Electric Potential in Insulated U-Tube

[PK_Kermit]

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

 

[member 553083]

The significance of defining the potential to be zero at infinity is that it allows you to calculate the potential at any given point using the principle of superposition. This means that you can break down the problem into parts, and sum up the resulting potentials at each part to get the total potential at the center of curvature. For this particular problem, you can calculate the potential at the center of curvature by breaking the problem into two parts: the potential due to the arc of the U-tube and the potential due to the two arms of the U-tube. The potential due to the arc can be calculated using the equation you provided (V = (1/4πε₀) (Q/r), with r being the radius of the arc and Q being the total charge). For the arms, you can use the same equation, but with the distance from the center of curvature to the end of the arm as the value for r. Once you've calculated the potential due to each part, you can add them together to get the total potential at the center of curvature.