How Do Electric Fields Behave at the Center of a Charged Triangle?

[beatbama85]

Here's my problem:

Three 10-cm-long rods form an equilateral triangle. Two of the rods are charged to + 10 nC, the third to -10 nC. What is the electric field strength at the center of the triangle?

Answer in back of book: 1.08 * 10^5 N/C

By looking at the corresponding section, it gives the formula for the electric field strength in the plane that bisects a charged rod as:
Q/(4*pi*e_0*r*sqrt(r^2 + (L/2)^2)),
where e_0 is given to be 8.85*10^-12.

I thought that you would add the vertical and horizontal components of the 3 electric fields in the center due to each rod to get the net electric field. After some calculations, I get that the final answer should be 2 times the formula I gave above, but this does not give me the correct answer. What am I doing wrong? Please respond, thanks!

 

[beatbama85]

Nevermind, I got it.

 

[quicknote]

Hello,

Thank you for sharing your problem with me. It seems like you have a good understanding of the formula for electric field strength in the plane that bisects a charged rod. However, in this case, we are dealing with three charged rods forming an equilateral triangle. This means that the distance from the center to any of the charged rods is not just r, but r*sqrt(3). This is because the center of the triangle forms a right triangle with one of the charged rods, and the hypotenuse of that triangle is r*sqrt(3).

So, the formula for the electric field strength at the center of the triangle would be:
Q/(4*pi*e_0*r*sqrt(3)*sqrt(r^2 + (L/2)^2))

Also, since two of the rods are positively charged and one is negatively charged, the net electric field at the center would be the sum of the electric fields due to each rod. This means that the final answer should be:
2*Q/(4*pi*e_0*r*sqrt(3)*sqrt(r^2 + (L/2)^2))

I hope this helps clarify your understanding. Keep up the good work!