Question

a 12 μc charge is place at the origin. A second -8.0 μc charge is placed 1.04cm to the right of the first. A third 6.0 μc charge is placed 1.0m above the first charge. What is the force on second charge due the other two?

Answer 1

Answer on Question #71640-Physics-Electromagnetism

A 12 µc charge is placed at the origin. A second -8.0 µc charge is placed 1.04cm to the right of the first. A third 6.0 µc charge is placed 1.0m above the first charge. What is the force on second charge due to the other two?

Solution

$$F_{21} = (9 \cdot 10^9) \frac{(8 \cdot 10^{-6})(12 \cdot 10^{-6})}{(0.0104)^2} = 7988 \, \text{N}.$$ $$F_{23} = (9 \cdot 10^9) \frac{(8 \cdot 10^{-6})(6 \cdot 10^{-6})}{(0.0104)^2 + (0.01)^2} = 2075 \, \text{N}.$$ $$F_X = 7988 + 2075 \cos\left(\tan^{-1}\left(\frac{1}{1.04}\right)\right) = 9484 \, \text{N}.$$ $$F_y = 2075 \sin\left(\tan^{-1}\left(\frac{1}{1.04}\right)\right) = 1438 \, \text{N}.$$ $$F = \sqrt{9484^2 + 1438^2} = 9.6 \, \text{kN}.$$

Answer: 9.6 kN.

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