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

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SUMMARY

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 a distance of r1=3m, where the electric field E is 4V/m. The calculated electric field at this new radial distance is 29.5V/m. The linear charge density (ρL) is determined to be 6.7×10-10 C/m. The integral used for calculating potential diverges, indicating a need for careful consideration of limits in the integral.

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  • Experience with electromagnetics, particularly from "Electromagnetics with Applications" by Kraus and Fleisch
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Students and professionals in physics, particularly those studying electromagnetism, as well as educators looking for practical examples of electric field calculations related to line charges.

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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

render001.png


(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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