# Why is the work through the element the same as the potential difference if we define potential difference this way?(lumped elements)

In the Feynman Lectures on Physics part ||, chapter 22-3 he defines potential differnce like this:

The picture is an element in a curicuit, the black lines on top and the bottom are the conductors, and the red lines are hypothetical lines.

He assumes that the induced magnetic field stays inside the elements (lumped element model). Outside the elements then Faradays law gives us that $$\nabla \times E = 0$$. So Stokes theorem tells us that $$\int E dl$$ is independent of the path from A to B if we integrate outside the element (the red lines are hypothetical lines outside the element), so we can define the potential this way ($$\int E dl$$ on any line outside the element).

But it seems very natural that the work that is done to the charge when moving through the element should be the potential difference. But how can we argue this? Strictly speaking we only have that this would be the work if we move the charge outside the element. What is the argument that this is also the work when the charge is moved inside the element from A to B?

The way to see this is through the application of Poynting’s theorem with the appropriate simplifying assumptions. This is shown in section 11.3 here:

http://web.mit.edu/6.013_book/www/book.html

With the circuit theory (lumped element) assumptions there exists a 2D closed surface $$a$$ such that on that surface the contribution of the magnetic induction to E is negligible as is the contribution of the displacement current. Then, with that simplification, the total power going into that circuit element is $$P=-\int_a \vec S \cdot d\vec a=-\int_a \Phi \vec J \cdot d\vec a = \Sigma v_i i_i$$

So the point is that we are not tracing the energy over the path of any electron. Indeed, the energy is not carried by the electrons so such path tracing is not relevant. We are tracing the energy that flows through the fields surrounding the lumped element. Under the circuit theory assumptions, that energy is equal to $$\Sigma v_i i_i$$.

Once the energy flows through $$a$$ we no longer know if the fields inside $$a$$ are conservative. But that is not required because we are only concerned about the currents, voltages, and power entering and leaving $$a$$.

• Thank you very much! The way you do it is probably the correct way to do it, but Feynman introduces Poynting much later, but in the same chapter as I talk about above he introduces power like $RI^2$, maybe he uses some sort of conservation of energy law, but it is not easy to understand when the potential is defined outside the path the charges actually take. Commented Sep 12, 2021 at 16:01
• But the point is that the path that the charges take is not important in this sense. The charges don’t carry the energy, the fields do. And even though the charges don’t follow that path, the fields exist all along that path and energy is transferred to the lumped element through the fields all across that boundary.
– Dale
Commented Sep 12, 2021 at 17:03