The following problem is on page 67 of Zangwill's electrodynamics:

An origin-centered spherical shell [of radius $a$] of infinitesimal thickness has uniform surface charge density $\sigma = {Q}/{4\pi\,a^2}$. A small hole of radius $b\ll a$ is drilled in the shell at the point $\mathbf{R}$. Find the electric field at all observation points r where $|\mathbf{r} − \mathbf{R}|\gg b$.

You can see this problem and solution on the previously stated page here.

I want to clarify my understanding/check if my understanding is correct. Let $\Theta$ be the step function. Zangwill obtains $$\mathbf{E}(r)=\frac{Q}{4\pi \epsilon_0}\frac{\hat{\mathbf{r}}}{r^2}\Theta(r-a)\,+\,\frac{Q'}{4\pi \epsilon_0}\frac{\mathbf{r}-\mathbf{R}}{|\mathbf{r}-\mathbf{R}|^3}\tag{1}\label{a}$$ with $Q'=-\pi b^2\sigma$.

My understanding:

Firstly, $\sigma = {Q}/{4\pi\,a^2}$ is just a consequence of Gauss' law and is obtained by using the spherical symmetry. Then $Q'=-\pi b^2\sigma$ is the "cut-out" of the sphere (which is why there is a $-\sigma$ term) and the $\pi b^2$ is simply the area of this cut-out. Now the term $\dfrac{\mathbf{r}-\mathbf{R}}{|\mathbf{r}-\mathbf{R}|^3}$ is a consequence of Coulomb's law, which tells us that the electric field is proportional to the inverse square of the distance in between the point charge and the observation point in the direction of the line of charge $\mathbf{r}-\mathbf{R}$. This deals with the term on the right hand side of \eqref{a}. As for the term on the left I am mostly unsure about the step function. The $\dfrac{\vec{\mathbf{r}}}{|\mathbf{r}|^2}=\dfrac{\vec{\mathbf{r}}}{r^2} $ is just as before, a product of Coulomb's law as with the proportionality constant. Then the field $\mathbf{E}$ is due to the superposition principle: we are taking an element of the surface away from the sphere (by adding the left piece of \eqref{a} to the right piece of the same equation) and therefore account for the hole.

So what is the step function doing there? Is it from one of the "much greater than conditions"? Any help would be much appreciated.


1 Answer 1


If you just have a spherical shell of constant charge density, there will be no electric field on the interior. You can see this by applying Gauss' Law to a sphere inside your shell,

$$\int_\text{over the sphere} E \cdot dA \simeq \int_\text{inside the sphere} dq = 0$$

But the same equation, for spheres outside the spherical shell, will yield $Q$ for the second integral. In other words, you have electric fields outside the spherical shell, but nothing on the inside. That's why you have the step-function.

On the other hand, the second term in your equation (1), comes from the little hole that you drilled in the shell (which you think of as an additional point-charge of the opposite sign); which contributes an electric field everywhere, both inside and outside the spherical shell.


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