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Introductory Physics Homework Help
Electric field in a semi-circle (non-uniformly charged)
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[QUOTE="BitterX, post: 3833455, member: 367626"] [h2]Homework Statement [/h2] A plastic semi-circle/arc with radius [B]R[/B] has a non-uniformly distributed charge upon it. The density of the charge/length is [itex]\lambda = \lambda_0 sin\theta[/itex] [B]θ[/B] is 0 in the middle of the arc find the electric field in the point where the radius come from [h2]Homework Equations[/h2] [itex]dE=\frac{kdQ}{R^2}[/itex] [itex]\phi = \int{Edr}[/itex] [h2]The Attempt at a Solution[/h2] charge differential to angle differential is : [itex]dQ=\lambda R = \lambda_0 sin\theta d\theta[/itex] for the field I separated for x and y [itex]E_x =\frac{k\lambda_0}{R} \int_{0}^{\pi}sin^2\theta d\theta=\frac{k\lambda_0}{R}=\frac{k\lambda_0\pi}{2R} [/itex] the same way I get [itex]E_y= \frac{k\lambda_0}{R}\int_{0}^{\pi}sin\theta cos\theta d\theta=0 [/itex] this is counter-intuitive for me, I thought the electric field should be pointing in the y direction. I guess I need to have my answers inverted, but would be glad to understand why, and maybe shed some light on my mistakes...edit: I think I've found my mistake... I should integrate once from 0 to [itex]\frac{\pi}{2}[/itex] and once from 0 to [itex]-\frac{\pi}{2}[/itex] and add what I get [/QUOTE]
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Introductory Physics Homework Help
Electric field in a semi-circle (non-uniformly charged)
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