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:$$I_0=fq=q\frac{\omega}{2\pi}$$
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1$\begingroup$ Could you add some references? $\endgroup$ – Tendero Jan 10 '17 at 15:01
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$\begingroup$ 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. $\endgroup$ – Frederic Thomas Jan 11 '17 at 12:33
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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 $I_0 = q\omega$.
