The electric force between two point charges each of the magnitude, q at a distance of r apart in air of permittivity is:
A
q2/4πεr
B
rε0/q2
C
4πq/ε0
D
qr2/ε0
E
\(\frac{q^2}{4\piεor^2}\)
Verified Correct Answer: E.
Physical Principles & Numerical Solution
According to Coulomb's law, the Force between two point charges \(Q_1\) and \(Q_2\) is equal to
F = \(\frac{Q_1 Q_2}{4\pi εor^2}\)
Source: Myschool Past Questions Archive (Record #23928)