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If we were not to neglect the fringing of the electric field lines in a parallel-plate capacitor, would we calculate a higher or lower capacitance? I thought in energy related terms, now after considering increease in total energy and the formula for energy approximately will be $Q^2/2C$ so $C$ should be lower but i dont get if this is a exact (correct) reason ?( This was seen to be present in HRK chap 30 , ques 11)

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Fools rush in where angels fear to tread...

Let the positive plate (carrying charge $Q$) be surrounded by an imaginary Gaussian box, one face of which is a plane between the plates, parallel to them and extending well out beyond their edges. All the electric flux ($Q/\epsilon_0$) coming out from the box will come out through this plane, if we assume (realistically) that all the flux from the positive plate crosses the gap to the negative plate.

With no edge effects, all the flux would pass through an area of the plane equal to the plate area, $A$, so the field strength over that area of surface would be $Q/\epsilon_0A$. But with the 'bulging' of the field due to edge effects, the same flux, $Q/\epsilon_0$, will be spread over a wider area, and so the mean field strength will be less than $Q/\epsilon_0A$.

The pd, $V$, between the plates is $\int_s \mathbf E. d\mathbf s$ over any path running from one plate to the other. Therefore we can choose the easiest path – from the centre of one plate, straight across the gap to the centre of the other. Even over this central path, the field strength will be reduced below $Q/\epsilon_0$ by the bulging of the lines, as argued earlier. Therefore, for a given $Q$, $V$ will be reduced. This implies that the capacitance, $Q/V$, will be greater than $\epsilon_0 A/d$.

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  • $\begingroup$ I just didnt get one thing the electric field value between the plates will have same value at any point on a plane parallel to it (plane doesnt go outside the area A) in case of edge effects too ? So as that PD will always be a constant calculated along any line joining the two plates ? If yes how one proves it? PD will be constant $\endgroup$
    – Orion_Pax
    Dec 6, 2021 at 3:21
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    $\begingroup$ You are, of course, quite right that the line integral of the field will be the same along whatever route we go from one plate to the other. In the central region of the gap, the field will be pretty uniform (provided that $\sqrt A$ is a few times larger than $d$), but I'm arguing that the value of $E$, even in this region, is lower than we'd calculate without taking account of edge effects. $\endgroup$ Dec 6, 2021 at 10:47
  • $\begingroup$ Hmm , well as in case of edge effects , area where field occupies more space no doubt in that gaussian part but they r vectors so ig they r at at some angles outside the plates , so how can we be sure that in central region E is less than we would get without considering edge effects? As those angles might cause mean field strength to get increase isnt? $\endgroup$
    – Orion_Pax
    Dec 6, 2021 at 12:11
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    $\begingroup$ (a) "cant this situation exist where inside field is more (parallel ones(uniform)) more than sigma/e° and outside is low((component one) (non uniform))". I don't think so. The fact that some flux from the top plate is going through the part of the gaussian surface outside area $A$ must reduce the flux that goes through area $A$, so that it's less than $Q/\epsilon_0$. (b) "And "√A few times larger than d" field will be pretty uniform do u have proof [...]?" No. It's clearly true in the limit $\sqrt A >> d$ (infinite plates) but for "a few times larger' you need to seek out advanced treatments. $\endgroup$ Dec 6, 2021 at 13:58
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    $\begingroup$ As soon as you leave the axis of symmetry (e.g. in order to prove my statement about field uniformity), things start to get much more difficult! Anyway, I'm glad to have been of some help. I applaud your tenacity in seeking to understand fully. $\endgroup$ Dec 6, 2021 at 14:26
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The actual capacitance is higher due to the edge effect because the field lines extend beyond the plates. See the following for a calculation https://www.calctown.com/calculators/edge-parallel

Hope this helps

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  • $\begingroup$ Conformal mapping i dont know what it means etc. I need a reason which is understandable $\endgroup$
    – Orion_Pax
    Dec 1, 2021 at 11:36
  • $\begingroup$ Then I suggest you do some research of your own $\endgroup$
    – Bob D
    Dec 1, 2021 at 11:43
  • $\begingroup$ Sir i tried online searching this was given in HRK as a question. $\endgroup$
    – Orion_Pax
    Dec 1, 2021 at 11:44
  • $\begingroup$ It must means there supposed to be a logical reason which helps as HRK is a book used by high school students $\endgroup$
    – Orion_Pax
    Dec 1, 2021 at 11:50
  • $\begingroup$ A simple search gives :sciencedirect.com/topics/engineering/fringe-effect and fma.if.usp.br/~nickolas/dps/pdf/20-09-19.pdf $\endgroup$
    – The Tiler
    Dec 5, 2021 at 18:43

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