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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?

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  • $\begingroup$ A thought: We could still define voltage as if the electric field wasn't changing, and I believe this is the case when looking at the electric field in terms of the scalar and vector potentials. But I'm not sure the implications of this on theoretical circuit design. $\endgroup$ – Sam Gallagher Jul 24 '18 at 0:50
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To answer your question, I am going to directly cite the definition of potential difference given in Engineering Electromagnetics by Hayt:

"..., we now define potential difference $V$ as the work done (by an external source) in moving a unit positive charge from one point to another in an electric field," $$ {\rm Potential} \,\, {\rm difference} = V = - \int_{{\rm init}}^{{\rm final}} {\vec E \cdot d\vec L} $$ As you can see, the definition is general in the sense that is valid for all electric fields, even non-conservative ones. In the special case where the electric field does not depend on time (is static), it exists a potential function $\varphi$ such that: $$ \vec E = \nabla \varphi $$ Furthermore, it can be shown that $V = - \varphi$, so for static electric fields: $$ \vec E = - \nabla V $$ Hence the name of potential difference. The main interest of this property of static electric fields is that it makes possible to calculate $\vec E$ given $V$ (and is easier than calculating $V$ given $\vec E$). However, in the general case of time-dependent electric fields, the potential function $\varphi$ is not guaranteed to exist. The line integral definition of $V$ is still valid, but the name potential difference does not make sense anymore; citing the book of Hayt textually: "In electrostatics, the line integral leads to a potential difference; with time-varying fields, the result is an emf or voltage."

One clear conclusion is that the voltage at a specific point depends on the path of integration chosen from the reference point. This is clearly seen by considering Faraday's law: $$ \oint_L {\vec E \cdot d\vec L} = - \int_S {\frac{{\partial \vec B}}{{\partial t}} \cdot d\vec S} = {\rm emf} $$ Where the closed path $L$ is the boundary of the surface $S$. Thus, if we change $L$, we change $S$, and the value of the integral changes. This seemingly illogic conclusion that the same point can have different voltages depending on the path to the reference can be understood by studying the emf in Farady's law. This emf is the voltage drop of the inductance of the integration path, so if we change the path, the inducatance changes, and the emf changes. From a circuit analysis perspective, this inductance is part of the circuit itself, so when we change the integration path, we change that inductance and we affect the operation of the whole circuit, in consequence the voltage at the considered point changes.

Finally, my answer is: you can use voltage for all your circuit calculations, you just have to include the effect of the path-dependent inductance.

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