Two particles, each of charge Q, are fixed at opposite corners of a square that lies in the plane of the page. A positive test charge q is p

Question

Two particles, each of charge Q, are fixed at opposite corners of a square that lies in the plane of the page. A positive test charge q is placed at a third corner. If F is the magnitude of the force on the q test charge due to only one of the other charges, what is the magnitude of the net force acting on the test charge due to both of these charges?

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Ben Gia 4 years 2021-07-15T04:42:30+00:00 1 Answers 6 views 0

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  1. Eirian
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    2021-07-15T04:43:46+00:00
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    Answer:

    The magnitude of the net force is √2F.

    Explanation:

    Since the two particles have the same charge Q, they exert the same force on the test charge; both attractive or repulsive. So, the angle between the two forces is 90° in any case. Now, as we know the magnitude of these forces and that they form a 90° angle, we can use the Pythagorean Theorem to calculate the magnitude of the resultant net force:

    F_N=\sqrt{F^{2}+F^{2}}\\\\F_N=\sqrt{2F^{2}}\\\\F_N=\sqrt{2}F

    Then, it means that the net force acting on the test charge has a magnitude of √2F.

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