# Electric field of infinite slab

When reading a book about basic electrodynamics (in a section about electrostatics), I came upon the following problem:

An infinite plane slab, of thickness $2d$, carries a uniform volume charge density $\rho$. Find the electric field, as a function of $y$, where $y=0$ at the center.

The slab parallel to the $x$-$z$ plane, and is thus perpendicular to the $y$-axis, contained between $y=-d$ and $y=d$ but reaching infinitely into the $x$ and $z$ directions.

The book I'm reading utilizes Gauss's law (using a "Gaussian pillbox" circling around the $y$-axis), but I was a bit confused by the method they used, and when doing the problem I instead thought like this:

If we place a test charge on the $y$-axis at $y=a$, then the charge experiences a positive force (pointing the positive $y$ direction) due to the volume charge behind it (from $y=-d$ to $y=a$) and experiences a negative force (gets "pushed backwards") due to the volume charge in front of it (from $y=a$ to $y=d$). So for $|y|<d$, the electric field would be the total field made up of a bunch of infinitesimally-thin charged planes behind the test charge minus the field made up of a bunch of infinitesimally-thin charged places in front of the test charge, or:

$$E(y) = \int_{-d}^y \frac{\rho}{2 \epsilon_0} \,dy \,\,\, - \,\,\, \int_{y}^d \frac{\rho}{2 \epsilon_0} \,dy = \frac{\rho}{\epsilon_0} y$$

Since the magnitude of a field of an infinitesimally-thin infinite plane is $\frac{\sigma}{2 \epsilon_0}$ and in this instance $\sigma = \rho dy$.

This is, in fact, the answer the book gives. Was my thought process incorrect? That is, can a volume of charge really "push" as if it were a point charge? I wasn't too sure if my logic was correct and that I could rely on this idea in the future.

If my intuition was wrong, could someone please explain how one would use Gauss's law in this problem?

## 2 Answers

Since in this problem you can exploit a symmetry (invariance by translation along the directions spanned by the slab) it is easier to compute the electric field by a wise use of Gauss' theorem. As your book says, this is quite a standard technique, which is used e.g. for the infinite plane, infinite cylinders (coaxial cables), etc.... The idea is that, as a consequence of this translation symmetry, the electric field at every point has component only along the direction perpendicular to the slab. Moreover the electric field figuring in Gauss' law is the total one. With all these hints you should now be able to solve your problem.

• I understand that it may be easier to use Gauss's law, but I was wondering whether it's possible to do as I thought I could do. Mar 8, 2015 at 21:26
• In that case I'd say your argument makes sense, since you are just exploiting the linearity of Maxwell's equations, i.e. the superposition principle. Mar 8, 2015 at 21:34

Example 7 in this site is exactly the same as your problem http://physicspages.com/2011/10/04/gausss-law-examples/

I would like to clarify something here. Magnitude of a field of an infinitesimally-thin infinite plane is $\frac{\sigma}{2 \epsilon_0}$. We use $\sigma$ here not $\rho$ because $\sigma$ indicate the planar charge density since the plane is very thin. This is different from that in your problem where the plane is having a thick, so the answer that was given by your book is correct. And a volume charge doesn't "push" as if it were a point charge because a volume charge contain infinite point charge

• I already know how to solve it using Gauss's law. I have the book's written-out solution. I was asking whether the way I obtained the answer was logical or not. And how does the fact that a volume charge contains infinite point charges change anything? (a sphere of volume charge acts like a point charge outside of the sphere, so does that kind of logic also apply here?) Mar 8, 2015 at 21:29