# Magnitude of Force

by Bama
Tags: force, magnitude
 P: 321 Magnitude of Force Your formula is correct but you forgot to inclue the sign of q3 in your calculation. Recall that the electrical force between two charges is given by: $$F = \frac {1}{4 \pi \epsilon_0} \frac {Q_1Q_2}{r^2}$$ So for part 1) Q1 = 4x10^-6 C and Q2 = -2.12x10^-6 C. Perhaps the choices for the possible solutions are wrong. Unless the question meant the unsigned magnitude of the force.
 P: 321 Well I used $Q_1 \text{ and } Q_2$ as variables in general for the formula. Applied in this case, $Q_1 = q_1$ and $Q_2 = q_3$. To answer you last question you can use vector addition to find the resultant force. The formula is $$\vec F = \frac {1}{4 \pi \epsilon_0} \frac {Q_1Q_2}{r^2} {\vec r}$$ However you could also just calculate the magnitudes of each force exerted on q1, then calculate the components of each vector and sum them all up.