Find the electric field of a long line charge at a radial distance

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The discussion focuses on calculating the electric field of a long line charge at a radial distance where the potential is 24V higher than at r_1=3m, with an initial electric field of 4V/m. The calculated electric field at the specified distance is 29.5V/m, and the linear charge density is determined to be 6.7*10^-10. The user encountered issues with an integral diverging while trying to calculate the potential, leading to confusion about the approach. Clarification is provided that the problem requires evaluating the potential difference over finite limits rather than considering an infinite point. The conversation emphasizes the importance of correctly setting up the integral for accurate results.
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TL;DR Summary: Find the electric field of a long line charge at a radial distance where the potential is 24V higher than at a radial distance r_1=3m where E=4V/m. Answer: 29.5V/m.

Never mind: I retract this question. The integral apparently is supposed to diverge! I apologize for not reading https://physics.stackexchange.com/questions/407797/potential-due-to-line-charge before I posted my question.

I am reading the book Electromagnetics with Applications by Kraus and Fleisch and have run into a snag with Problem 2-3-4.

Find the electric field of a long line charge at a radial distance where the potential is 24V higher than at a radial distance r_1=3m where E=4V/m. Answer: 29.5V/m.

For a line charge, the electric field is

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(rho_L is the linear charge density). Since we know E_r=4 at r=3, we can calculate rho_L=6.7*10^-10. To calculate the potential at r=3, I use

render002.png


but this integral diverges ... where did I go wrong? \hat{r} is the unit vector orthogonal to the line.
 
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Why did you even bother to consider the point at infinity? The problem is essentially asking you to do an integral over finite limits that you will give the potential difference of 24 V.

If, as your title implies, you have a solution, please post it here so that others can profit from it.
 
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