- #1

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[tex]dE_{rad}=-\frac{1}{4\pi\epsilon_0}\frac{r\lambda d\theta}{(r^2+z^2)}\hat{\textbf{r}}[/tex]

[tex]E_{rad}=-\frac{1}{4\pi\epsilon_0}\frac{r\lambda}{(r^2+z^2)}\int_0^{2\pi} \hat{\textbf{r}} d\theta[/tex]

???

- Thread starter rick_2009
- Start date

- #1

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- 0

[tex]dE_{rad}=-\frac{1}{4\pi\epsilon_0}\frac{r\lambda d\theta}{(r^2+z^2)}\hat{\textbf{r}}[/tex]

[tex]E_{rad}=-\frac{1}{4\pi\epsilon_0}\frac{r\lambda}{(r^2+z^2)}\int_0^{2\pi} \hat{\textbf{r}} d\theta[/tex]

???

- #2

Born2bwire

Science Advisor

Gold Member

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So the electric field element points in the r-hat direction, convert that to cylindrical coordinates and then also provide the appropriate translation since you will be observing it at a point offset from the origin of the coordinate system of the electric field element.

- #3

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I don't get your answer

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