Charge 1 is -30 μC, charge 2 is 5 μC, the distance between the charges is 8 cm, and they are located in a vacuum. If the third charge of 3 μC would be placed into position B, i. e., exactly between the two other charges, what would be the magnitude of the net force it experiences?

F₃ = 590.625 N

Explanation:

We have a situation where, we have 3 Charges, as follows:

Charge 1 = q₁ = - 30 μC

Charge 2 = q₂ = 5 μC

Charge 3 = q₃ = 3 μC

Distance between Charge 3 and Charge 1 = r₃₁ = 4 cm = 0.04 m

Distance between Charge 3 and Charge 2 = r₃₂ = 4 cm = 0.04 m

So, in order to find net force on the charge 3:

F₃ = F₁₃ + F₂₃

where,

F₃ = Net force on charge 3

F₁₃ = Force on Charge 3 due to Charge 1

F₂₃ = Force on Charge 3 due to Charge 2

We are adding the magnitudes of forces, because their direction will be same. Since, charge 1 is negative and charge 3 is positive, so direction of force will be towards charge 1. And charge 2 is positive thus direction of its force will also be towards charge 1.

Now, using Coulomb Law:

F₃ = kq₁q₃/r₁₃² + kq₃q₂/r₁₂²

F₃ = kq₃ (q₁/r₁₃² + q₂/r₁₂²)

F₃ = (9 x 10⁹ N.m²/C²)(3 x 10⁻⁶ C)[(30 x 10⁻⁶ C)/(0.04 m)² + (5 x 10⁻⁶ C)/(0.04 m)²

F₃ = 590.625 N

the answer is a. hope this .