Homework Help: Eletric Field of a Ring

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1. Sep 25, 2016

Diego Rolim Porto

• 1. The problem statement, all variables and given/known data
• 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)?
2. Relevant 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?
3. 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.

2. Sep 26, 2016

Simon Bridge

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).

... 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.

Looks like you have done the calculus wrong.

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