Physics help :)

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smooticus

Active member
Hey guys, lets see if you guys can help me before the guys over at physicsforums

I have a semi circle of uniform charge (Q) and length (L). I need to find the equation for the electric field at what would be the center of the full circle. I just need some help getting started, I know that by symmetry the y components will cancel (i have the semicircle upright on the left side of my y axis). but other than that i dont know how to move forward.

Am I looking to put the electric field equation in terms of r and \theta and then integrate twice? or can I put this in terms of x and integrate that way? BTW I do know the answer just need to know exactly how its done.

 
physics forums is cool sometimes, they talk about smart stuff. unfortunately, i only know chem and bio so i'm of no help.
 
copied this off yahoo answers, I am decent at college physics, but I seem to have forgot this

lamda=Q/pi R = charge density on the semicircle

a semicircle with a radius R

In differential notation we have at the center of the s-circle

dE=kdq/R^2, here dE is a vector opposite to the radius

we have dq=lamda dl

so

dE=k lamda dl /R^2

Since dE is a vector, all the horizontal components of dE neutralised themselves. Only the vertical components add up .Then we write

dEv=(k lamda dl) cos teta /R^2, where teta is the angle between the vertical and the vector dE. In this equation the only variable is teta but before the integration we have to change dl=???

since we know that teta=l/R, then l=teta R

then

dl=R d(teta)

then: dEv=(k lamda ) cos teta R d(teta)/R^2

dEv= (k lamda) cos(teta) d (teta)/R

Let's rearrange

dEv= ((k lamda)/R)cos(teta) d (teta)

we are ready now to integrate the above equation with teta varying from pi/2 to - pi/2

Ev=((k lamda)/R) integral (cos(teta) d (teta))

Ev=((k lamda)/R )sin (teta) with teta varying from pi/2 to - pi/2

Ev=((k lamda)/R)( 1- -1)

Ev= (2 k lamda)/R where lamda is equal to Q/pi R

Ev= 2 k Q/pi R^2 whith k is the constant that you will find in your textbook

The electric field is in the -y direction
 
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