Use direct integration over the charge on a wire segment to find the potential V(z) on the z-axis due to a finite segment of wire with linear charge density lambda extending along the x-axis from x = - a/2 to +a/2
The electric potential dV at some point P due to the charge element dq is
dV = ke dq / r
where r is the distance from the charge element to point P.
Thus the total potential can be found by integrating:
V = ke integral of dq / r
The Attempt at a Solution
I'm sort of lost here, but I have a few thoughts. I'd appreciate some critique of my work here and hopefully some help...unless, of course, I'm right. Then letting me know that would surely help :D.
I assume the wire has length a (since it extends from - a/2 to +a/2). Using right-triangle geometry, I believe the distance from the x-axis to any element on the z-axis will be equal to the square root of a^2 + b^2 (where b is the distance on the z-axis). So, I'll integrate this over the limits x = o to x = a and I come up with V = ke lamdba integral of dx / square root of a^2 + b^2. Such an integral appears in the table in my book and so V becomes equal to
ke lamda ln(a + (square root of a^2 + b^2).
Am I close here...or way off? Thanks a bunch for any help!