- #1

- 32

- 0

## Homework Statement

3 rods each of length 1 meter form an equilateral triangle. Two rods have a uniform charge distribution of [tex]8\times 10^{-6}[/tex] C and the third of [tex]-8\times 10^{-6}[/tex] C. What is the electric field strength at the center of an equilateral triangle

## Homework Equations

[tex]\vec{E} = \frac{8.99\times 10^{9} \times q}{r^{2}}[/tex]

## The Attempt at a Solution

I worked out the magnitude of the electric field strength at the center due to only one of the rods.

I got the following integral:

[tex]\vec{E} = 2\times 8.99\times 10^{9}\int_{0}^{l}(\frac{y\lambda dx}{(x^{2}+y^{2})^{\frac{3}{2}}})\widehat{\vec{y}} = \frac{2\times 8.99\times 10^{9}\lambda l}{y\sqrt{y^{2}+l^{2}}}\widehat{\vec{y}}[/tex]

After that I substituted the values given in the question and got:

[tex]\vec{E} = \frac{2\times 8.99\times 10^{9}\times 8\times 10^{-6}\times 0.5}{(\frac{0.5}{\sqrt{3}})\sqrt{(\frac{0.5}{\sqrt{3}})^{2}+(0.5)^{2}}} = 431520[/tex]

Since the horizontal components of the positively charged rods are equal but opposite, they cancel out so we have the following:

[tex]\vec{E} = (2\times 431520 \times sin(30))+(431520)) = 863040 N/C[/tex]

But this answer is wrong.

I have been trying to get this question done since yesterday and never got the right answer. Would very much appreciate your help.