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

In summary, the conversation is about finding 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. The answer to this problem is 29.5V/m. The integral to calculate the potential at r=3 diverges, so the individual apologizes for not reading a related article before posting the question. The solution involves calculating the linear charge density and using it to do an integral over finite limits in order to find the potential difference of 24V.
  • #1
noowutah
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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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  • #2
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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Likes member 731016 and SammyS

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

What is the electric field of a long line charge at a radial distance?

The electric field (E) of a long line charge at a radial distance (r) can be determined using Gauss's Law. For a line charge with linear charge density λ (charge per unit length), the electric field at a distance r from the line is given by the formula: \( E = \frac{\lambda}{2 \pi \epsilon_0 r} \), where \( \epsilon_0 \) is the permittivity of free space.

What assumptions are made when calculating the electric field of a long line charge?

When calculating the electric field of a long line charge, several assumptions are typically made: (1) The line charge is infinitely long, (2) The charge distribution is uniform along the length of the line, and (3) The effects of any nearby objects or boundaries are negligible, allowing the use of symmetry.

How does the electric field vary with distance from a long line charge?

The electric field of a long line charge decreases with increasing radial distance from the line. Specifically, the electric field is inversely proportional to the radial distance (r), as given by the formula \( E = \frac{\lambda}{2 \pi \epsilon_0 r} \). This means that if the distance from the line charge doubles, the electric field strength is halved.

What role does the linear charge density (λ) play in determining the electric field?

The linear charge density (λ) directly affects the magnitude of the electric field generated by the line charge. A higher linear charge density results in a stronger electric field. The relationship is linear, meaning that if the linear charge density doubles, the electric field strength also doubles, as shown in the formula \( E = \frac{\lambda}{2 \pi \epsilon_0 r} \).

How can Gauss's Law be applied to find the electric field of a long line charge?

Gauss's Law can be applied to find the electric field of a long line charge by considering a cylindrical Gaussian surface with radius r and length L, centered on the line charge. The symmetry of the problem allows us to assume that the electric field is radial and uniform over the curved surface of the cylinder. By applying Gauss's Law, \( \oint \mathbf{E} \cdot d\mathbf{A} = \frac{Q_{\text{enc}}}{\epsilon_0} \), and solving for E, we find that \( E = \frac{\lambda}{2 \pi \epsilon_0 r} \).

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