# Change in Electric Field inside a conductor

I was reading Purcell's E&M and the author was showing how the force on the charge distribution per unit area for a thin spherical shell with surface charge density of $$\sigma$$ is proportional to the average of the electric field just above and just below the surface. He tried to show something more general to prove it. He wanted to show that the thin slab of thickness $$dx$$ in the diagram has force per unit area proportional to $$(E_1 + E_2)/2$$ where $$E_1$$ and $$E_2$$ are electric fields just to the left and right of the slab respectively. The diagram is a picture of a cross section near the surface of a charged object (like maybe the thin shell itself). Here he uses $$E_2 - E_1 = 4 \pi \sigma = 4 \pi \rho dx$$ which comes from applying Gauss's law near the slab (cgs units). My problem with this is that he is claiming that the electric field changes by $$4 \pi \sigma$$ as we move from left to right. But that's only if we count the electric field for the charges enclosed by the Gaussian surface. What about the charges outside the Gaussian surface? I know that their flux will be zero but that doesn't mean that the electric field due to them at $$x$$ and $$x+dx$$ will be the same.

• The expression $E_2-E_1=4\pi\rho dx$ Is wrong and not mention in the book. It is $dE=4\pi\rho dx$ that is a true expression. – Young Kindaichi Oct 22 '20 at 4:06
• Yes I'm sorry for the confusion. By $E_1$ and $E_2$ I don't mean the field outside at the two ends like in the diagram. I mean the field just outside the slab which is inside the surface. In that case dE is the same as what I mean by $E_2 - E_1$. I don't quite follow why $dE=4 \pi \rho dx$ is a true expression. Isn't this $dE$ only due to the charges inside the slab. How do I know that this change in electric field due to charges inside the slab is the same as the change in the total electric field? – Brain Stroke Patient Oct 22 '20 at 12:18
• $dE=4\pi\rho dx$ is derived from the gauss law. You should note that. – Young Kindaichi Oct 22 '20 at 12:28
• Yes that's why I said it's due to the charges inside the slab. Gauss law gives the electric field due to charge enclosed. Only in very special cases does it give the total electric field. – Brain Stroke Patient Oct 22 '20 at 12:40
• You confuse yourself with the gauss law all the time. Please note that it's never mentioned in gauss's law that the field we taking about is due to the charge encloses the volume.The quantity on the left was the total flux due to field,we don't care from where this field come from. – Young Kindaichi Oct 22 '20 at 12:50

Yeah, you are right Purcell didn't consider the possibility of charges other than the spherical shell. So you can do this it in a more general way without taking a spherical shell.

Suppose you have Electric field $$\mathbf{E}$$ in space due to some charge distribution. The electric field always undergoes a discontinuity when you cross a surface charge $$\sigma$$. In fact, it is a simple matter to find the amount by which $$\mathbf{E}$$ changes at such a boundary. Gauss's law says in SI units $$\oint_S\mathbf{E}\cdot d\mathbf{a}=\frac{Q_{enc}}{\epsilon_0}=\frac{\sigma A}{\epsilon_0}$$ Diagram from Griffiths, Introduction to Electrodynamics

where $$A$$ is the area of the pillbox lid. (If $$\sigma$$ varies from point to point or the surface is curved, we must pick $$A$$ to be extremely small.) Now, the sides of the pillbox contribute nothing to the flux, in the limit as the thickness $$\epsilon$$ goes to zero, so we are left with $$E^{\perp}_{above}-E^{\perp}_{below}=\frac{\sigma }{\epsilon_0}$$ The normal component of $$\mathbf{E}$$ is discontinuous by an amount $$\sigma/\epsilon_0$$ at any boundary.

The tangential component of $$\mathbf{E}$$, by contrast, is always continuous. For if we apply $$\oint \mathbf{E}\cdot d\mathbf{l}=0$$

or $$E^{||}_{above}=E^{||}_{below}.$$

Explanation For Purcell's idea

We consider a continuous change in the electric field going from one side to the other side. The figure shows how the density changes (figure shows a cross section of surface.) Now see how the electric field change going from $$x$$ to $$x+dx$$. $$E(x+dx)-E(x)=\frac{\rho(x) dx}{\epsilon_0}$$ $$dE=\frac{\rho(x) dx}{\epsilon_0}$$ $$\int_{E_1}^{E_2}dE=\int_0^{x_0}\frac{\rho(x) dx}{\epsilon_0}$$ $$E_2-E_1=\int_0^{x_0}\frac{\rho(x) dx}{\epsilon_0}$$

That's the exact equation that are used in the whole discussion.

