# Magnetic and Electric energy for AC fields

In "Applied Frequency Domain Electromagnetics" (here the page) there are these two equations for the computation of the magnetic and electric energies stored in a certain volume $$V_0$$:

• $$W_{e}= \frac{1}{4} \cdot \int_{V_0} D^* \cdot E \,\,\,dV$$
• $$W_{h}= \frac{1}{4} \cdot \int_{V_0} B^* \cdot H \,\,\,dV$$

Where E,D,H,B are phasors.

I do not understand:

• why there is "1/4" instead of "1/2";
• the meaning of "stored energy" in case of purely sinusoidal fields. I think that stored energy is the integral between 0 and infinite of instant power as function of time, but I do not know what the result represents in this case: should it be 0 because of zero mean value of fields? Or infinite because of their infinite duration in time?
• Usually if the energy density is something like $E^* \cdot E/4$ means there was an average over the oscillation cycles. Does a cycle-averaged energy make sense in your context? Jun 30, 2020 at 15:31
• Yes, it may have been evaluated on a oscillation period. Jun 30, 2020 at 15:35

As I mentioned, looks like the author averaged over the oscillation cycle. This is useful when the instantaneous energy is not important. The form of the cycle-average he uses, $$E^* E/2$$, appears as a 'trick' when dealing with complex oscillating fields. Since the energy is quadratic in the fields (and thus nonlinear) one should use: $$u = \frac 1 2 \langle\text{Re}(D)\cdot\text{Re}(E)\rangle$$ for the cycle averaged energy density because only the real part is physically measurable. When both fields evolve with the time factor $$e^{-i\omega t}$$, the cycle average is easy to calculate: $$\langle\text{Re}(D)\cdot\text{Re}(E)\rangle = \frac 1 T \int_T \frac 1 4 (DE + D^* E^* + D^* E + DE^*) \, dt$$ The terms $$DE$$ and $$(DE)^*$$ oscillate with twice the frequency while $$D^* E$$ and $$DE^*$$ don't oscillate at all, thus only the last two terms survive the cycle average. The result is $$\langle\text{Re}(D)\cdot\text{Re}(E)\rangle = \frac{ D^* E + DE^*}{4} = \frac 1 2 \text{Re}(D^* E)$$ and that's where the extra $$1/2$$ comes from.

This stored energy is the average energy the fields carry around time $$t$$. If the fields are purely oscillatory, the average is the same for all times. But if there's absorption or energy pumping into the fields the cycle-average increases over time. Integrating the instantaneous power will give you the instantaneous net energy in your interval. For instance, integrating from $$t = 0$$ to $$\infty$$ may give you zero net energy because on average the energy remains in the field (the same amount is traded back and forth, so the average change is zero).

• Hello, I have another question which is related to this topic. We have seen that in time domain the average electrical energy density is u=1/2 <D,E>, while in frequency domain it is u=1/4 <D*,E>. Is this true also for the energy stored in a capacitor? I have always seen that u = 1/2 CV^2? Does it become u = 1/4 C V^2 (with V = peak voltage value)? Jul 6, 2020 at 16:24
• @Kinka-Byo yes, some formulae also change. That's why some prefer to use the rms value of voltage/current. For the case of sinusoidal AC lines, the peak and rms values are related by $V_\text{rms} = V_\text{peak}/\sqrt 2$, and thus the formulae remains the same as in the static case if you agree to use rms values instead Jul 7, 2020 at 15:52
• perfect! Thank you very much Jul 7, 2020 at 15:58