The Electric potential due to a charge distribution on a surface is :
$\Phi \left ( x \right )=\int \frac{\sigma \left ( {x^{}}' \right )dx{}'}{\left \| x-x{}' \right \|}da$ I want to show that it's a continuous function everywhere.
[Physics] continuity of the electric potential due to a surface charge
electromagnetismelectrostaticsmathematics
Related Solutions
This problem can be solved by a conformal inversion, but the answer becomes more difficult when the plates are not infinitely far apart.
Quick sketch
Choose the point of inversion to be the contact point of the plane and the sphere, the constant electric field at the left infinite half-plane becomes a dipole field right at the origin, while the sphere becomes a second parallel plane.
The method of images then produces reflected dipoles with the same orientation, with a uniform spacing along the x axis. The field is given by a superposition of these dipole fields, which allows you to solve for the electric field on the plane corresponding to the inverted sphere, and inverting back gives the charge.
The sum over the n-th image dipole is complicated before integrating for the total charge, but the total charge contributed by the n-th image on the right (plus its partner on the left) surprisingly falls off as $1/(n+1)^2$, making a sum which can be expressed completely in terms of $\zeta(2)={\pi^2\over 6}$.
The answer
The charge induced on the sphere when the plates are infinitely far away is equal to
$$ Q= (4\pi \sigma R^2) {\pi^2\over 12}$$
Where $\sigma$ is the initial charge density on the plates. The sphere acquires a charge which is about .82 of the charge it would have if it got a uniform density of charge equal to the density on the plates.
Electrostatic conventions
Normalize the electrostatic Laplace equation so that a point source has a field
$$ Q \over 4\pi r$$
With this convention, you can find the field of a dipole of magnitude D by superposing two opposite charges infinitesimally close, so that the answer is
$$ D x\over 4\pi r^3 $$
A conducting surface in this convention has a surface charge equal to that electric field outside coming in perpendicularly, so I will use $\sigma$ and surface $E$ interchangably.
Conformal inversion
In 3d, a solution $\phi$ to Laplaces equation has conformal dimension 1/2, which means that it can be inverted around the origin about a circle with radius a to give another solution to Laplace's equation:
$$ \phi'(x,y,z) = {a\over r} \phi({a^2 x \over r^2}, {a^2 y \over r^2}, {a^2 z\over r^2}) $$
The multiplying factor in front comes form the linear scale factor of an inversion, which is $a^2/r^2$ at a point a distance r, but with a square root to account for the 1/2 scale dimension of $\phi$.
The best way to prove that inversions are a symmetry is to check that the inversion takes a fundamental solution (1/r potential) to another fundamental solution at a different location. This means that inversion will move source points to source points under inversion, and keep source-free the images of those regions where Laplace's equation is satisfied source-free.
When you have a conductor, you can add a constant to the potential to make it be at zero potential. In this case, it stays a conductor under the inversion formula given above (it is easy to check that surfaces with $\phi=0$ are transformed into new surfaces with $\phi=0$). The inversion of a charged spherical conducting shell is another charged spherical conducting shell, or a conducting plane, in the case that the sphere passes through the origin of the inversion. If you invert a plane in a point on the plane, you get the same plane back.
When you invert a conductor at zero potential, the surface charge distribution is changed. The electric field is still perpendicular to the transformed surface, because a zero-potential surface transforms to a zero potential surface, but the electric field is modified by the 3/2 power of the conformal scale-factor. The reason is that the electric field is the inverse distance between equipotential lines, and the distance is increased by the scale factor, while the potential itself is decreased by the square-root of the inversion factor.
Inverting a sphere in contact with a conductor
Let the radius of the sphere be R, and place the point of contact of the sphere and the plane at the origin. Invert the origin with an inversion radius 2R, and the plane conductor stays the same plane, and the sphere becomes another parallel plane at zero potential at a distance 2R along the x-axis.
The constant electric field E which was at left infinity before inverts into a point dipole of dipole moment $E(2R)^3$ sitting right on top of the origin. The problem is to determine the field of a charge between two conducting plates, touching one of the two plates, making a dipole with its image charge, right at this point.
This problem is simple by images. If you place a dipole with the same orientation at every point on the x axis where $x=2nA$, where n is an integer, it is easy to check that symmetry (and convergence) guarantee that the solution is constant potential on the conducting planes.
This infinite sum, when centered on the inversion image-plane of the sphere, gives the charge density of the surface.
Summing up the image dipoles at the location of the plane gives the surface charge
$$ \sigma(u) = 2E \sum_{n \;\;\;\mathrm{odd}} { u^2 - 2n^2 \over (u^2 + n^2)^{5/2}} $$
where $u={\rho\over 2R}= \tan({\theta\over 2})$, where $\rho$ is the distance from the origin of the plane, and $\theta$ is the angular location on the sphere, with $\theta=0$ the contact point of the sphere and the image plane, and the sum is over all positive odd numbers.
The surface charge on the sphere
If the charge density on the image plane of the sphere at distance $\rho$ from the origin is $\sigma(\rho)$, the charge density on the sphere at any $\theta,\phi$ coordinates (where $\theta=0$ is along the axis linking the origin to the inversion plane) is given by
$$ \sigma(\theta) = {\sigma(\rho)\over cos({\theta\over 2})^3 } $$
and the total charge on the sphere is given by
$$ Q = \int \sigma(\theta) d\cos\theta d\phi = 2\pi R^2 \int \sigma(\theta) d\cos(\theta) $$
This is a consequence of the inversion formula for electric fields.
