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Eclair_de_XII
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Homework Statement
"A straight, nonconducting plastic wire ##x=9.50_{10^{-2}}m## long carries a charge density of ##λ=1.3_{10^{-7}} C/m## distributed uniformly along its length. It is lying on a horizontal tabletop. If the wire is now bent into a circle lying flat on the table, find the magnitude and direction of the electric field it produces at a point ##y=6_{10^{-2}}m ## directly above its center."
Homework Equations
##dQ=λdx##
##x=9.50_{10^{-2}}m##
##λ=1.3_{10^{-7}} C/m##
##y=6_{10^{-2}}m ##
The Attempt at a Solution
##|E|=(\frac{1}{4\pi ε_0})⋅\int \frac{dQ}{(x^2+y^2)}##
Let ##x=2\pi r##. Then ##dQ=λdx=2\pi λ dr##
##|E|=(\frac{1}{4\pi ε_0})⋅\int \frac{dQ}{(x^2+y^2)}=(\frac{2\pi λ}{4\pi ε_0})⋅\int \frac{dr}{(r^2+y^2)}##
##|E_y|=(\frac{1}{4\pi ε_0})⋅\int \frac{dQ}{(x^2+y^2)}=(\frac{2\pi λ}{4\pi ε_0})⋅\int \frac{ydr}{(r^2+y^2)^{\frac{3}{2}}}=(\frac{2\pi yλ}{4\pi ε_0})⋅\int \frac{dr}{(r^2+y^2)^{\frac{3}{2}}}##
Let ##r=ytan\theta##. Then:
##dr=y\sec^2\theta d\theta##
##(r^2+y^2)^\frac{3}{2}=y^3sec^3\theta##
##\theta = tan^{-1}(\frac{r}{y})##
##|E_y|=(\frac{2\pi yλ}{4\pi ε_0})⋅\int \frac{y\sec^2\theta d\theta}{y^3sec^3}=(\frac{1}{4\pi ε_0})(\frac{2\pi λ}{y})⋅\int cos\theta d\theta=(\frac{2\pi yλ}{4\pi ε_0})⋅(sin\theta)=(\frac{1}{4\pi ε_0})(\frac{2\pi λ}{y})(\frac{r}{\sqrt{r^2+y^2}})=(\frac{1}{4\pi ε_0})(\frac{2\pi λ}{y})(\frac{x}{2\pi })(\frac{1}{\sqrt{\frac{x^2}{4\pi ^2}+y^2}})##
Thus, ##|E_y|=(\frac{1}{4\pi ε_0})(\frac{2\pi λ}{y})(\frac{x}{\sqrt{x^2+4\pi ^2y^2}})##.
Can someone check my work, or point out a better way to do this problem? Thanks.
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