Consider a point charge $q$ at $(0,0,0)$, the potential at $\bf{r}$ is
given by $V(\bf{r})$ $= \frac{q}{4\pi\epsilon_0r}$. If you consider a
path through $(0,0,0)$, you encounter a discontinuity in electric
field and the direction of field changes. Is the potential for the
field of point charge discontinuous at the location of the point
charge?
If you specify the potential as a function from real numbers to real numbers then the function is discontinuous at zero by virtue of the fact that it is not defined at zero. I.e., the definition of continuity
$$
\lim_{r\to 0}V(r)\to V(0)
$$
does not hold because V(0) is not defined (it is not a real number).
I'm facing this confusion because, in the book by Griffiths, it is
mentioned that the electric field is discontinuous at the surface of
charge, with charge density $\sigma$ i.e
$E_{above}-E_{below}=\frac{\sigma}{\epsilon_0}\hat{n}$. It is also
mentioned that the potential is continuous across a boundary like
that. Is my analogy between point and surface charge wrong?
The analogy seems somewhat imperfect to me because both $E$ and $V$ go to infinity in the first example, but neither does in the second example. If E remains finite, phi will be continuous (as discussed below).
Also, is potential for an electric field continuous everywhere? Or is
it continuous across field discontinuities?
Typically $\phi$ is continuous since, for example, if $\phi$ had a discontinuity $\Delta\phi$ at some point $x_0$ then
$$
\Delta\phi=\lim_{\delta\to 0}\phi(x_0+\delta)-\phi(x_0)=\lim_{\delta\to 0}\int_{x_0}^{x_0+\delta}\frac{d\phi}{dx}dx=\lim_{\delta\to 0}\int_{x_0}^{x_0+\delta}E_j(x)dx
=\lim_{\delta\to 0}E_j(x_0)\delta\;,
$$
which is zero unless E(x_0) blows up. I.e., for finite electric fields (even discontinuous finite electric fields) the potential remains continuous.
But, remember, in general, the potential does not have to be continuous everywhere (as evidenced by the case mentioned above of a point charge). This is because the governing equation for $\phi$ is
$$
\nabla^2 \phi=-\rho/\epsilon_0\;,
$$
where $\rho$ is the charge density. And, I'll bet that you can come up with some wacky charge densities (a single point charge is one case) where $\phi$ is discontinuous.
Charges could be distributed along only one side of the infinite conducting plate, with one of two Cases:
- The conductor has no thickness, you just have a theoretical distribution of charges on a plane, or
- The conductor has thickness, and the charges have to be perfectly straight for electrostatic equilibrium.
Since you said the E field could "go through the sheet", I'm gonna assume you meant Case 2.
In Case 2, you could try to keep all charges in only one plane. However, any charge not perfectly in line with the other charges will cause other charges to move, breaking the electrostatic equilibrium. So we call the configuration of charges along only one side of the conductor an unstable equilibrium.
Generally we don't consider unstable electrostatic distribution of charges as being under electrostatic equilibrium. For example, the excess charges on a conducting sphere at electrostatic equilibrium is always on the outer surface of the sphere, even though having all charges in the exact center of the sphere is also a valid (but unstable) electrostatic equilibrium.
My above reasoning sounds sketchy, and could be wrong. But I was just trying to get you a feel of the Uniqueness Theorem. If you can understand the math, you will know why there is only one possible distribution of charges for a conductor in electrostatic equilibrium.
So by the Uniqueness Theorem, the excess charges are distributed on both sides of the infinite conducting plate.
A positive test charge on the other side feels a repulsion away from the conducting plate only because of E field generated by excess charges on the side closer to the positive test charge.
The E field generated by excess charges on the further side do not affect the positive test charge. This is because within the conductor, the E field generated by excess charges on the further side is cancelled out by the E field generated by excess charges on the closer side.
Thus there is no net E field within the conductor. You can also prove there is no net E field in the conductor by drawing a Gaussian surface inside the conductor (excluding the surface charges). Since there is no net charge enclosed, electric flux, and thus electric field, is zero.
P.S. you don't need to put the positive test charge near the infinite conducting plate for repulsion. The E field is uniform, thus does not vary with distance from the plate. A positive test charge very far away will experience the same repulsion.
Best Answer
I think there is a good conceptual question here. I'm going to try to pitch this explanation at a level that is probably slightly higher than what is covered in AP Physics C, but that hopefully you can follow (let me know if not). This is an area where pursuing your question can lead you deeper into the subject than what you are expected to know on the test.
In a sense you are right. If you remove the sheet itself at the origin, then you can think of the two sides of the sheet separately. The potential is only defined up to a constant, so you could add different constants to the two sides, and the potential would not be continuous at the sheet. But the field would be the same everywhere on either side of the sheet, as it would be if the potential were continuous.
So, clearly, grokking the difference between these cases relies on understanding what's happening exactly at the sheet, which is quite a singular place and tricky to get a handle on
One physical way to understand what is going on is to imagine fuzzing out the infinitely thin sheet into a slab with a finite thickness. It is possible to solve for the field and potential inside a slab, although I don't remember if you do it in AP Physics C or not. The field continuously goes to zero at the center of the slab, and then changes direction. Because the field is continuous, the potential is also continuous (and differentiable). Then, in the limit that the slab becomes infinitely thin and becomes a sheet, the potential will remain continuous. Since the infinitely thin sheet is really just an idealization for a charge distribution with a small-but-finite thickness, we should expect on physical grounds that the potential of a sheet should be continuous, based on this limit.
We can also probe the sheet more directly, at the cost of some mathematical abstraction. In particular, the underlying mathematical fact here is that the derivative of a discontinuous function is more singular at the discontinuity, than the derivative of a function with a non-differentiable kink is at the kink. This can be made precise in terms of the Dirac delta function, which I suspect you haven't met but you will if you continue studying physics. The derivative of a step function is a delta function. The delta function is a "function" (actually a distribution, but that's a detail at this level) which is zero everywhere except a single point, where it is infinitely large. If the potential is discontinuous at a point, then the electric field will be proportional to a delta function at that point. This means that there would be an infinitely strong electric field at the surface of the sheet, which is a singularity pointing to a breakdown of Maxwell's equations.
Even without knowing about Dirac delta functions, you can detect the difference between a continuous and a discontinuous potential, by integrating the electric field, as suggested in the comments by hypnortex. If you integrate the electric field of a sheet, you will get a continuous potential. In order for the integral to be discontinuous at the origin, the field must be infinitely large there.