# Calculating electric fields due to continuous charge distributions

1. Mar 30, 2013

### al_famky

calculating electric fields due to continuous charge distributions?

a question I came across doing some electric field questions, and the answer was really confusing.

1. The problem statement, all variables and given/known data
Charge is distributed along a linear semicircular rod with a linear charge density λ as in picture attatched. Calculate the electric field at the origin

2. Relevant equations

3. The attempt at a solution
for the first question, the magnitude of E was calculated with dE=$\frac{adθλ}{4\pi\epsilon_{0}a}$, but why is there only one "a" in the denominator, not $a^{2}$? The final result was this equation $\int$$^{\phi}_{-\phi}$$\frac{λcosθdθ}{4\pi\epsilon_{0}}$=$\frac{λsin\phi}{2\pi\epsilon_{0}}$ which makes sense only if the "a" was dropped.
Thanks to anyone who'd be willing to explain the answers!

Last edited: Mar 30, 2013
2. Mar 30, 2013

### krome

Did you forget to attach the picture? Maybe I'm just stupid and don't know how to find it. I would want to make sure I understand the question correctly. The integral in your "attempt" goes from $-\phi$ to $+\phi$. That makes me think that the rod is only a semicircle when $\phi = \pi / 2$. I'm guessing that's why it's called "semicircular" rather than "a semicircle". Is that correct?

Other than that, you're absolutely correct. It should be $a^2$ in the denominator of $dE$ because otherwise the units are not even correct. As it is, the answer you've quoted has units of charge per unit length, which is the unit of electric potential, not electric field.

Finally, I'm assuming you understand where the factor of $cos\theta$ in the integral comes from since you didn't ask about it specifically.

3. Apr 4, 2013

### al_famky

my apologies, I thought i did upload the picture.
But you got the gist of the problem from the without the picture anyways.
Unit check, the fundamental approach to going over problems---why didn't I think of that?
I do understand the cosθ, thx for asking

Thank you for taking the time to reply!

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