Physics Asked by Dr. Momo on June 4, 2021
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:
$$mathbf{E_{2D}}(z)=frac{sigma}{4piepsilon_0}int_{-a/2}^{a/2}int_{-a/2}^{a/2}frac{z}{(x^2+y^2+z^2)^{3/2}}dxdy=frac{sigma}{2epsilon_0}left{left(frac{4}{pi}right)arctan{sqrt{1+left(frac{a^2}{2z^2}right)}}-1right}$$
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):
$$mathbf{E_{1D}}(z)=frac{1}{4piepsilon_0}frac{4lambda az}{(z^2+a^2/4)sqrt{z^2+a^2/2}}$$
He integrates this expression to find the solution to the original problem:
$$mathbf{E_{2D}}(z)=frac{2sigma z}{4piepsilon_0}int_{0}^{a}frac{a}{(z^2+a^2/4)sqrt{z^2+a^2/2}}da$$
where he casually claims that
$$lambda rightarrowsigma frac{da}{2}$$
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 🙂
$sigma$ 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.
Answered by Brain Stroke Patient on June 4, 2021
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 $$Q=4xlambda.$$
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 $A=(x+dx)^2-x^2=2xdx+dx^2approx2xdx$, where we only take the contribution at first order in $dx$. The total charge on the frame is $$Q=sigma A=2xsigma dx.$$
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 $$lambdatosigmafrac{dx}{2}.$$
Answered by Urb on June 4, 2021
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