General question pertaining to a thin rod and E

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The discussion centers on calculating the electric potential due to a uniformly charged thin rod extending along the z-axis. The potential at point P1 on the z-axis is derived to be λ ln(3), while the potential at point P2 on the x-axis is also explored to match this value. There is confusion regarding whether the problem is asking for electric potential difference or potential energy, with an emphasis on the integration method used. The original formula for the electric field of a wire is questioned, particularly why it seems to be disregarded in the book's solution. The participant expresses frustration over the complexity of the concepts involved in the calculations.
enkerecz
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I was under the assumption that the electric field for an arbitrary length of wire was 2\lambda/r.

In this problem

A thin rod extends along the z-axis from z = -d to z = d. The rod carries a charge
uniformly distributed along its length with linear charge density lamda. By integrating over
this charge distribution, calculate the potential at a point P1 on the z-axis with coordinates (0,0,2d). By another integration find the potential at a point P2 on the x-axis and locate this point to make the potential equal to the potential at P1.

The potential at (0,0,2d) is \lambda ln(3)...

Of course, the potential is the integral of E\bulletds

the second part follows naturally from the first, so I'm not concerned with it.

Am I missing something here? Do that want potential difference or potential energy? I'm assuming it is electric potential difference, but their methods confuse me.. The book completely ignores the formula it gave for a wire and decided the answer to this was only the integral of (lamda/r) dr from point d to d+2d... How can they just drop off a scalar like that?
 
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Lambda = charge/length;

\lambdar/r^2=E=\lambda/r
I feel like a complete idiot for struggling with this for so long..
 
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