Charge Distrubution evenly on Arc (Radius R)

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SUMMARY

The discussion focuses on calculating the electric field at the center of an arc of radius R with a uniformly distributed charge Q. The derived mathematical expression for the electric field E as a function of the angle θ is E = 2kQ/(πR²), where k represents Coulomb's constant. The integration limits for the electric field calculation should be adjusted from -θ0 to θ0, rather than from -π/2 to π/2. Additionally, participants express confusion regarding the graphical representation of the electric field as a function of θ.

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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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