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Eletric Field of a Ring

  • #1
  • Homework Statement

  • A Ring with center in (0,0) and R radius.
  • The charge distribution from the ring is: ψ(θ) = ψo*Sin(θ), where θ is the angle from the x axis (counterclowise).
    Negative values of the sine determine negative charge, and 0 no charge at all.
  • What is the field (E) created from this distribution (only x and y components)?

Homework Equations


  • If there we introduce a constant eletric field, F, in the positive direction of the y axis. How can i find the
    equipotentials lines from the sum of the fields?

The Attempt at a Solution


I tried to find the dEx and dEx as a function of dθ, so i could integrate from 0 to 2*π and get the field vector as function of x and y.
I found:
  • dEx = K*ψ(θ)*[x-R*Cos(θ)]*dθ/[ (y-R*Sin(θ))2+(x-R*Cos(θ))2 ]3/2
  • dEy = K*ψ(θ)*[y-R*Sin(θ)]*dθ/[ (y-R*Sin(θ))2+(x-R*Cos(θ))2 ]3/2
  • where K = 1/(4*π*εo) or about 9*10^9 N*m2*C-2
But when i calculate the integration i get 0 for both. Any ideas?
I can add a picture later, i can't now.
 

Answers and Replies

  • #2
Simon Bridge
Science Advisor
Homework Helper
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There is too much missing from the problem statement to complete the assignment.
ie.
What is the orientation of the ring? Is it in the x-y plane?
Where are we supposed to find the ##\vec E## field? only in the x-y plane? (we are only asked for x,y components).

I tried to find the dEx and dEy as a function of dθ,
... you mean you tried to find ##dE_x## and ##dE_y## in terms of ##\theta##
The reasoning usually goes like this: The electric field element ##d\vec E## due to the charge element ##dq## on the ring between ##\theta## and ##\theta +d\theta## is given by ... etc.

But when i calculate the integration i get 0 for both. Any ideas?
Looks like you have done the calculus wrong.
 

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