Question

1. given that q = +18 μc and d = 21 cm, find the direction and magnitude of the net electrostatic force exerted on the point charge, q1, in the figure below.

q1 = +q
q2 = -2.0q
q3 = +3.0q

Explanation:

Step1: Recall Coulomb's law

The force between two point - charges is given by $F = k\frac{q_1q_2}{r^2}$, where $k = 9\times10^{9}\ N\cdot m^{2}/C^{2}$, $q_1$ and $q_2$ are the charges, and $r$ is the distance between them.

Step2: Calculate the force due to $q_2$ on $q_1$

$q_1=q = 18\times10^{- 6}\ C$, $q_2=-2.0q=-2\times18\times10^{-6}\ C$, and $r = d=0.21\ m$.
$F_{12}=k\frac{q_1|q_2|}{d^{2}}=(9\times10^{9})\frac{(18\times10^{-6})\times(2\times18\times10^{-6})}{(0.21)^{2}}$
$F_{12}=(9\times10^{9})\frac{18\times10^{-6}\times36\times10^{-6}}{0.0441}=\frac{9\times18\times36\times10^{-3}}{0.0441}\ N$
$F_{12}=\frac{5832\times10^{-3}}{0.0441}\ N = 132.245\ N$ (attractive, towards $q_2$).

Step3: Calculate the force due to $q_3$ on $q_1$

$q_3 = 3.0q=3\times18\times10^{-6}\ C$, and $r = 2d = 0.42\ m$.
$F_{13}=k\frac{q_1q_3}{(2d)^{2}}=(9\times10^{9})\frac{(18\times10^{-6})\times(3\times18\times10^{-6})}{(0.42)^{2}}$
$F_{13}=(9\times10^{9})\frac{18\times10^{-6}\times54\times10^{-6}}{0.1764}=\frac{9\times18\times54\times10^{-3}}{0.1764}\ N$
$F_{13}=\frac{8748\times10^{-3}}{0.1764}\ N = 49.592\ N$ (repulsive, away from $q_3$).

Step4: Calculate the net force

The net force $F_{net}=F_{12}-F_{13}$.
$F_{net}=132.245 - 49.592=82.653\ N$ towards $q_2$.

Answer:

The magnitude of the net electrostatic force is $82.7\ N$ (rounded to one - decimal place) and the direction is towards $q_2$.