Potential for two semicircular rods

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SUMMARY

The discussion centers on calculating the electric potential at point P, located at the center of two semicircular rods with radii R1 and R2, where R2 > R1, each carrying a linear charge density of lambda coulombs per meter. The relevant equations include V = KQ/R and V = E*s. The user inquires whether the electric fields from the two horizontal rods cancel out and seeks confirmation on the expression for the electric field at point P, proposed as 2k(lambda)*(1/R1 + 1/R2). Additionally, the user questions the limits of integration for calculating the potential.

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Homework Statement


Two semicircular rods, one of radius R1 and the other of radius R2 where R2 > R1, are joined by two horizontal, straight rods. The rods carry a charge of lambda coulombs per meter. Calculate the potential at point p at the center of the two semicircles. (P would be the center of the two concentric circles if the rods were full circles)

This configuration looks like a rainbow basically and p is on the ground halfway between each side of the rainbow.


Homework Equations


V= KQ/R
V= E*s

The Attempt at a Solution



If I calculate V using the electric field, does the electric field between the two rods cancel so I would just have to use the electric field due to the inner rod? I know the electric fields of the two horizontal rods cancel.
 
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OK I suppose my real question is, is the electric field at point P

2k(lambda)*(1/R1 + 1/R2)

And then if that is so, to find the potential does the integral go from 0 to R1?
 

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