The problem is to find the E field a distance [tex]z[/tex] above the centre of a square sheet of charge of side length [tex]a[/tex] and charge density [tex]\sigma[/tex].(adsbygoogle = window.adsbygoogle || []).push({});

The solutions use the result for a square loop and integrate to get the result for the square sheet, however it's the change [tex]\lambda \to \sigma \frac{da}{2}[/tex]

In the diagram on the right, since we have [tex]a+da[/tex], widths of [tex]\frac{da}{2}[/tex] make sense. But what if we instead chose to do [tex]a+2da[/tex], so that we would have widths of [tex]da[/tex]? That would increase the integrand by a factor of 2.

Also, I still can't get my head around how the 1-dimensional charge density [tex]\lambda[/tex] could be converted into a 2-dimensional surface charge density [tex]\sigma[/tex]. Could someone please explain the reasoning behind it?

Thanks

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# Infinitesimals - finding the E field

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