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 Electricity and Magnetism Electricity and Magnetism Physics Help Forum Dec 3rd 2009, 03:07 PM #1 Member   Join Date: May 2009 Posts: 54 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 Dec 5th 2009, 08:37 AM #2 Member   Join Date: May 2009 Posts: 54 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/R sin 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?   Dec 5th 2009, 09:36 AM #3 Senior Member   Join Date: Sep 2009 Location: india Posts: 409 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.   Dec 5th 2009, 10:01 AM #4 Member   Join Date: May 2009 Posts: 54 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).   Dec 6th 2009, 08:39 AM   #5
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 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
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.   Dec 6th 2009, 09:19 AM #6 Senior Member   Join Date: Sep 2009 Location: india Posts: 409 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)   Dec 6th 2009, 12:43 PM #7 Member   Join Date: May 2009 Posts: 54 Is it possible for you to show what that graph would look like? Thanks.   Dec 6th 2009, 02:18 PM #8 Member   Join Date: May 2009 Posts: 54 would it look like this? http://i848.photobucket.com/albums/a...g?t=1260137870  Tags arc, center, electric, field Search tags for this page
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# prove equation E=kqcos(theta)/r^2

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