Question

Point charge 3.0 μC is located at x = 0, y = 0.30 m, point charge -3.0 μC is located at x = 0 y = -0.30 m. What are (a)the magnitude and (b)direction of the total electric force that these charges exert on a third point charge Q = 4.5 μC at x = 0.40 m, y = 0?

Answer 1

The total electric force that these charges exert on a third point charge is 11.9 N and the direction is towards positive y axis.

Magnitude of the electric force between the charges

The force between the first charge and the third charge is calculated as follows;

F₁₃ = kq₁q₃/r₁₃²

F₁₃ = (9 x 10⁹ x 3 x 10⁻⁶ x 4.5 x 10⁻⁶) / (0.1)²

note: distance between charge 3.0 μC and Q = 4.5 μC  = 0.4 m - 0.3 m = 0.1 m

F₁₃ = 12.15 N

The force between the second charge and the third charge is calculated as follows;

F₂₃ = kq₂q₃/r₂₃²

F₂₃ = (9 x 10⁹ x -3 x 10⁻⁶ x 4.5 x 10⁻⁶) / (0.7)²

note: distance between charge -3.0 μC and Q = 4.5 μC  = 0.4 m - (-0.3 m) = 0.7 m

F₂₃ = -0.248 N

Net force between the charges

F(net) = F₁₃ + F₂₃

F(net) = 12.15 - 0.248

F(net) = 11.9 N

Thus, the total electric force that these charges exert on a third point charge is 11.9 N and the direction is towards positive y axis.

Learn more about electric force here: brainly.com/question/17692887

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