Electric Fields in Equilateral Triangle

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SUMMARY

The discussion focuses on calculating the electric field at the midpoint of a side of an equilateral triangle with point charges. A charge of +2.0 µC is placed at each corner, and the electric field is determined using the formula E=KQ/R². The participant correctly identifies that the electric fields from the two opposite charges cancel out, leaving only the contribution from the charge at the apex. The calculated electric field magnitude is 113,422 N/C, using R = 0.23√3 m.

PREREQUISITES
  • Understanding of electric fields and point charges
  • Familiarity with Coulomb's Law and the formula E=KQ/R²
  • Knowledge of geometry, specifically properties of equilateral triangles
  • Ability to manipulate square roots and basic algebraic expressions
NEXT STEPS
  • Study the concept of superposition of electric fields in multiple charge systems
  • Learn about electric field lines and their representation for point charges
  • Explore the application of electric fields in different geometric configurations
  • Investigate the effects of varying charge magnitudes and distances on electric field strength
USEFUL FOR

Students in physics, particularly those studying electromagnetism, as well as educators and anyone interested in understanding electric fields in geometric arrangements.

longcatislong
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Homework Statement



A point charge q = +2.0 µC is placed at each corner of an equilateral triangle with sides 0.23 m in length.
What is the magnitude of the electric field at the midpoint of any of the three sides of the triangle?


3. I want to know if I did this correctly, or totally screwed it up

At point p, the midpoint between on a side, the electric fields will cancel from the two point charges directly opposite each other. The only electric field will be felt from the point charge that's on the tip of the triangle.

So, I did E=KQ/R^2. I found R=.23Sqrt2, by knowing I have a 30-60-90 triangle if a bisect it.

E=(9e9)(2e-6)/(.23Sqrt3)^2=113422 N/C
 
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I don't get the same value for R, but it might just be me.
 

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