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Electric Potential in Insulated U-Tube

by PK_Kermit
Tags: electric, insulated, potential, utube
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PK_Kermit
#1
Mar7-06, 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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