# Can electric potential be discontinuous?

I am studying the following problem, which is 9.11 of "Modern Electrodynamics" by Zangwill.

The idea is that you have a wire attached to a perfectly conducting sphere (radius $$a$$) buried into the ground (where the Earth has conductivity $$\sigma_E$$); in between, there is a layer of conductivity $$\sigma_2$$ (perhaps the Earth was churned up or something, resulting in different electrical properties). We need to find the resistance between the end of the wire and a point deep (i.e. infinitely far away) in the ground. I understand that we need to find basically the voltage, that is, the potential difference between the end of the wire and the point deep in the ground. I'm imagining a current $$I$$ through the wire that spreads over the surface of the conducting sphere, and from there, radially into the ground. The voltage is constant over the sphere's surface, which simplifies things. Then we say, on one hand, that the current $$I$$ spreads uniformly over the hemisphere below the ground at radius $$r$$: $$j = I/2\pi r^2$$. On the other hand, in any region,

$$\vec{j} = \sigma \vec{E} \Longrightarrow j = -\sigma \frac{\partial \phi}{\partial r}$$

since all the current is radial. Putting these together, one would think

$$\phi(r) = \frac{I}{2\pi r \sigma}.$$

My issue is that $$\sigma$$ is obviously discontinuous since $$\sigma_2 \neq \sigma_E$$ in general. This implies a discontinuous electric potential. I learned that potential is always continuous in electrostatics, and indeed precisely because the voltage is an integral of electric field, the voltage should be continuous even across dielectric boundaries, unless there's an infinite jump in $$\vec{E}$$. Is this true? If not, why is a discontinuous potential acceptable?

Note: The solution manual for Zangwill agrees with this method of doing the problem, and treats the discontinuous potential without comment.

Edit: The text of the solution as I have it reads as follows: • How did you get from $j=-\sigma\frac{\partial \phi}{\partial r}$ to $\phi(r)=\frac{I}{2\pi r \sigma}$? – The Photon Dec 23 '18 at 17:47
• My idea was to set the two expressions for $j$ above to be equal, and then take an integral: $j = I/2\pi r^2 = -\sigma \frac{\partial \phi}{\partial r}$, then the integral of the middle is $-I/2\pi r$, so you get $\phi(r) = \frac{I}{2\pi r\sigma}$. I guess I'm implicitly integrating from $\infty$ to $r$... but I don' think that introduces any issues. – flevinBombastus Dec 23 '18 at 18:01
• I think the issue is you can't take an integral from infinity to r in region 2, because region 2 does not extend to infinity. You have to take the integral from the boundary with region E, and use the solution for the potential in region E and the boundary conditions to get the c.o.i. for the integral in region 2. – The Photon Dec 23 '18 at 18:19
• Since we can write $\sigma$ in a piecewise fashion, why couldn't we write the integral in a piecewise fashion also? – flevinBombastus Dec 23 '18 at 18:27
• You can, that's another way of saying the same thing I suggested. You won't get the result that $\phi(r)=\frac{I}{2\pi r \sigma_2}$ in region 2. – The Photon Dec 23 '18 at 18:29

First, since

$$V_{ba} = \int_b^a -\vec{E}\cdot{\rm d}\vec{\ell}$$

if the E-field is finite, there can be no discontinuity in the potential.

Second, the boundary value that will actually help solve your problem is derived from Gauss' Law (as shown here). The condition is that the normal component of E-field at the interface between two materials has a discontinuity proportional to the sheet charge present on the interface.

There will be a sheet charge on the surface of your perfectly conducting sphere, so that discontinuity will appear there.

But at the interface between your finite-conductivity regions, the sheet charge will be determined by the polarizability of the two materials, and the boundary condition will be that the normal component of the displacement field (usually denoted $$\vec{D}$$) is continuous. This is shown here.

• I see what you are saying, that $\vec{D}$ has the discontinuity proportional to the free charge on the interface. However, shouldn't that still give a continuous potential? – flevinBombastus Dec 23 '18 at 18:12
• Yes, it will still give a continuous potential. There will be a discontinuity in the slope of the potential, plotted against r. – The Photon Dec 23 '18 at 18:13
• So this doesn't quite answer my question, right? The potential here appears to have $\sigma$ in it, which is discontinuous at the interface, making $V$ itself discontinuous. I agree that what you've said above is a proof that $V$ is continuous; I guess I don't see what I'm doing wrong in the above. – flevinBombastus Dec 23 '18 at 18:16
• Correct, I'm not trying to answer the whole problem, just get you on the right track to solve it yourself. – The Photon Dec 23 '18 at 18:16
• I really appreciate the discussion. But it feels like we're back at the same place, because your answer seems to show that $V$ is continuous, while the $V$ of the solution is discontinuous, and I don't see any good reason why that is. I'm not really sure of any other way to calculate $V$, since there's current through the space in which I need $V$, preventing use of Laplace's equation or Poisson's equation. – flevinBombastus Dec 23 '18 at 18:25