Take one end of the rod as the origin, establish the X axis along the direction of the rod, and take the infinitesimal dx on the rod.

Then dq=λdx

Is the potential of infinitesimal at point p dU = dq/4ωω(a+L-x)=? λdx/4ωω(a+L-x)？

P point potential

U=(？ λ/4ωω)∫[ 1/(a+L-x)]dx

Substitute the upper limit of integral l? Is the lower limit 0 integral available?

U=？ (λ/4ωω)ln( 1+L/a)