How Do You Find the Electric Field at the Origin From a Semi-Ring of Charge?

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SUMMARY

The discussion focuses on calculating the electric field at the origin due to a semi-ring of charge with total charge Q and radius R. The electric field E is determined using the equation E = k∫(r - r')ρ_l(r')/|r - r'|³ dl', where the integration limits are from -π/2 to π/2. The charge density is defined as λ = Q/(2πR). The user is advised to incorporate cos(θ) into the numerator of the integral to account for the direction of the electric field.

PREREQUISITES
  • Understanding of electric fields and charge distributions
  • Familiarity with calculus, particularly integration techniques
  • Knowledge of Gauss's Law and its limitations
  • Basic concepts of cylindrical coordinates
NEXT STEPS
  • Study the derivation of electric fields from continuous charge distributions
  • Learn about the application of Gauss's Law in non-symmetrical charge configurations
  • Explore the use of cylindrical coordinates in electrostatics
  • Investigate the effects of charge density variations on electric field calculations
USEFUL FOR

This discussion is beneficial for physics students, electrical engineers, and anyone involved in electrostatics or electric field analysis, particularly in scenarios involving non-uniform charge distributions.

seang
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Homework Statement


How do I find the electric field from a semi-ring of charge at the center of the would-be circle? Does that make sense? Or another way: If the semi-circle of charge is centered at the origin, what is the electric field at the origin? I hope you know what I mean now.

the total charge of the ring is Q, the radius is R.

Oh yeah, the semi circle is in the left hand plane

Homework Equations


[tex] E = k\int_{}^{l}\frac{(r-r')\rho _l(r')}{|r-r'|^3} dl'<br /> [/tex]

The Attempt at a Solution



[tex] E = k\int_{\pi/2}^{-\pi/2}\frac{(-r')\rho _l(r')}{|-r'|^3} d\theta[/tex]

Is converting to cylindrical coordinates the right move? Is this even the equation I should be using? What should I use as the charge density?
 
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Try applying Gauss's Theorem. Assume a gaussian surface enclosing the ring and [tex]\int E.ds=\frac{q}{\epsilon}[/tex]
 
Gauss won't work here because there is not full circular symmetry
There is enough symmetry to see the direction of the E field.
Just put cos\theta into the numerator of your integral.
rho is really a linear density lambda=Q/2pi R.
 

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