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Electric field on a charge some height above the center of a disc
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[QUOTE="vela, post: 4504358, member: 221963"] You need to clean up your math. Your carelessness is leading to dumb mistakes. Think about what Q and A represent here. You can't have lone differentials sitting around. You should have written \begin{align*} dA &= 2\pi r\,dr \\ dQ &= \sigma\,dA = 2\pi \sigma r \, dr \\ dE_y &= \frac{k\,dQ}{h^2+r^2} \frac{h}{\sqrt{h^2+r^2}} = \frac{2\pi hk \sigma r \, dr}{(h^2+r^2)^{3/2}} \end{align*} If you have an infinitesimal quantity on one side of the equation, you have to have one on the other side as well. Look at those two lines. According to what you wrote, $$\frac{kh2\pi \sigma dr}{(h^2 + r^2)^{\frac{3}{2}}} = kh\pi \sigma \int_{0}^{R} \frac{2r dr}{(h^2 + r^2)^{\frac{3}{2}}}$$ because both sides of the equation equal ##E_y##. The integration apparently does absolutely nothing. You need to think about what Q and A represent. Q is the charge of what? A is the area of what? [/QUOTE]
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Electric field on a charge some height above the center of a disc
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