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Goshia [24]
3 years ago
7

Three charges are arranged as shown in the diagram.

Physics
2 answers:
KIM [24]3 years ago
7 0

Force due to -8 uC charge is given as

F_1 = \frac{kq_1q}{r^2}

F_1 = \frac{(9\times 10^9)(8\times 10^{-6})(6 \times 10^{-6})}{0.3^2}

F_1 = 4.8 N

Force due to -5 uC charge is given as

F_2 = \frac{kq_2q}{r^2}

F_2 = \frac{(9\times 10^9)(5\times 10^{-6})(6 \times 10^{-6})}{0.3^2}

F_2 = 3 N

Now we know that two forces on 6 uC charge is perpendicular to each other

so we will have

F = \sqrt{F_1^2 + F_2^2}

F = \sqrt{3^2 + 4.8^2}

F = 5.66 N

so net force due to both charges on 6 uC is 5.66 N

Klio2033 [76]3 years ago
4 0

Answer: 5.7

Explanation: on edge its 5.7

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