Electric field and an equilateral triangle

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Homework Help Overview

The problem involves calculating the electric field strength at the center of an equilateral triangle formed by three charged rods. Two rods are positively charged, and one is negatively charged, with specific lengths and charge values provided.

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • Participants discuss the calculation of the triangle's dimensions, including the radius and height, and question the correct interpretation of the variables used in the electric field equation. There is uncertainty about the correct value for L and how to apply it in the formula.

Discussion Status

The discussion is ongoing, with participants attempting to clarify the dimensions of the triangle and how they relate to the electric field calculation. Some guidance has been offered regarding finding the height, but there is no explicit consensus on the correct approach yet.

Contextual Notes

Participants are working under the constraints of the problem statement and are trying to reconcile their calculations with the physical setup of the triangle. There is a focus on ensuring the correct interpretation of the triangle's geometry and the application of relevant equations.

abeltyukov
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Homework Statement



Three 8 cm long rods form an equilateral triangle. Two of the rods are charged to +8 nC, the third to -8 nC. What is the electric field strength at the center of the triangle?


Homework Equations



E = 4piKλL

The Attempt at a Solution



I found the radius of the triangle to be 4.2688 cm. I tried doing E = 4(pi)(9 x 10^9)(100 nC/m) (.042688 m) but that seems wrong. Is L not the radius of the triangle?


Thanks!
 
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Whats the height of the triangle? Divide that by 2
 
I still don't get the right answer when I plug in 0.0472 m for L. Any ideas?

Thanks!
 
Thats still not it...start by finding the height
 
The height is 4.72 cm, right? sqrt (8^2 + 4^2). Where do I go from there?

Thanks!
 
This is not resolving vectors if that's what youre thinking. The hypotenuse is 8 cm and one of the sides is 4 cm. Then you need to cut that height in half
 

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