Question

What is the electric force between a +2 µC point charge and a –2 µC point charge if they are separated by a distance of 5.0 cm? Show your work. (µC = 1.0 × 10–6 C)

Answer 1

The electrostatic force between two point charges is given by:
$F=k_e \frac{q_1 q_2}{r^2}$
where
$k_e = 8.99 \cdot 10^9 Nm^2 C^{-2}$ is the Coulomb's constant
q1 and q2 are the two charges
r is the distance between them

In our problem, charge 1 is
$q_1 = +2 \muC = +2.0 \cdot 10^{-6}C$
while charge 2 is
$q_2 = -2 \mu C=-2.0 \cdot 10^{-6} C$
and their distance is
r=5.0 cm=0.05 m

So, the electrostatic force between them is
$F=(8.99 \cdot 10^9 Nm^2C^{-2}) \frac{(+2\cdot 10^{-6}C)(-2 \cdot 10^{-6}C)}{(0.05 m)^2} =-14.4 N$

And the negative sign means the force is attractive, because the two charges have opposite sign.