Voltage on a point charge from a stick of charge

AI Thread Summary
The discussion focuses on calculating the electric potential energy of a point charge near a charged rod with a linear charge density of ßx. The integral approach is emphasized, with the correct formulation being the second integral, which accounts for the charge distribution along the rod. The voltage is derived from the charge density rather than the point charge itself, as the point charge does not affect the voltage generated by the rod. The integral is set up from 0 to L, with the expression involving ßx in the denominator. Understanding the relationship between the charge distribution and the resulting electric potential is crucial for solving the problem.
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Homework Statement


In the figure the rod has a charge density of ßx, where x=0 is the left end of the rod, etc. Find the electric potential energy of the point charge q a distance a from the end of the rod and lying along the rod's axis.
http://www.usi.edu/science/physics/pickett/206/4p4f1.jpg

Homework Equations


V=\frac{q}{4\pi\epsilon r}


The Attempt at a Solution


Since it is voltage from the whole rod it should be an integral, but where I'm not sure at is what should be on top of the fraction

The integral is from 0 to L
\int\frac{q}{4\pi\epsilon (a+L-x)}dx
or
\int\frac{\beta x}{4\pi\epsilon (a+L-x)}dx
 
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To calculate the voltage where the point charge is, the point charge itself is irrelevant. You want the voltage generated by the charge distribution on the rod, so it's the second integral.
 
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