Question

Three identical point charges of 3.5 μC are placed on a horizontal axis. The first charge is at the origin, the second at x2 = 6 cm, and the third is at the x3 =7 cm. What is the magnitude and direction of the electrostatic force which acts on the charge at the origin? Round your answer to the nearest tenth of a Newton. For this question and your answer, a positive (+) will indicate to the right, and a negative (-) will indicate to the left.

Answer 1

We have

We have

Q = magnitude of charge at each of the three locations A, B and C=3.5 x 10⁻⁶ C

r₁ = distance of charge at origin (A) from charge at B = 6cm = 0.06 m

r₂ = distance of charge at origin from charge at C = 7cm=0.07m

F₁ = magnitude of force by charge at B on charge at origin

F₂ = magnitude of force by charge at C on charge at origin

Let's calculate F₁

`F_1=(kQ^2)/(r^2_1)`

k is a constant that is 9x10^9

we substitute the values

`F_1=((9*10^9)(3.5*10^(-6))^2)/(0.06^2)`

`F_1=30.625N_{}`

then for the force with charge C

`F_2_{}=(kQ^2)/(r^2_2)`
`F_1=((9*10^9)(3.5*10^(-6))^2)/(0.07^2)=22.5N`

Then we calculate the net force

`F=F_1+F_2`
`F=30.625+22.5`
`F=53.125N`

As we can see in the diagram the forces go to the left therefore the solution is

`F=-53.1N`