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Qmechanic
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How is voltage defined when E$\vec{E}$ changes?

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Sam Gallagher
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How is voltage defined when E changes?

Short version: If voltage is defined for conservative vector field $\vec{E}$ only, at what point in a changing electric field does voltage become undefined?

Long Version: Voltage is typically defined as the change in potential energy between two points in an electrostatic field. $$ V = -\int_C \vec{E}\ \circ d\vec{l}$$ This brings about the question of the existence of a function $\phi$ such that $\nabla \phi = \vec{E} $, and in order for this to be true, we require that: $$ \frac{\partial E_x}{\partial y} = \frac{\partial E_y}{\partial x} $$ $$ \frac{\partial E_x}{\partial z} = \frac{\partial E_z}{\partial x} $$ $$ \frac{\partial E_y}{\partial z} = \frac{\partial E_z}{\partial y} $$ In addition to this requirement, we see that: $$ \nabla \times \vec{E} = \vec0 $$ $$ \oint_C \vec{E} = 0 $$ However, these requirements are broken if $\vec{E}$ is changing, as $\vec{E}$ is no longer static, and an associated magnetic field affects the electric field by supplying a curl component.

I am an electrical engineer, and I've studied a fair amount of microwave engineering enough to know that waveguides which support TE and TM modes of electromagnetic wave transmission have 'voltages' and 'currents' which are defined in a different way from those we use in circuit theory. However, we continue to use voltage and current with TEM waves, low frequency design, and RF design. These voltages cannot be a potential function of $\vec E$, so the natural question is, how are they defined?

A simple resolution is that they are defined for quasi-static fields, i.e. the fields change slowly so we have no issue. Another resolution is a different definition of voltage.

My question is, at what point (in terms of amplitude, frequency, slew rate) in circuit theory does $\nabla V \neq \vec{E}$ become a problem for calculations involving voltage defined by potential energy in an electrostatic field?