Net electric field of a charged arc

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SUMMARY

The discussion focuses on calculating the net electric field at the origin due to charged arcs, specifically using the formula Enet=λ1(2sin45°)/(4πε0r1)+λ2(2sin45°)/(4πε0r2)+λ3(2sin45°)/(4πε0r3). The term 2sin(45°) arises from the geometry of the electric field vectors created by the arcs, which are quarter circles in the second quadrant. Participants express confusion about the derivation of this term and the setup of the integral needed to compute the net electric field. The conversation emphasizes the importance of understanding vector components in electric field calculations.

PREREQUISITES
  • Understanding of electric field concepts and vector addition
  • Familiarity with calculus, particularly integration techniques
  • Knowledge of electrostatics, including charge distribution and linear charge density (λ)
  • Basic trigonometry, specifically sine functions and their application in physics
NEXT STEPS
  • Review the derivation of electric fields from continuous charge distributions
  • Learn how to set up and evaluate integrals for electric fields in different geometries
  • Study the application of trigonometric functions in vector analysis within physics
  • Explore the concept of linear charge density (λ) and its implications in electric field calculations
USEFUL FOR

Students in physics, particularly those studying electromagnetism, as well as educators and anyone seeking to deepen their understanding of electric fields generated by charged arcs.

Jrlinton
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Homework Statement


So this was a problem worked in class by the professor in class.
Find the net electric field at the origin due to the arcs
upload_2017-2-17_14-44-14.png


Homework Equations


L=2πr/4
λ=q/L
E=kQ/r2

The Attempt at a Solution


So the professor gave the answer using the fromula
Enet1(2sin45°)/(4πε0r1)+λ2(2sin45°)/(4πε0r2)+λ3(2sin45°)/(4πε0r3)

I can understand the formula except for where the 2sin45° comes from. I mean the arc is a quarter circle of 90° in the second quadrant but I am unsure where this term comes from and how it would change if the parameters of the problem were to change.
 
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See if this diagram helps
upload_2017-2-17_15-10-28.png
 
So if U am understanding this correctly then it is the added sine values of the two angles created when drawing the vector for the field?
 
Jrlinton said:
So if U am understanding this correctly then it is the added sine values of the two angles created when drawing the vector for the field?
I'm not following what you are saying here. In order to see why there is a factor of 2sin(45o) in the answer, you need to go through the derivation. This means setting up and evaluating the integral for the net electric field. Are you having trouble setting up the integral?
 
Yes, I am having trouble coming up with the integral.
 
Jrlinton said:
Yes, I am having trouble coming up with the integral.
OK. Show us your attempt at setting up the integral and we can go from there. It helps to first consider the direction of the net electric field.
 

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