Electric Field inside a charged ring

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SUMMARY

The electric field inside a charged ring is determined to be zero due to the symmetry of the system. The potential within the ring remains constant, leading to the conclusion that the electric field, which is the gradient of potential, is also zero. This conclusion is supported by vector calculus, where integrating the electric field contributions from symmetrical points on the ring results in equal and opposite fields that cancel each other out. The discussion emphasizes the importance of symmetry in simplifying the analysis of electric fields in charged configurations.

PREREQUISITES
  • Understanding of electric fields and potentials
  • Familiarity with vector calculus
  • Knowledge of symmetry in physics
  • Basic concepts of integration in physics
NEXT STEPS
  • Study the concept of electric potential and its relationship to electric fields
  • Learn about vector calculus techniques for evaluating electric fields
  • Explore symmetry in electrostatics and its implications for field calculations
  • Investigate the method of integrating electric fields from continuous charge distributions
USEFUL FOR

Students studying electromagnetism, physicists interested in electrostatics, and educators teaching electric field concepts.

yeezyseason3
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Homework Statement


Given a charged ring in 2-d, what is the e-field inside the ring?

Homework Equations


Epoint = kq/r^2

The Attempt at a Solution


This isn't a homework question, but more of a problem I keep running into whenever I think about it. I assumed it was 0. I came to this conclusion because the gradient of potential is electric field, and potential within a ring is constant, therefore electric field is 0.
 
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You can check by considering that the electric field is a vector... the field dead center is easy, you are wondering about off-center right?
 
Simon Bridge said:
You can check by considering that the electric field is a vector... the field dead center is easy, you are wondering about off-center right?
Yea I am, I tried integrating and it got really messy, I assumed I was doing something wrong and instead assumed that potential is constant and hence e field is 0.
 
You "assumed" the potential was constant?
Didn't you calculate it?

Trying to do the vector calculus is nasty - try exploiting the line of symmetry through the center of the ring and the point you want to find the field for ... if you integrate equal angles either side of that line for a short arc, you should be able to find another similar arc on the opposite side that will integrate to equal and opposite field.
 
yeezyseason3 said:
I tried integrating and it got really messy
If you show what you did, perhaps we can help ?
 

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