# Conversion of 1D charge density to 2D charge density via integration

I'm self-studying EM (using the third edition of Griffiths) and have a quick question. Problem 2.41 states:

Find the electric field at a height $$z$$ above the center of a square sheet (side a) carrying a uniform surface charge density $$σ$$.

With the power of Mathematica, the problem is straightforward enough, though I did have to fight with it a bit:

$$\mathbf{E_{2D}}(z)=\frac{\sigma}{4\pi\epsilon_0}\int_{-a/2}^{a/2}\int_{-a/2}^{a/2}\frac{z}{(x^2+y^2+z^2)^{3/2}}dxdy=\frac{\sigma}{2\epsilon_0}\left\{\left(\frac{4}{\pi}\right)\arctan{\sqrt{1+\left(\frac{a^2}{2z^2}\right)}}-1\right\}$$

I got the problem right, so "all is well," but in reviewing Griffiths' solution, I came across an oddity that has been bothering me for a little while. He uses the result of the electric field due to a 1-D square charge density (Problem 2.4): $$\mathbf{E_{1D}}(z)=\frac{1}{4\pi\epsilon_0}\frac{4\lambda az}{(z^2+a^2/4)\sqrt{z^2+a^2/2}}$$

He integrates this expression to find the solution to the original problem: $$\mathbf{E_{2D}}(z)=\frac{2\sigma z}{4\pi\epsilon_0}\int_{0}^{a}\frac{a}{(z^2+a^2/4)\sqrt{z^2+a^2/2}}da$$

where he casually claims that $$\lambda \rightarrow\sigma \frac{da}{2}$$

There lies my confusion. How can an expression involving a differential (da) equal an expression that does not contain a differential? This feels like abuse of notation! I am having trouble understanding the theoretical basis for this conversion, and would love for somebody to talk me through it.

In trying to solve this problem, I've thought about expressing 1- and 2D charge densities as the product of the 3D charge density and various dirac delta functions, but the units don't work out unless I multiply by a differential. So... I'll let the experts explain it to me :)

• It would be good to give different names to the upper limit of the last integral and the variable where the integration is performed. – Urb Mar 19 at 17:41
• That's my bad! Griffith's taught me to prime my variables :) – Dr. Momo Mar 20 at 0:27

It's not that the expressions $$\lambda$$ and $$\sigma da/2$$ are "equal", but rather, one needs to establish an equivalence in order to use the one-dimensional result to solve the two-dimensional case.
In the first case, a square of side $$x$$ and line density $$\lambda$$ has a total charge $$Q=4x\lambda.$$
To solve the sheet case, one needs to give the square a little width $$dx$$, so that it turns into a thin frame. I should assign to this frame a surface charge density $$\sigma$$, because it's no longer a line. What $$\sigma$$ should I use? The one that gives me the same total charge.
Imagine the inner square of the frame with side $$x$$ and the outer one with side $$x+dx$$. They enclose an area $$A=(x+dx)^2-x^2=2xdx+dx^2\approx2xdx$$, where we only take the contribution at first order in $$dx$$. The total charge on the frame is $$Q=\sigma A=2x\sigma dx.$$
So as you integrate over the square sheet from $$0$$ to $$a$$, each frame of side $$x$$ and width $$dx$$ contributes to the electric field as if it were a line of charge density $$\lambda\to\sigma\frac{dx}{2}.$$
$$\sigma$$ can be written as $$dq/(dxdy)$$ in cartesian coordinates. So when you take the limit that the rectangle of area $$dxdy$$ approaches a line by for example taking the limit $$dx \to 0$$, $$\sigma$$ blows up as it should. But $$\sigma dx$$ does not because as $$dx$$ gets smaller and smaller $$\sigma$$ gets larger and larger such that their product approaches a finite value.