Question

Three point charges are on the x-axis; q1 = 4.0 μc is at x = -3.0 m, q2 = 4.0 μc is at the origin, and q3 = -6.0 μc is at x = 3.0 m. find the force on q1.

Answer 1

The last part is your answer

Answer 2

By using Coulomb's law, we will see that the force is 0.01N (away from the other two)

How to use Coulomb's law?

Coulomb's law says that the force between two charges q₁ and q₂ is given by:

$F = k*\frac{q_1*q_2}{r^2}$

Where k is a constant and r is the distance between the charges. If F is negative, the force is attractive, if F is positive, the force is repulsive.

Here we want to get the force on q₁ due to the other two charges, so we will have:

$F = k*q_1*(\frac{q_2}{r_{12}^2} + \frac{q_3}{r_{13}^2} )$

Now we know that:

• k = 8.99*10^9 N*m^2/C^2
• q₁ = 4.0*10^(-6) C
• q₂ = 4.0*10^(-6) C
• q₃ = -6.0*10^(-6) C

• r₁₂ is the distance between charges 1 and 2, this is 3m
• r₁₃ is the distance between charges 1 and 3, this is 6m

(you get these two by taking the difference between the x-position of the charges).

Then the force acting on q₁ is:

$F = (8.99*10^9 N*m^2/C^2)*(4.0*10^{-6} C)(\frac{4.0*10^{-6} C}{(3m)^2} + \frac{-6.0*10^{-6} C}{(6m)^2} ) = 0.01 N$

If you want to learn more about Coulomb's law you can read:

https://brainly.com/question/24743340