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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 :

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

[tex]F_1=\frac{kQ^2}{r^2_1}[/tex]

k is a constant that is 9x10^9

we substitute the values

[tex]F_1=\frac{(9\times10^9)(3.5\times10^{-6})^2}{0.06^2}[/tex]

[tex]F_1=30.625N_{}[/tex]

then for the force with charge C

[tex]F_2_{}=\frac{kQ^2}{r^2_2}[/tex][tex]F_1=\frac{(9\times10^9)(3.5\times10^{-6})^2}{0.07^2}=22.5N[/tex]

Then we calculate the net force

[tex]F=F_1+F_2[/tex][tex]F=30.625+22.5[/tex][tex]F=53.125N[/tex]

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

[tex]F=-53.1N[/tex]

View image LynoxL422751

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