• But Purcell's Gaussian surface did not enclose just part of the shell, it was inside the thin surface of the shell. – Brain Stroke Patient Oct 21 '20 at 8:55
• See Purcell's proof for the force in not general, He proof it specifically for the spherical shell, but you can extend this to other types of surfaces as I did. – Young Kindaichi Oct 21 '20 at 13:33
• You should probably cite the source of that diagram. I can't be the only one who recognized it from a very well-known electrodynamics textbook. – Michael Seifert Oct 21 '20 at 13:42
• Yeah of course, why not. – Young Kindaichi Oct 21 '20 at 13:43
• My question is not regarding the validity of that equation when you take the Gaussian surface to be covering part of the surface. I could just as well do it with your example. Suppose your arbitrary surface has some thickness $t$ and I take the Gaussian surface of thickness $dx$ inside your surface. If the volume charge density inside is $\rho(x)$, then for sufficiently small dx, Purcell said $E_1 - E_2 = 4 \pi \rho dx = 4 \pi \sigma(x)$ . Purcell didn't do it for a specific spherical shell case. He did it for the general case and then came back to the spherical shell as an example. – Brain Stroke Patient Oct 21 '20 at 15:26

I don't know whether this would help... But this is how I kind of figured it out and skipped when I read that.

If we consider Gauss's law in differential form :

$$\vec \nabla \cdot \vec E = \frac{\rho} {\varepsilon_0}$$

and Imagine the slab to be broad enough that variations in $$E$$ about $$Z$$ and $$Y$$ are negligible, Then the equation becomes :

$$\frac{\partial E} {\partial x} = \frac{\rho} {\varepsilon_0}$$

$$\frac{E(x+dx) - E(x) } { dx} = \frac{\rho} {\varepsilon_0}$$

EDIT Also, I don't find any inconsistency when applying gauss's law directly. Could you please clarify?

I assume the electric Field from the slab is along X alone partly due to symmetry and partly because it is an element of a broad slab.

Let $$\vec E(x)$$ be the net electric field inside the slab (due to charges outside and inside).

We know that $$\int \vec E\cdot d\vec s= \frac{q_{in}} {\varepsilon_0}$$ is valid if we consider or not consider Electric field from outside charges.

So we can very well say that : $$E(x+dx) \cdot A - E(x) \cdot A + \text{flux from sides} = \frac{\rho \cdot A dx}{\varepsilon_0}$$

And for a slab of limiting thickness, flux from sides can be neglected and A can be cancelled to get the result.

Here I believe I did consider contribution to $$\vec E$$ from all charges.

• If $x = 0$ is the coordinate of one end of the surface, then suppose at $x=0$ there is a point charge on the surface. Then on the Gaussian surface at $x$ or $x+dx$, it's electric field isn't constant (because it spreads out from the point charge and so is constant only over a spherical surface centered on the point charge). But when you write $E(x) \cdot A$ or $E(x+dx) \cdot A$, you're assuming that the field is constant over the surface so that the flux is just the field times the area. – Brain Stroke Patient Oct 23 '20 at 15:22
• That certainly works if the field is due to inside charges only due to symmetry like you said, but I don't understand why the net electric field will also be constant over that surface because of that example with point charge at $x=0$ I gave. That is pretty much the source of all my confusion. – Brain Stroke Patient Oct 23 '20 at 15:23
• Yes that's right. I am assuming the field to be perpendicular to the surface ( It being constant is okay because A is small). You know that right... We apply gauss law only in case of symmetries. If we have a point charge at origin which changes my perpendicular field assumption, then It would not work. Adding that charge destroyed the symmetry. But such a charge would not exist here right... charges would be uniformly distributed over the surface, giving us a uniform electric field (for small A) . – Rishab Navaneet Oct 23 '20 at 15:30
• Yes you're right I forgot the charges were uniformly distributed over any $x =$constant surface. My bad. But I have another related question now. – Brain Stroke Patient Oct 23 '20 at 15:45
• Hey... firstly, we cannot expect the inside of a gaussian surface to be weirdly different from the outside. We won't find just a single charge outside and a charge distribution just as we enter your gaussian surface. The charges outside would also be in a distribution. If we have a symmetry, It doesn't mean that only inside the surface the field is symmetric. It should be symmetric outside as well. Now, The field I considered is due to all charges. But of course I didn't imagine Isolated charges outside or Variations in Direction of $\vec E$* Because there won't be such an isolated charge – Rishab Navaneet Oct 23 '20 at 15:49