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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 :)

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  • $\begingroup$ It would be good to give different names to the upper limit of the last integral and the variable where the integration is performed. $\endgroup$ – Urb Mar 19 at 17:41
  • $\begingroup$ That's my bad! Griffith's taught me to prime my variables :) $\endgroup$ – Dr. Momo Mar 20 at 0:27
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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 $$\lambda\to\sigma\frac{dx}{2}.$$

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$\sigma$ 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.

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