The resulting infinite sum of integrals is
$$ Q = 4\pi R^2 \sum_{n=1,3,5,...} \sqrt{2} \int_{-1}^1 {(2n^2-1) + (2n^2+1) z \over [(n^2+1) + (n^2-1) z]^{5/2}} dz $$
Where z is $\cos(\theta)$, performing the integral is elementary (but tedious), and is easily done by a version of partial fractions, as follows:
$$ {A + Bz\over (C+Dz)^{5/2}} = {A- {BC\over D}\over (C+Dz)^{5/2}} + {{B\over D} \over (C+Dz)^{3/2}}$$
Where $A=2n^2-1$, $B=2n^2 +1$, $C= n^2+1$ and $D=n^2-1$. Once the integrand is decomposed this way, the elementary integral (including the $\sqrt{2}$ in front) over each part gives
$$ \sqrt{2} {3\over 2} {AD - BC\over D^2} ( {1\over (C+D)^{3/2}} - {1\over (C-D)^{3/2}}) + \sqrt{2} {2B\over D^2} ( {1\over (C+D)^{1/2}} - {1\over (C-D)^{1/2}}) $$
There is a nontrivial check on the algebra in the limit $D\rightarrow 0$, where the pole terms of the two parts must cancel, because there is no real divergence in this limit before the partial-fracton decomposition. Substituting for A,B,C,D is made easier by first noting that $C+D=2n^2$ and $C-D=2$, and the determinant like thing is $AD-BC= -6n^2$. The result simplifies absurdly (it must cancel the pole-parts at n=1, but the cancellation of the pole at n=0 is a surprise)
$$ {2\over (n+1)^2}$$
This essentially finishes the computation. The sum is
$$ \sum_{n=1,3,5,7} {2\over (n+1)^2} = {1\over 2} \sum_{n=1,2,3..} {1\over n^2} = {\pi^2\over 12}$$
And this multiplies the factor $4\pi R^2\sigma$, which is the charge on the sphere if it were uniformly charged with the same density as on the infinite plates to begin with.
Corrections for finite size of capacitor
When the second plate is not infinitely far away, the inversion process produces a small sphere with charge instead of a point dipole. The corrections here can be computed to leading order.
The problem is that you're ignoring the angular dependence of your probe point $\mathbf r$, and that is messing with your integral. If your probe point has cylindrical coordinates $(s,\phi,z)$ and your integration variables are $(s',\phi')$, then the distance between the two is $$\frac 1 {|\mathbf r-\mathbf r'|}=\frac{1}{\sqrt{s^2-2ss'\cos(\phi-\phi')+s'^2+(z-z')^2}}$$ by the cosine rule (draw it!). If you put this into your integral it will no longer vanish.
(A few pointers on the new integral: the new dependence on $\phi'$ is a bit more complicated. The standard practice is to change variables to $\varphi=\phi'-\phi$. This will leave a simpler denominator, and a factor of $\cos(\varphi+\phi)=\cos(\varphi)\cos(\phi)-\sin(\varphi) \sin(\phi)$ on the numerator. One of the two terms will vanish and the other will yield to a change of variables to $u=\cos(\varphi)$.)
Best Answer
The answer is not in general. There exist pathological charge densities for which the electric potential has discontinuities. Fortunately, these can always be ruled out on physical grounds: they usually have finite amounts of opposite charge infinitely close to each other, which allows for arbitrarily high electric fields, and therefore divergent energy densities and infinite configuration energies.
How does this work? Take two conducting plates and put their edges in contact. I'm quite fond of using the north and south hemispheres of a given sphere, separated at the equator (which has the advantage of not introducing any extraneous infinities), but for simplicity let's use two semi-infinite planes, in the $x,z$ plane and separated by the $x$ axis.
Now, connect both plates to a voltage source and set one to a potential $+V_0$ and the other to $-V_0$. The potential at the $x$ axis is (of course!) discontinuous, by construction.
If that makes you uneasy, consider spreading uniform but opposite surface charges $\pm \sigma_0$ on the $\pm z$ plates. Fix some positive coordinate $z_0$. Then if you zoom in more and more towards the plates (i.e. as you take $x\rightarrow 0$) the plates loom larger and larger, and when $x\ll z_0$ the $-$ plate looks so far away that the local potential has to be positive, and bounded away from zero. Similarly, if $z_0<0$ the potential close enough to the plates must be negative. Thus you have points arbitrarily near the $x$ axis which have potentials bounded away from zero, both positive and negative, so the potential is discontinuous there.
So why is this not a problem? In these pathological examples you are putting a lot of negative charge, in conducting surfaces, next to a lot of negative charge. This assumes that you have separated them with an insulator of zero thickness which can withstand large (albeit finite) potential differences (so infinite electric fields). This is just not physical: any dielectric will break down or spark as you approach these conditions. As soon as you introduce a finite separation, the discontinuity goes away.
More generally the electrostatic potential can have discontinuities. These are most usually considered as surface discontinuities, in which case they are known as dipole layers, which you can think of as infinitely thin capacitors set at constant potential. (Therefore, they must have two layers of opposite but infinite charge, which is again not physical.)
A real dipole layer would be something like a lot of polar molecules, all pointing the same way normally to the surface. We know that the potential inside is not discontinuous and that the electric fields inside are large but finite, but if you're outside then all you see are two plane layers of charge which look for all the world like they are setting up a discontinuous potential, so you might as well put that into your math and assume it is infinitely thin.