Charge Distrubution evenly on Arc (Radius R)

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The discussion revolves around calculating the electric field at the center of an arc with a uniformly distributed charge Q. The mathematical expression derived for the electric field E as a function of the angle θ is E = 2kQ/(πR²). There is confusion about the integration limits, with a suggestion to integrate from -θ0 to θ0 instead of -π/2 to π/2. Participants also express uncertainty about how to sketch the electric field as a function of θ, indicating that completing the solution is necessary before creating the plot. Overall, the focus is on refining the mathematical approach and understanding the graphical representation of the electric field.
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Homework Statement



A charge Q is distributed evenly on a wire bent into an arc of radius R, as shown in
the figure below.What is the mathematical expression that describes the electric field
at the center of the arc (point P indicated) as a function of the angle θ? Sketch
a graph of the electric field as a function of θ for 0 < θ < 180.

I added the figure for the question as an attachment.



The Attempt at a Solution



lambda=Q/pi R

dE= kdQ/R^2
dE= (kdQ/R^2) cos θ

dQ=lambda dl
dl= Rd theta
dQ=lambda R dθ

dE=(k[lambda R d θ]/R^2)cos θ


E=(k lambda R cos θ/R^2)d θ(from pi/2 to -pi/2)
E=k lambda/R cos θ dθ
E=k lambda/Rsin θ
E=k lambda/R[sin(pi/2)-sin(-pi/2)]
E=k lambda/2R
E=k(Q/piR)/2R=2kQ/piR^2
E= 2kq/piR^2

Is this right? I used K for 1/4pi(E) to make it easier to type.
Also i am kind of lost on what is expected for the sketch.
 

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looks good ... might make more sense if the E-field component was upward (to balance gravity, if you put a charged item in the center, say)
 
Why are you integrating from -π/2 to π/2 ?

Integrate from -θ0 to θ0 or similar.
 
i didn't notice that. i was integrating from the wrong limits. Thank you for pointing that out. I am still confused on the sketch of the electric field as a function of theta if anyone could point me in the right direction with this it would be much appreciated.
 
Finish working out the solution before you can do the plot.
 
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