Question

What's the force on charge Q3?

Answer 1

`394"N"` to the left.

Explanation:

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

`color(blue)(F=1/(4piepsilon_0)*(abs(q_1)abs(q_2))/r^2)`

where `r` is the distance between the charges

To find the net force on `Q_3`, we will need to calculate the force exerted on it by both `Q_1` and `Q_2`.

We have the following information:

• `Q_1=-50muC`
• `Q_2=100muC`
• `Q_3=-20muC`
• `r_(12)=20.0"cm"`
• `r_(13)=40.0"cm"`
• `epsilon_0=8.85 10^(−12)" C"^2//"Nm"^2`

Let's start with `Q_1`.

• `Q_1` and `Q_3` are both negative charges. Therefore the charges repel, and the force exerted on `Q_3` by `Q_1` points to the right.

`F_(13)=1/(4piepsilon_0)*(abs(-50*10^-6C)abs(-20*10^-6C))/(40.0*10^-2m)^2`

`=>~~56"N"` (to the right)

Now we can calculate the force between `Q_2` and `Q_3`:

• `Q_2` is a positive charge, whereas `Q_3` is a negative charge. Therefore the charges attract, and the force exerted on `Q_3` by `Q_2` is to the left.

`F_(23)=1/(4piepsilon_0)*(abs(100*10^-6C)abs(-20*10^-6C))/(20.0*10^-2m)^2`

`=>~~450"N"` (to the left)

• Then, since the charges are all placed along a straight line, we can simply add these values for force together to get the net force.
• However, note that these are forces and do have directions associated with them. We can already see that the net force will be to the left, as this is is a much greater value than that to the right.
• We will therefore subtract `F_(13)` from `F_(23)`.

`F_(" on "Q_3)=450"N"-56"N=394"N"`

That is, `394"N"` to the left.

Alternatively, you can define left as the negative direction and make the calculated force net negative. Then upon adding the two forces you would obtain a negative net force, which indicates that the net force is to the left.