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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! )

[ATTACH=CONFIG]92[/ATTACH]

Also, let it be known that the rod as a uniform linear charge density of $\lambda$

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 $\theta _{1}$ 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 $\theta _{2}$. Finally, the far left side of the rod is $x_{1}$ and the far right side of the rod is $x_{2}$

Step 1: $\lambda = \frac{Q}{L}$

Step 2: $dQ=\lambda dx$

Step 3: $E=\frac{1}{4\pi \epsilon _{0}}\frac{Q}{r^{2}}$

$\underset{dE}{\rightarrow}=\frac{1}{4\pi \epsilon _{0}}\frac{dQ}{r^{2}}$

$\underset{dE}{\rightarrow}=\frac{1}{4\pi \epsilon _{0}}\frac{\lambda dx}{r^{2}}$

$cos\theta =\frac{y}{r}$

$\underset{dE}{\rightarrow}=dEcos\theta$

$dE=\frac{1}{4\pi \epsilon _{0}} (\frac{\lambda dx}{r^{2}})(\frac{y}{r})$

$\int dE=\int_{x_{1}}^{x_{2}} \frac{1}{4\pi \epsilon _{0}} (\frac{\lambda dx}{r^{2}})(\frac{y}{r})$

$E=\frac{y\lambda }{4\pi \epsilon _{0}}\int_{x_{1}}^{x_{2}} \frac{dx}{r^{3}}$

$x=ytan\theta$

$r^{3}=\frac{y^{3}}{cos^{3}\theta }$

$dx=ysec^{2}\theta d\theta$

$E=\frac{y\lambda }{4\pi \epsilon _{0}}\int_{\theta _{1}}^{\theta _{2}} \frac{ysec^{2}\theta d\theta }{y^{3}}$

$E=\frac{y\lambda }{4\pi \epsilon _{0}}\int_{\theta _{1}}^{\theta _{2}} \frac{cos\theta d\theta }{y^{2}}$

$E=\frac{\lambda }{4\pi \epsilon _{0}y}\int_{\theta _{1}}^{\theta _{2}} {cos\theta d\theta}$

$E=\frac{\lambda }{4\pi \epsilon _{0}y}[sin\theta _{2}-sin\theta _{1}]$

$r=\sqrt{(\frac{L}{2})^{2}+y^{2}}$

$sin\theta _{2}=\frac{\frac{L}{2}}{\sqrt{(\frac{L}{2})^{2}+y^{2}}}$

$sin\theta _{1}=\frac{\frac{-L}{2}}{\sqrt{(\frac{-L}{2})^{2}+y^{2}}}$

$E=\frac{\lambda }{4\pi \epsilon _{0}y}\frac{L}{\sqrt{(\frac{L}{2})^{2}+y^{2}}}$

$\lambda =\frac{Q}{L}$

$E=\frac{Q}{4\pi \epsilon _{0}y\sqrt{(\frac{L}{2})^{2}+y^{2}}}$

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 $\theta _{2}$ and$\theta _{1}$ and plug in $\frac{\pi }{2}$ and $\frac{-\pi }{2}$ for the angles, because as $L \to \infty$ , $\theta \to \frac{\pi }{2}$ aka $90^{\circ}$

$E=\frac{\lambda }{4\pi \epsilon _{0}y}\int_{{\frac{-\pi }{2}}}^{\frac{\pi }{2}} {cos\theta d\theta}$

$E=\frac{\lambda }{4\pi \epsilon _{0}y}[sin\frac{\pi }{2}-sin\frac{-\pi }{2}]$

$E=\frac{2\lambda }{4\pi \epsilon _{0}y}$

$E=\frac{\lambda }{2\pi \epsilon _{0}y}$

$E=\frac{Q }{2\pi \epsilon _{0}yL}$

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

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

I am absolutely amazed... this is tremendous!!!

haha thanks alot

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

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