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Niles
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[SOLVED] Electric field of a plate
Find the electric field a distance z above the center of a square loop (side s) carrying a uniform line charge λ (see Figure 1) - it's problem 2 on this page:
http://teacher.pas.rochester.edu/PHY217/Homework/HomeworkSet3/HomeworkSet3.html
Ok, first I use symmetry, so the horizontal components must cancel for the 4 edges. The vertical component of one edge is given by:
\frac{2}{{4\pi \varepsilon }}\int\limits_{x1 }^{x2 } {\frac{\lambda }{{\left( {x^2 + z^2 } \right)}}} \cos \phi dx,
where cos(\phi) = z / sqrt(x^2+z^2) so x1 = s/2 and x2=s/sqrt(2) since these are the minimal and maximal values of x.
Pay attention to that I have multiplied with 2, so we find the one whole edge, not just the half of one edge. I integrate, and multiply it all with 4 to give the correct result.
Am I correct?
Homework Statement
Find the electric field a distance z above the center of a square loop (side s) carrying a uniform line charge λ (see Figure 1) - it's problem 2 on this page:
http://teacher.pas.rochester.edu/PHY217/Homework/HomeworkSet3/HomeworkSet3.html
The Attempt at a Solution
Ok, first I use symmetry, so the horizontal components must cancel for the 4 edges. The vertical component of one edge is given by:
\frac{2}{{4\pi \varepsilon }}\int\limits_{x1 }^{x2 } {\frac{\lambda }{{\left( {x^2 + z^2 } \right)}}} \cos \phi dx,
where cos(\phi) = z / sqrt(x^2+z^2) so x1 = s/2 and x2=s/sqrt(2) since these are the minimal and maximal values of x.
Pay attention to that I have multiplied with 2, so we find the one whole edge, not just the half of one edge. I integrate, and multiply it all with 4 to give the correct result.
Am I correct?
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