# Electric field strength at the center of an equilateral triangle

Questions:

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?

Attempt:

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?

• 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. Aug 10, 2014 at 15:00

Electric field is a vector quantity. So, treat them as vectors and find the vector sum of the electric fields. $$\vec{E}_{net}=\vec{E}_{1}+\vec{E}_{2}+\vec{E}_{3}$$
• 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.
• 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. Aug 11, 2014 at 4:36
• 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. Aug 11, 2014 at 5:03