Question

Two charges of `6 C` and `-3 C` are positioned on a line at points `7` and `2`, respectively. What is the net force on a charge of `-1 C` at `-2`?

Answer 1

`F_3=4.13*10^8N`

Explanation:

Consider the figure. Let the charges `6C, -3C` and `-1C` be denoted by `q_1,q_2` and `q_3` respectively.
Let the positions at which charges are placed be in the units of meters.

Let `r_13`be the distance between the charges `q_1 and q_3`.
From figure
`r_13=7-(-2)=7+2=9m`

Let `r_23`be the distance between the charges `q_2 and q_3`.
From figure
`r_23=7-2=5m`

Let `F_13` be the force due to charge `q_1` on the charge `q_3`
`F_13=(kq_1q_3)/r_13^2=(9*10^9*(6)(1))/9^2=0.667*10^9N`
This force is attractive and is towards charge `q_1`.

Let `F_23` be the force due to charge `q_2` on the charge `q_3`
`F_23=(kq_2q_3)/r_23^2=(9*10^9*(3)(1))/5^2=1.08*10^9N`
This force is repulsive and is towards charge `q_3`.

The total force or net force on charge `q_3` is the sum of above two forces.
Since the above two forces `F_13` and `F_23` are in same direction therefore we have to subtract smaller force from larger force.
Let `F_3` be the total force on the charge `q_3`.
`implies F_3=F_23-F_13=1.08*10^9-0.667*10^9=0.413*10^9=4.13*10^8N`
`implies F_3=4.13*10^8N`
Since The larger force `F_23` is towards charge `q_3`.
Therefore the force `F_3` is also towards charge `q_3`.