Coulomb's law:

$\overline{){{\mathbf{F}}}_{{\mathbf{E}}}{\mathbf{=}}\frac{\mathbf{k}\mathbf{q}\mathbf{Q}}{{\mathbf{r}}^{\mathbf{2}}}}$

From the figure, each charge produces a vertical component. The vertical components are in the same axis as the weight. The weight is directed downward while the repulsive electric force is directed upward.

F_{E} makes the **hypotenus****e**. To solve for the **opposite** (vertical component of the electric force) we'll use the 'sin' trigonometric ratio. **opp = hyp•sinθ**

ΣF = 4F_{E}•sinθ - mg = 0

Four equal positive point charges q are fixed to the corners of a horizontal square. A fifth positive point charge Q is on a free particle of mass m positioned directly above the center of the square, at a height equal to the length d of one side of the square. Refer to the figure.

Determine the magnitude of q in terms of defined quantities, the Coulomb constant, and g, if the system is in equilibrium.

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