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Electric Potential in Insulated UTube 
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#1
Mar706, 10:35 AM

P: 1

Hey, I have this question, and I'm having quite a bit of difficulty figuring it out.
"An insulating rod is bent into a semicircular arc of radius a, and a total electric charge Q is distributed uniformly along the rod. Calculate the potential at the center of curvature of the arc if the potential is assumed to be zero at infinity." Which i see to be as so: ..................................................................... ..... ..................................................................... ....... ..............o..........< the 'o' represents the center of curvature of ...........\...... /............................................the arc.... .............\.../......< this part is round......................... ...............u....................................................... ...... There is a similar question in my textbook, where tehy work out the potential at a point from a ring of charge. they use the equasion: V= (1/4[pi]E0) [integral] dq/r [where dq would be infinitismal elements of charge about the ring] for the ring, the r distance is always the same; so they remove r and integrate dq, to end with: V= (1/4[pi]E0) ( Q / r ) now, my question is the U tube. I figure, since under the center of curvature lies a semicircle, maybe you could take half the potential found from the previous equasion; i.e., for the curvature part, you could use: 0.5(1/4[pi]E0)( Q / r ) but, there is nothing in my textbook to say what I would do in terms of the extending arms of the U tube; as well, the question doesn't say anything about the length of the arms. what is the significance of defining V to be zero at infinity, and where should I go from here? Thanks, PK_Kermit 


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