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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    $\begingroup$ Could you add some references? $\endgroup$
    – Tendero
    Jan 10, 2017 at 15:01
  • $\begingroup$ 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. $\endgroup$ Jan 11, 2017 at 12:33

1 Answer 1


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 $I_0 = q\omega$.


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