Four point charges have equal magnitudes. Three are positive, and one is negativ

Question

Four point charges have equal magnitudes. Three are positive, and one is negative, as the drawing shows. They are fixed in place on the same straight line, and adjacent charges are equally separated by a distance d. Consider the net electrostatic force acting on each charge. Calculate the ratio of the largest to the smallest net force magnitude.

Explanation / Answer

The magnitude of the electrostatic force between two charge : F = kQ1*q2/r^2 ---> Q1 = Q2 =Q = k Q^2/d^2 The force experienced positive charge : Ft0tal = v(F^2 + F^2 + 2F^2 * cos ?) -----> ? The force experienced negative charge : Ftotal = v(F^2 + F^2 + 2F^2 * cos ?) -----> ? plsz rate me 1st

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