)(-2))/(3^2)=1.798*10^(10)N`

For `-9C` and `3C`:
`F=((8.99*10^9)(-9)(3))/(4^2)==-1.5170625*10^(10)N`

Net force = `(1.798*10^(10))-(1.5170625*10^(10))=2.809375*10^(9)~~2.81*10^9N`