Question

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

Answer 1

`vecF_(n e t)=6.76*10^9` `N` (to the right)

Explanation:

The force between two charges is given by Coulomb's law:

`|vecF|=k(|q_1||q_2|)/r^2`

where `q_1` and `q_2` are the magnitudes of the charges, `r` is the distance between them, and `k` is a constant equal to `8.99xx10^9Nm^2//C^2`, sometimes referred to as Coulomb's constant.

Here is a quick diagram of the situation:

To find the net force on the `3C` charge, we consider the force exerted on it by the `9C` and `2C` charges. Let's call the `3C` charge `q_1`, the `9C` charge `q_2` and the `2C` charge `q_3`. Then:

`vecF_(n e t)=sumvecF=vecF_(2 on 1)+vecF_(3 on 1)`

Recall that like charges repel, while opposite charges attract. Each of our charges are positive. Therefore, the force vectors can be drawn in as follows:

23 should say 21 in the above diagram

where `21 (23)` is the force of `q_1` on `q_2` and `31` is the force of `q_3` on `q_1`

So we can calculate `vecF_(31)` and subtract from that `vecF_(21)` to find the net force on the `q_1` charge.

`|vecF_(31)|=k*(|q_1||q_3|)/(r_(31)^2)`

`=(8.99xx10^9Nm^2//C^2)*(3C*2C)/(2)^2`

`=1.35*10^(10)` `N`

`|vecF_(21)|=k*(|q_1||q_2|)/(r_(21)^2)`

`=(8.99xx10^9Nm^2//C^2)*(3C*9C)/(6)^2`

`=6.74*10^9` `N`

Therefore, we have:

`vecF_(n e t)=6.76*10^9` `N` (to the right)