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E&M fun (the derivation for electric field from a charged rod to a point charge)

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(First off, I know Mr. F probably has this derivation on the course notes, but I figured it would be great practice to write it out, and also people might be more likely to look at it if it is typed out.)

We just started Electricity & Magnetism in Physics C, and right off the bat the difficulty is enormous. Now I understand why college engineers always complain about electrostatics. At the moment we are derriving the electric field on a point charge, from uniformly charged objects. We've done a rod, a ring, and a disc, perpendicular to a point charge. They are quite difficult, because they involve differentials and require you to manipulate the givens to a integrable point. I figured for my blog post this week i would go through the steps and derrive the electric field on a point charge from a uniformly charged rod.

Here is the picture I drew in photoshop to represent the problem I will be derriving (this picture is much better if you keep in mind that I figured my way around an advanced photoshop that I've never used! :P )

[ATTACH=CONFIG]92[/ATTACH]

Also, let it be known that the rod as a uniform linear charge density of png.latex? \lambda

Since the rod is equal on either side of the y axis, the x components of the electric field will cancel, so we only need to worry about the y components. Also, I left it out, but the angle between r and y is png.latex?\theta _{1} and the the equal angle that would be on the the other side of y if i had made another vector to the end of the rod would be png.latex?\theta _{2}. Finally, the far left side of the rod is png.latex?x_{1} and the far right side of the rod is png.latex?x_{2}

Step 1: png.latex? \lambda = \frac{Q}{L}

Step 2: png.latex? dQ=\lambda dx

Step 3: png.latex? E=\frac{1}{4\pi \epsilon _{0}

png.latex? \underset{dE}{\rightarrow}=\f

png.latex?\underset{dE}{\rightarrow}=\fr

png.latex? cos\theta =\frac{y}{r}

png.latex?\underset{dE}{\rightarrow}=dEc

png.latex? dE=\frac{1}{4\pi \epsilon _{0

png.latex? \int dE=\int_{x_{1}}^{x_{2}}

png.latex? E=\frac{y\lambda }{4\pi \epsi

png.latex? x=ytan\theta

png.latex? r^{3}=\frac{y^{3}}{cos^{3}\th

png.latex? dx=ysec^{2}\theta d\theta

png.latex? E=\frac{y\lambda }{4\pi \epsi

png.latex? E=\frac{y\lambda }{4\pi \epsi

png.latex? E=\frac{\lambda }{4\pi \epsil

png.latex? E=\frac{\lambda }{4\pi \epsil

png.latex? r=\sqrt{(\frac{L}{2})^{2}+y^{

png.latex? sin\theta _{2}=\frac{\frac{L}

png.latex? sin\theta _{1}=\frac{\frac{-L

png.latex? E=\frac{\lambda }{4\pi \epsil

png.latex? \lambda =\frac{Q}{L}

png.latex? E=\frac{Q}{4\pi \epsilon _{0}

And there you have it! (Jeez that equation editor takes so much time to use...)

If your problem using a rod that has an infinite length... you can back up a few steps before you finally solve your last integral in terms of png.latex?\theta _{2} andpng.latex?\theta _{1} and plug in png.latex?\frac{\pi }{2} and png.latex?\frac{-\pi }{2} for the angles, because as png.latex? L \to \infty , png.latex? \theta \to \frac{\pi }{2} aka png.latex?90^{\circ}

png.latex? E=\frac{\lambda }{4\pi \epsil

png.latex? E=\frac{\lambda }{4\pi \epsil

png.latex? E=\frac{2\lambda }{4\pi \epsi

png.latex? E=\frac{\lambda }{2\pi \epsil

png.latex? E=\frac{Q }{2\pi \epsilon _{0

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im wondering if any can see the picture I posted. Its just showing up as a red X on my computer, and I dont know if that is just because I dont have the right program to view it, or because i don't know how to properly post a picture

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nevermind, i fixed it. Mr. F has a blog to help solve every problem possible... :P

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haha thanks alot :)

i'm lucky this is coherent... it was pretty late last night!

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