# Is poyinting vector theorem somehow related to PE and KE?

I'm trying to understand a DC circuit with only battery and light bulb component. In this scenario, electric potential energy as electrons move from negative to positive terminal is converted to kinetic energy of the electrons. So as they move more, their kinetic energy increases. So, we have an energy transfer because electrons when they move and their kenetic increases, the energy actually can be seen as in "transfering mode".

While this is clear to me, I learned now that there exists poynting vector theorem and we also have it here. While it applies to EM waves, we have it in DC circuit as well and this energy doesn't transfer through wire.

Is this energy($$\frac{1}{\mu_r} ExB$$) completely different from potential and kinetic energy or is it related to them somehow and would we have this energy even if electrons in a circuit travel with constant speed(though I'm not sure how they'd travel with constant speed since each time potential energy transforms into kinetic energy so kinetic energy must be increasing ?)

• I suppose PE and KE are potential and kinetic energy? Maybe you should clarify in your question. Jun 16 at 12:02
• Yes.. true, they're. updated now Jun 16 at 12:13

In its most basic form, Poynting's theorem states that for any volume $$\mathcal{V}$$ (and assuming that there are no charges entering or leaving $$\mathcal{V}$$), you have $$\frac{d}{dt} {\text{KE of charges} \choose \text{in } \mathcal{V}} + \frac{d}{dt} {\text{PE of charges} \choose \text{in } \mathcal{V}} + {\text{flux of } \vec{S} \choose \text{out of } \mathcal{V}} = 0$$ So the flux of the Poynting vector tells you the rate at which the total energy (kinetic plus potential) is changing. If $$\vec{S}$$ has a net flux out of your volume, then the total energy in your volume is decreasing; if $$\vec{S}$$ has a net flux out, then the volume is decreasing. The assumptions that go into this are that the only forces acting on the charges are the Lorentz force due to an $$\vec{E}$$ and $$\vec{B}$$ field, and that Maxwell's equations hold.
• if you're envisioning a volume that contains that battery but not the load, then charges will be leaving and entering $$\mathcal{V}$$.
But these extensions can be made, and they lead to basically the same conclusion: the rate of change of total energy inside $$\mathcal{V}$$ (taking into account external sources of energy and charges entering & leaving the volume) is strictly related to the flux of $$\vec{S}$$ through the surface of $$\mathcal{V}$$.