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9 May, 03:02

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?

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  1. 9 May, 06:53
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    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
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