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Old Dec 3rd 2009, 03:07 PM   #1
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electric field at the center of an arc

A charge Q is arranged evenly on a wire bent into an arc or radius R as shown in the Figure. What is the electric field at the center of the arc as a function of the opening angle theta? Sketch a graph of the electric field as a function of theta for 0<theta<180 degrees.

http://i848.photobucket.com/albums/a...g?t=1259862072

E= KQ/R^2 S cos theta (the S is suppose to be an integral symbol)


Is the above equation correct? If not, I'm not really sure what to do, also I'm not sure how the graph should look.
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Old Dec 5th 2009, 08:37 AM   #2
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lambda=Q/pi R

dE= kdQ/R^2
dE= (kdQ/R^2) cos theta

dQ=lambda dl
dl= Rd theta
dQ=lambda R d theta

dE=(k[lambda R d theta]/R^2)cos theta


E=(k lambda R cos theta/R^2)d theta(from pi/2 to -pi/2)
E=k lambda/R cos theta d theta
E=k lambda/Rsin theta
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?
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Old Dec 5th 2009, 09:36 AM   #3
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yes, but only if that arc is a semicircle.(which i think it is not, otherwise why will they ask u to plot it versus theta???)

Last edited by r.samanta; Dec 6th 2009 at 08:47 AM.
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Old Dec 5th 2009, 10:01 AM   #4
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It appears to be a semicircle from the figure given with the problem, which I provided a link for in my original post ( or at least the crappy version that I made in paint).
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Old Dec 6th 2009, 08:39 AM   #5
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Originally Posted by tag16 View Post
A charge Q is arranged evenly on a wire bent into an arc or radius R as shown in the Figure. What is the electric field at the center of the arc as a function of the opening angle theta? Sketch a graph of the electric field as a function of theta for 0<theta<180 degrees.

http://i848.photobucket.com/albums/a...g?t=1259862072

E= KQ/R^2 S cos theta (the S is suppose to be an integral symbol)


Is the above equation correct? If not, I'm not really sure what to do, also I'm not sure how the graph should look.
no dear.. if you consider this wire to be a straight wire then the electric field due to this wire must be

E=Q/4*pi*e*x^2*(x^2+a^2)^1/2
where
E=electric field,Q=charge,e=8.85x10^-12,x=length of wire,a=distance of wire and a test charge..
where in your case a=R(radius of the semicircle)..
just try over it

Last edited by jawad khan; Dec 6th 2009 at 08:41 AM.
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Old Dec 6th 2009, 09:19 AM   #6
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but the wire isnt straight!! please tell me what u think of the following
E= 2k lambda/R * sin theta where lambda= linear charge density=Q/2r theta
so E= k Q/R^2 sin theta/theta
as a check, put theta=0 , take limit theta tends to zero, u get Kq/R^2, just what we expect(wire becomes a point charge)
for theta= pi/2, u will get 2 k Q/ pi R^2 what u found.
graph will start from kQ/R^2 for theta=0, decrease to 2kQ/pi R^2 for theta = pi/2 and further decrease to zero as theta=pi (when wire becomes circular, fields cancel from all sides)
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Old Dec 6th 2009, 12:43 PM   #7
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Is it possible for you to show what that graph would look like? Thanks.
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Old Dec 6th 2009, 02:18 PM   #8
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would it look like this?

http://i848.photobucket.com/albums/a...g?t=1260137870
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