Three $x$ m long rods form an equilateral triangle. Two of the rods are charged to $+q$ C and the third to $-q$ C. What is the electric field strength at the center of an equilateral triangle?


I know how to find the electric field strength due to one rod, I just divide the rod into pieces, each of length $dx$ and charge $dq$ then integrate from 0 to x.

After I find the electric field at the center due to each rod, how do I find the resultant electric field?

  • 1
    $\begingroup$ Can you show a few more steps...around here people generally respond better to homework questions if you've done a few more steps and have a specific question where you are stuck. $\endgroup$
    – paisanco
    Commented Aug 10, 2014 at 15:00

1 Answer 1


Electric field is a vector quantity. So, treat them as vectors and find the vector sum of the electric fields. $$\vec{E}_{net}=\vec{E}_{1}+\vec{E}_{2}+\vec{E}_{3}$$

  • $\begingroup$ After doing the integration I get $\vec{E} = \frac{2\times 8.99\times 10^{9}\times \lambda \frac{x}{2}}{y\sqrt{y^{2}+\frac{x}{2}^{2}}} \times \vec{r}$. What is the direction here? I am having trouble determining the unit vector. Also, does my work look right? x in this case is the length of each rod and y is the distance from middle of rod to center of triangle. $\endgroup$
    – user57015
    Commented Aug 11, 2014 at 2:00
  • $\begingroup$ I am not checking your calculations, but assuming they are right. For a positively charged rod the Electric field will be perpendicular to the rod, pointing away from the rod. hyperphysics.phy-astr.gsu.edu/hbase/electric/elelin.html read this. You will see that they have only calculated $E_{z}$. That is the only surviving component of Electric field and it has the same sign as the charge. $\endgroup$ Commented Aug 11, 2014 at 4:36
  • $\begingroup$ Now, about you calculation don't use a arbitrary $\times$ anywhere, it looks like a cross product. It is right but a bit confusing. Your calculation seems correct for the rod along y axis. As you want to find out the E field at the center replace $\vec{r}$ with the exact vector and you are there. Do the same for the other two rods. I guess the rest you can do yourself. If needed, I will help. $\endgroup$ Commented Aug 11, 2014 at 5:03
  • $\begingroup$ Thank you for your help. I was able to get the right answer. $\endgroup$
    – user57015
    Commented Aug 14, 2014 at 0:57

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.