The electric force between two point charges each of the magnitude, q at a distance of r apart in air of permittivity is:
- q2/4ΟΞ΅r
- rΞ΅0/q2
- 4Οq/Ξ΅0
- qr2/Ξ΅0
-
\(\frac{q^2}{4\piΞ΅or^2}\)
The correct answer is: E
Explanation
According to Coulomb's law, the Force between two point charges \(Q_1\) and \(Q_2\) is equal to
F = \(\frac{Q_1 Q_2}{4\pi Ξ΅or^2}\)