Three point charges have equal magnitudes, two being positive and one negative.

Question

Three point charges have equal magnitudes, two being positive and one negative. These charges are fixed to the corners of an equilateral triangle, as the drawing shows. The magnitude of each of the charges is 4.9 µC, and the lengths of the sides of the triangle are 3.9 cm. Calculate the magnitude of the net force that each charge experiences.

(a) FA =



(b) FB =

(c) FC =


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Explanation / Answer

The magnitude of the electrostatic force between two charge : F = kQ1*q2/r^2 ---> Q1 = Q2 =Q = k Q^2/r^2 = 9 x 10^9 * (5 x 10^-6)^2/9 x 10^-4 =250 N The force experienced positive charge : Ftatal = v(F^2 + F^2 + 2F^2 * cos ?) -----> ? =120^ o = 250 N The force experienced negative charge : Ftotal = v(F^2 + F^2 + 2F^2 * cos ?) -----> ? = 60^ o = 433 N

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