Electric Flux Through Spherical Surface Centered at Origin

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Homework Help Overview

The discussion revolves around calculating the electric flux through a spherical surface centered at the origin, given a uniform linear charge density along the x-axis. The problem involves understanding the relationship between charge, linear charge density, and the geometry of the situation.

Discussion Character

  • Conceptual clarification, Assumption checking

Approaches and Questions Raised

  • Participants discuss the calculation of charge using the formula Q=λL and question the definition of L in this context. There is a focus on whether L should be defined as 2πr or 2R, considering the charge distribution along the axis.

Discussion Status

Some participants have provided guidance on the interpretation of L, suggesting that it should be 2R instead of 2πR. There is an acknowledgment of differing interpretations regarding the geometry involved in the problem.

Contextual Notes

The original poster indicates that their calculated flux does not match the expected answer, prompting a review of the assumptions and definitions used in the calculations. The discussion is framed within the constraints of a homework assignment, which may limit the information shared.

rizamadiyar
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A uniform linear charge density of 4.0 nC/m is distributed along the entire x axis. Consider a spherical (radius = 5.0 cm) surface centered on the origin. Determine the electric flux through this surface.

Homework Equations


L=2rπ
φ=Q/ε
λ=Q/L

The Attempt at a Solution


I found the charge by substituting values into Q=λL, then I found the flux, but the answer I get is incorrect, the correct one is 45 N m2/C

What am I missing?
 
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rizamadiyar said:
I found the charge by substituting values into Q=λL,
Where L is what? You wrote L=2πr. Is that what you used? On what basis?
 
haruspex said:
Where L is what? You wrote L=2πr. Is that what you used? On what basis?
Oh, I got this, L=2R, not 2πR, since the charge is distributed on the axis. Am I right?
 
rizamadiyar said:
Oh, I got this, L=2R, not 2πR, since the charge is distributed on the axis. Am I right?
Yes.
 

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