Where is the maximum electric field on the axis of a charged ring?

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SUMMARY

The maximum electric field (Emax) along the axis of a uniformly charged ring occurs at the position x = a/(√2) and has a magnitude of Emax = Q/(6√3πε0a²). This conclusion is derived from the application of the electric field equation E = keΔq/r², where ke represents Coulomb's constant. The vertical components of the electric field from opposite sides of the ring cancel out, leaving only the horizontal component along the axis.

PREREQUISITES
  • Understanding of electric fields and Coulomb's law
  • Familiarity with the concept of uniformly charged rings
  • Knowledge of vector components in physics
  • Basic calculus for deriving maximum values
NEXT STEPS
  • Study the derivation of electric fields from charged geometries
  • Learn about the properties of electric fields in three-dimensional space
  • Explore the concept of electric field lines and their significance
  • Investigate applications of electric fields in real-world scenarios, such as capacitors
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Students studying electromagnetism, physics educators, and anyone interested in understanding electric fields generated by charged objects.

leroyjenkens
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Homework Statement


Show that the maximum magnitude Emax of the electric field along the axis of a uniformly charged ring occurs at x=a/(sqrt2) and has the value Q/(6(sqrt3)πε0a2)


Homework Equations


E=keΔq/r2


The Attempt at a Solution


I made the vertical components cancel along the axis from the top and bottom of the ring so that I just have the x-axis field, but I really don't know what else to do. I'm not even sure if that equation is the right one to use. This question seems to be made deliberately confusing.
 
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Show your work in detail, please.

ehild
 

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