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Electric Field due to a disk of radius R in the xy-plane
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[QUOTE="BvU, post: 6864948, member: 499340"] Ok, and I suppose [ATTACH type="full" width="66px" alt="1678728290829.png"]323561[/ATTACH] is the Coulomb constant, a.k.a. ##\displaystyle {1\over 4\pi\varepsilon_0}## :smile: . ( re notation: I use separate symbols for the vector in the disk and the vector from ## dA## to point P ) You have ##\vec r = \vec p -\vec r' ## with ##\vec p =<0,0,z>## and ##\vec r'= <r'\cos\theta,r'\sin\theta,0>## and ##\hat r = {\vec r \over |\vec r|} ## And then I can follow $$d\vec E = {1\over 4\pi\varepsilon_0} \; {1\over r^2}\;{\vec p - \vec r' \over |\vec r| } \; dQ$$ where ## dQ = \sigma dA = \sigma r' dr'\, d\theta ## (*) Note that ##dQ## is a scalar ! [INDENT=2]( (*) I have the impression you overlooked the factor ##r'## ? Nice to see vela agrees...)[/INDENT] In sort, I agree with your [ATTACH type="full" width="103px" alt="1678730089811.png"]323562[/ATTACH] (quoting ##\LaTeX## really is a lot more comfortable :wink:). Except that it looks as if you think ##dA## is a vector and then in the subsequent [ATTACH type="full" width="112px" alt="1678730174238.png"]323563[/ATTACH] you come up with dot product ##\vec r \cdot \vec n## I don't think it's valid, but here it works because you pick out the ##z## component, which is the only one that doesn't cancel from symmetry. ##\ ## [/QUOTE]
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Electric Field due to a disk of radius R in the xy-plane
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