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I was reading a paper on electrohydrodynamics which has the following sentence (in my own words):

At the interface/boundary, the requirement of continuity of the tangential component of the electric field is described by: $$V_1=V_2$$

We know that continuity of tangential component of electric field implies $\mathbf n\times(\nabla V_2-\nabla V_1)=0$ where $\mathbf n$ is the normal to the interface and $V$ is the potential. I am unable to understand how this implies $V_1=V_2$. Thank you.

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Short answer, you can get the result in electrostatics, for simple (path) connected surfaces, if the electric field isn't singular on the surface.

How? For simple surfaces were can adjust the whole potential in one surface by a uniform amount and it doesn't change the electric field in that region. So pick a point A on the interface surface and why not assume the potentials match there? (If they didn't adjust the potential uniformly in that region until they do.)

Now, since we are in electrostatics, potential differences can be uniquely determined by integrating the electric field along a path between the two points. So given another point on the interface surface, there is a path between them that stays in the surface, so we can integrate along that path and get a potential difference. Since potential differences are unique, that is the potential difference. So potential differences between two points on a path connected surface are uniquely determined by the electric field components in the surface (the tangential ones). In your problem, the two potenetials agree on those, they agree on the potential differences, so they differed by at most a uniform constant.

If they differed by a uniform constant, then there would be a singular electric field at the surface pointing orthogonal (by singular in strength) to the surface. This is because you get a finite change in potential in a nonfinite amount of change in the z direction.

The surface also has to be simple enough that it locally has tangential components, so don't have a cone with a sharp point for instance.

So it works in electrostatics for simple surfaces that are path connected surfaces, if the electric field isn't singular on the surface.


Does setting $V_1=V_2$ remove the freedom to have a normal electric field component?

Not at all. It removes the freedom to have the normal component of the electric field be infinitely large on the infinitely thin interface surface. The electric field can have a normal component and the normal component can be discontinuous, it just can't be infinite. Consider the potential $V(x,y,z)=az$, it is continuous across the surface $z=0$ yet the electric field is normal to the surface. Now consider the potential $V(x,y,z)=a|z|$, it is continuous across the surface $z=0$ yet the electric field is normal to the surface, and this time the normal component of the electric field is discontinuous. Now consider the potential $V(x,y,z)=a|z|+(bz/|z|)$. It is discontinuous across the surface $z=0$, so when you compute the gradient of $V$ on the actual surface itself you get an infinite electric field. This is completely unrelated to the electric fields on either side of the surface, because $V(x,y,z)=a|z|+(bz/|z|)$ and $V(x,y,z)=a|z|$ have the exact same electric fields on either side of the surface $z=0$, they only have different fields on the surface itself.

Can you elaborate on when $\hat n\times\left(\vec{\nabla} V_2-\vec{\nabla} V_1\right)=\vec{0}$ implies $V_1=V_2$?

To be totally honest, it doesn't imply it at all, you need to assume that the electric field on the surface is not infinite. However I still worry that you don't understand the statement actually being said, so the following argument should hopefully explain what it is that we are showing.

Assume the surface has at least one point, P, where there is a normal vector to the surface. Assume there is a potential from side 1 of $V_1(P)$ and there is a potential from side 2 of $V_2(P)$, these might actually be different. If so, then the electric field right on the surface has an infinitely large component normal to the surface. If the electric field right on the surface does not have an infinitely large component normal to the surface, then $V_1(P)=V_2(P)$, so they agree at that point. We could apply this argument at every point on the surface so really the result $V_1=V_2$ has nothing to do with whether $\hat n\times\left(\vec{\nabla} V_2-\vec{\nabla} V_1\right)=\vec{0}$. However, we can use $\hat n\times\left(\vec{\nabla} V_2-\vec{\nabla} V_1\right)=\vec{0}$ and $V_1(P)=V_2(P)$ to show that $V_1=V_2$, and this will show what we mean by $V_1=V_2$, which is that $V_1(Q)=V_2(Q)$ for every Q on the surface.

Now consider another point on the surface Q. Assume there is a path from $P$ to $Q$ that stays entirely on the surface. Then you can compute the potential difference $V_1(P)-V_1(Q)$ by integrating $\int_P^Q\vec{E}\cdot d\vec{r}$ entirely by staying just a bit on side1. Similarly you can compute $V_2(P)-V_2(Q)$ be integrating $\int_P^Q\vec{E}\cdot d\vec{r}$ entirely by staying just a bit on side2. So each integral $\int_P^Q\vec{E}\cdot d\vec{r}$ started and stopped on the surface and stayed entirely on the surface so $d\vec{r}$ was entirely tangential, so $\vec{E}\cdot d\vec{r}$ only used the tangential part of $\vec{E}$. But the two potentials have the same tangential part of $\vec{E}$ since $\hat n\times\left(\vec{\nabla} V_2-\vec{\nabla} V_1\right)=\vec{0}$. Thus $V_1(P)-V_1(Q)$=V_2(P)-V_2(Q)$. If $V_1(P)-V_1(Q)$=V_2(P)-V_2(Q)$ then we can rearrange to bring all the Ps to one side and all the Qs to the other side and get $V_1(P)-V_2(P)=V_1(Q)-V_2(Q)$. But the left hand side is zero since $V_1(P)=V_2(P)$, so the right hand is zero, so $0=V_1(Q)-V_2(Q)$, so $V_1(Q)=V_2(Q)$.

So all we needed was a path from a single point to a single other point, and one place where the electric field wasn't infinite.

But having the electric field not be infinite is all you really needed, since that by itself gets you continuity at any point. Unless you want infinite electric fields you always want the potential to be continuous everywhere and in all directions.

All that is meant by $V_1=V_2$ is that they agree on the surface. Each one is only defined on their own side of the surface, so the surface is the only place where both exist and that is where we insist they are equal. And we insist they are equal for reasons unrelated to the tangential component of the electric field. Having electric fields with equal tangential components is actually a consequence of having the potentials agree on the surface. So maybe it is better to think that $V_1=V_2$ on the surface implies that $\hat n\times\left(\vec{\nabla} V_2-\vec{\nabla} V_1\right)=\vec{0}$ on the surface.

To summarize, you have it backwards. $V_1=V_2$ on the surface implies that $\hat n\times\left(\vec{\nabla} V_2-\vec{\nabla} V_1\right)=\vec{0}$ on the surface.

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  • $\begingroup$ I think this can be proved nicely using complex analysis? $\endgroup$
    – Ali Moh
    Apr 11 '15 at 20:32
  • $\begingroup$ I didn't understand much by your answer (and would appreciate if your could elaborate) but setting $V_1=V_2$ removes the freedom to have a normal electric field component. $\endgroup$
    – zed111
    Apr 12 '15 at 11:09
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It doesn't imply that directly, you still have gauge freedom. I think what they mean is that if you set V1=V2 at one point on the boundary, it will hold true for all the boundary because the tangential component is identical.

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