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I asked the question previously here - Link. Look for the accepted answer.

I have 2 questions and since they both relate to the answer for my previous question, I will include both here. Note that for Q1, I'm looking for the intuitive answer without integral, while in second, integral explanation.

While I almost understood everything, there's something thats bugging me. So on the plate, there're patches where $x>d\sqrt{2}$ and when we step away, those patches contribute to the increase in ⊥ while patches where $x<d\sqrt{2}$ contribute to the decrease in ⊥, so they cancel out each other, leaving E unchanged everywhere.

Q1: What I wonder is that since plate is infinite, there will be more numbers of patches for $x>d\sqrt{2}$ than for $x<d\sqrt{2}$ so their increase effect must be bigger then than decrease effect of patches $x<d\sqrt{2}$. This comes from the fact that there's finite patches from 0 to $d\sqrt{2}$ and infinite patches from $d\sqrt{2}$ to infinity. What am I missing ?

Q2: I wonder in the integral calculation, what area d ⊥ gives us and why is there extra x in there ?

Qmechanic
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Giorgi
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  • As to your second question concerning $x \mathrm{d}x$ is straightforward, it's polar coordinates, the area element of which usually looks like $ \mathrm{dr}(r \mathrm{d}\phi)$ – Triatticus Jun 12 '23 at 02:34

1 Answers1

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What you're missing is that although there are infinitely many patches, the total contribution from these patches is finite. The fact that there are so many more patches that are far away is balanced by the fact that they have a weaker influence on the field.

I think the most compelling case for this is to look at the graph of $\delta E_\perp$: enter image description here

While the positive region is infinitely wide, it is also infinitely thin and has finite area.

While convincing, this depiction is slightly deceptive. As you noticed (in your Q2), we add a factor of $x$ when integrating. This is because we care about area. The region which has the change in contribution $$ \delta E_\perp = \frac{x^2-2d^2}{r^5} \delta d $$ is the infinitely thin ring of radius $x$. Since it's infinitely thin, we can say it's basically a rectangle with sides $dx$ and $2\pi x$ (the circumference of the ring in question). So the actual change in $E_\perp$ includes an (infinitesimal) area: $$ \delta E_\perp = \frac{x^2-2d^2}{r^5} \delta d~\mathrm dA = \frac{x^2-2d^2}{r^5} \delta d (2\pi x)\mathrm dx $$ The graph of this one is less extreme, but still gets the point across and is not deceptive (integrating this curve yields the actual change in $E_\perp$ when we take a tiny step away): enter image description here

The red curve is $\delta E_\perp$. The black curve is $2\pi/x^2$, which is only there to show that the positive region has finite area; the black curve is always greater than the red, and the area under the black curve is finite.

In this graph, the area of the negative region is equal to the area of the positive region, so the change in $E_\perp$ is zero.

When there is a finite region with finite values and an infinite region with infinitesimal values, I'd caution against reasoning about which has greater area without doing an explicit integral.

For an explicit counterexample, if we had a strip of uniform charge $\sigma$ that is $W$ wide and infinitely long, we get to use the same ideas. For a point directly in the middle of the strip ($\frac{W}{2}$ on either side), and $d$ distance away from the strip, we get the same symmetry argument showing that the parallel components all cancel out. We also get the same formulas for the perpendicular component that J. Murray showed: $$ E_\perp = \frac{dA}{r^2}\cos(\theta)=dA \frac{d}{(x^2+d^2)^{3/2}} $$ and when $d\to d+\delta d$: $$ \delta E_\perp = \frac{x^2-2d^2}{r^5} \delta d $$ so patches with $x > d\sqrt{2}$ increase their contribution when we step away. Also, we still have infinitely many patches with $x > d\sqrt{2}$ and finitely many with $x < d\sqrt{2}$. But in this scenario, the net electric field decreases as we move away from the strip (I'll leave the integration to you as an exercise).

Jacob Stuligross
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  • Good answer. Thank you, but I have a hard time imagining exactly what thin ring of radius x are you talking about. can you show on the graph (maybe mark it) which area are we calculating ? – Giorgi Jun 12 '23 at 09:45
  • The thin ring is the set of all the points on the plane that are $r$ away from the point we're calculating for. It's the intersection between a thin spherical shell of radius $r=\sqrt{x^2 + d^2}$ and the plane where the charge is. That makes a thin circular ring with radius $x$. – Jacob Stuligross Jun 12 '23 at 15:14
  • The thin ring is not on the graphs that I included. It's a region of the plane of charge, not the axes I'm plotting on. The second graph plots the radius of the ring (horizontal) vs the change in electric field due to that ring (vertical). Integrating this curve gives you the rings of all radii, which covers the entire plane of charge. – Jacob Stuligross Jun 12 '23 at 15:25
  • For infinite line of charge, we integrate over $$ \delta E_\perp = \frac{x^2-2d^2}{r^5} \delta d $$ from -infinity to infinity but this gives me -2/d^2. but we know that it should give me -2/d and not d^2. where am I making a mistake then ? – Giorgi Jun 13 '23 at 09:30
  • I think what we do here is integration actually gives us -2/d^2. Then we can transform to d. is on the same axis as d, so it can definitely be expressed as = z * d. Then we got -2 * z * d / d^2 = -2 * z / d. Hence we got 1/d dependence and not 1/d^2 – Giorgi Jun 13 '23 at 10:55
  • But they say dependence is 1/r and not 1/d. Do they mean 1/r in which r is presumed to be d(perpendicular) ? – Giorgi Jun 13 '23 at 11:11
  • Please ask a new question if you have a new question. I won't try to write a complete answer to this in the comments. – Jacob Stuligross Jun 13 '23 at 14:22
  • Hey. DONE https://physics.stackexchange.com/questions/768124/infinite-line-of-charge-1-r-dependence. - Thanks for the help on this question – Giorgi Jun 13 '23 at 14:45