Original question:

The figure shows a plastic rod of length L and total charge Q having a uniformly distributed charge. The charge per unit length of the rod is $\lambda$. Find the electric field at a point P at a distance of r from the rod.

Here is the figure: enter image description here

And this is my attempt at a solution:

enter image description here

Let dL be a infinitesimally small element of the rod making an angle $\theta$ with the vertical connecting point P and the rod, and subtending an angle d$\theta$ at P. The electric field due to dl at P is dE.

I've resolved dE into two components :



And dE therefore would be :


where $r_{eff}$=$\frac{r}{cos{\theta}}$


dE$_(parallel)$=$\frac{kdqcos\theta sin\theta}{r}$

dE$_(perpendicular)$=$\frac{kdqcos\theta cos\theta}{r}$

now to integrate dE$_(parallel)$ within limits -$\alpha$ and $\beta$,I substitute dq as follows

since $\lambda$=$\frac{dq}{dl}$


and then use the formula dl=$r_{eff}$d$\theta$,

and finally obtain the expression:

dE$_(parallel)$=$k\lambda d\theta sin\theta$

and finally get the answer E$_(parallel)$=k$\lambda (cos\alpha - cos\beta)$

However the answer is actually

E$_(parallel)$=k$\frac{\lambda (cos\alpha - cos\beta)}{r}$

According to me, the answer should have been the same,but I think that my substitution of dl=$r_{eff}$d$\theta$ is wrong, as it seems that all other steps are correct(including the integration). My assumption was that the relation dl=$r_{eff}$d$\theta$ should hold since a infintesimally small part a circle would appear as a line, but this does not seem to be the case. If this is true, does the relation never hold, apart from when it concerns larger(not infinitesimal) parts of circles?

Thank you for taking the time to read this, any help is appreciated

  • 3
    $\begingroup$ Hi and welcome to the Physics SE! Please note that we don't answer homework or worked example type questions. Please see this Meta post on asking homework/exercise questions and this Meta post for "check my work" problems. $\endgroup$ Commented Sep 25, 2017 at 7:56
  • $\begingroup$ Hi, thank you for bringing that to my attention. I've edited my question and hope that it now meets the site's requirements. Please let me know if there are any further changes to be made or if the question is completely off topic for this site,thanks! $\endgroup$
    – GreenApple
    Commented Sep 25, 2017 at 11:39

1 Answer 1


The relation $\delta l = r_{eff}\delta \theta$ is not correct because it does not take into account the orientation of the line element $\delta l$ relative to the line element $r_{eff}$. For example, if $\delta l$ was parallel to $r_{eff}$ then $\delta \theta=0$ whereas $\delta l \ne 0$.

  • $\begingroup$ Thank you for taking the time to comment! My understanding is that since I'm integrating $\sin\theta d \theta$ in the end; when d$\theta$ is equal to zero, the entire term would be left out of the integration, and this would imply the element parallel to r(eff) would not contribute to the electric field at P, which is not true. Thank you for clearing that up! However is there any way I can add those terms separately again? $\endgroup$
    – GreenApple
    Commented Sep 25, 2017 at 15:32
  • $\begingroup$ I am sorry, I do not understand your comment or your additional question. You need to substitute $\delta l \cos\theta = r_{eff}\delta \theta$. $\endgroup$ Commented Sep 25, 2017 at 15:49
  • $\begingroup$ Could you please explain why you've used δlcosθ=reffδθδlcos⁡θ=reffδθ ? $\endgroup$
    – GreenApple
    Commented Sep 25, 2017 at 18:16
  • $\begingroup$ $\delta l =r_{eff}\delta\theta$ only applies when $\delta l$ is perpendicular to $r_{eff}$ - ie when $\theta=0$. $\delta l \cos\theta$ is the component of $\delta l$ which is perpendicular to $r_{eff}$. This relation applies for all values of $\theta$. $\endgroup$ Commented Sep 25, 2017 at 22:01

Not the answer you're looking for? Browse other questions tagged or ask your own question.