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 (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)) ((+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.