# Radiation of Electric Dipole - Antenna

In the theory regarding Radiation of an electric dipole, an example using an antenna is shown in which the antenna is considered as a certain length of wire with a sinusoidal current inside it. For the following calculations the relationship below is used: $$I_0=q\omega$$ I do not understand this formula. For me, the current would be given by:$$I_0=fq=q\frac{\omega}{2\pi}$$

• Could you add some references? – Tendero Jan 10 '17 at 15:01
• I would suggest that you show the full calculation in order to see $I_0=q\omega$ in the context. It is possible that it means something completely different to what you suggested. But from 3 letters it cannot be guessed. – Frederic Thomas Jan 11 '17 at 12:33

If $I(t) = I_0 \cos \omega t = dq/dt$, then integration leads to $$Q(t) = \frac{I_0}{\omega} \sin \omega t$$ (with a constant of integration that can be chosen to be zero).
If we now define $q$ as the amplitude of the time-varying charge, then it is clear that $I_0 = q\omega$.