Stuck on this one electric field problem

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SUMMARY

The discussion focuses on solving an electric field problem using the equation E=k*Q/r^2. The user attempts to calculate the components of the electric field from two charges (E1 and E2) and a third charge (E3), applying trigonometric functions to resolve the components. The user correctly identifies that E1 and E2 are equal and simplifies the y-component of E3. The method is validated by another participant, who suggests correcting the radius calculation from L^4 to L^2.

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  • Understanding of electric field concepts and equations, specifically E=k*Q/r^2
  • Basic knowledge of trigonometry, including sine and cosine functions
  • Familiarity with vector components in physics
  • Ability to manipulate algebraic expressions and equations
NEXT STEPS
  • Review vector addition in electric fields
  • Study trigonometric identities and their applications in physics
  • Learn about electric field superposition principles
  • Explore advanced electric field problems involving multiple charges
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Students studying physics, particularly those focusing on electromagnetism, as well as educators looking for examples of electric field calculations.

vitaebellaa
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Homework Statement


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Homework Equations


E=k*Q/r^2


The Attempt at a Solution



Ey=E1sin30 + E2sin30 + E3; where E1=E2 (I'm denoting the 2 bottom charges as E1,E2, the negative as E3)
Ex=E1cos30 - E2cos30, this cancels.
from E=k*Q/r^2
I calculated r^2 to be L^4/4 (don't know if this is right... trig. and I haven't really been on speaking terms since high school)
since sin30=1/2... E1(1/2) + E2(1/2) just equal (1)E1; E1=k*Q/(L^4/4)
I guess from there I would have to add the y component of E3, which would be k*Q/(L^2/4)
Is my math/'plan of attack' okay so far? Is the next step simply adding E1 + E3 and simplifying?

THANK YOU! :smile:
 
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hi vitaebellaa! :smile:

yes that's the correct method (obviously, you meant L2 not L4) …

for r, draw a right-angled triangle with length L/2 :wink:
 

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