Answer:
[tex]F_e = \frac{1}{4 \pi \epsilon_0} \frac{q_1 q_2}{r^2}[/tex]
Explanation:
The electrostatic force (also called Couloumb force) between two electrical charge is given by the following formula:
[tex]F_e = \frac{1}{4 \pi \epsilon_0} \frac{q_1 q_2}{r^2}[/tex]
where:
[tex]\epsilon_0=8.85\cdot 10^{-12} Fm^{-1}[/tex] is the vacuum permittivity
[tex]q_1[/tex] is the magnitude of the first charge
[tex]q_2[/tex] is the magnitude of the second charge
[tex]r[/tex] is the distance between the two charges
This formula gives only the magnitude of the force. The direction depends on the relative sign of the two charges:
- If the two charges have same sign, the force is repulsive (this corresponds to a positive sign in the formula)
- If the two charges have opposite sign, the force is attractive (this corresponds to a negative sign in the formula)
