Let's say I have a uniformly-charged wire bent into a semi-circle around the origin. How can I find the electric field (magnitude and direction)

I'm not even sure if I should use Coulomb's or Gauss' law either. Kind of stuck setting up a solution and starting somewhere.

  • $\begingroup$ Is this point generic? Or for example on the same plane of the wire? $\endgroup$
    – Rik Poggi
    Jan 29, 2012 at 9:45
  • $\begingroup$ Did you mean that you have to find the field at the origin? $\endgroup$ Jan 29, 2012 at 10:12
  • $\begingroup$ A wire can only stay uniformly charged if bent to a full circle. $\endgroup$ Jan 29, 2012 at 13:32
  • $\begingroup$ @Debangshu Currently looking for the field at the origin but I would like to know how to do it for any arbitrary point as well. $\endgroup$
    – n0pe
    Jan 29, 2012 at 18:11

2 Answers 2


Let us say you have a charged unit semi-circle ($\rho=1,-\frac{\pi}{2}\leq\varphi\leq\frac{\pi}{2}$) and are trying to calculate the potential in a point with coordinates $(x,y,z)$. You get (I neglect all kinds of constant factors): $V(x,y,z)=\int^{\frac{\pi}{2}}_{-\frac{\pi}{2}}d\varphi\frac{1}{\sqrt{(x-\cos(\varphi))^2+(y-\sin(\varphi))^2+z^2}}=\int^{\frac{\pi}{2}}_{-\frac{\pi}{2}}d\varphi\frac{1}{\sqrt{x^2+y^2+z^2+1-2 x\cos(\varphi)-2 y\sin(\varphi)}}$. I guess this may be an elliptic integral. To get the field, you need to take a gradient of this integral.

EDIT: To answer MaxMackie's question in the comment: a point charge in a point $\overrightarrow{r'}=(x',y',z')$ creates a field described by potential $V(\overrightarrow{r'},\overrightarrow{r})=\frac{1}{|\overrightarrow{r}-\overrightarrow{r'}|}$, where $\overrightarrow{r}=(x,y,z)$ is the field point (again, constant factors are neglected). If you have a charge distribution $\rho(\overrightarrow{r'})$, the new potential $V_\rho(\overrightarrow{r})=\int d\overrightarrow{r'}\rho(\overrightarrow{r'})V(\overrightarrow{r'},\overrightarrow{r})$. To get the electric field, take a gradient of the potential with respect to $\overrightarrow{r}$.

  • $\begingroup$ What's the general rule to something like this though? Like the formula to use in this situation that can be applied to any problem. $\endgroup$
    – n0pe
    Jan 29, 2012 at 18:10
  • $\begingroup$ See the addition in my answer $\endgroup$
    – akhmeteli
    Jan 29, 2012 at 18:41

If you are just looking for the field at the origin, it does have a closed form. The situation is somewhat as follows: enter image description here

We choose a differential element which subtends as angle $d\theta$ at the origin. Correspondingly, we choose an element in the diametrically opposite quadrant i.e a element which subtends angle $d\theta$ at the origin, but that differential angle is $-\theta$ away from the negative $x$-axis. So, the components of the forces along the $x$-axis cancel out each other (the cosine components) and the sine components exist. I mean, if the differential element produce a field $E$ at the origin, then the field from two opposite elements which are at angle $\theta$ away from the positive and negative $x$-axis gives a field of $2E\sin \theta$ along the negative $y$-axis. If, the charge density is $\rho$, then, \begin{equation} E = \frac{1}{4\pi \epsilon_{0}} \frac{\rho R d\theta}{R^2} \end{equation} Hence, the net electric field is along the negative $y$-axis and the magnitude is \begin{equation} E= \frac{1}{4\pi \epsilon_{0}} \frac{2\rho}{R} \int_{0}^{\pi/2}\sin \theta d\theta =\frac{\rho}{2\pi \epsilon_{0}R} \end{equation} Now, finding the field at an arbitrary point may not be that easy. In fact, akhmetali has given an answer but it seems that the integral is elliptic in nature. Neither, could I see any obvious symmetry in the problem (spherical, azimuthal etc.). There are standard methods to deal with difficult situations like method of images, solving the Poisson Equation etc. These are discussed at length in J.D. Jackson's book. I hope you can find something there regarding your problem.


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