How Is the Electric Potential Difference Calculated Near an Infinite Wire?

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SUMMARY

The electric potential difference near an infinite wire can be calculated using Gauss's Law. The surface charge density, denoted as ρ_S, contributes to the electric field, which radiates outward from the wire. The correct formula for the electric field E at a distance b from the center of the wire is E = ρ_Sa / (ε₀b). It is crucial to include the permittivity of free space, ε₀, in the calculations to ensure accuracy.

PREREQUISITES
  • Understanding of Gauss's Law
  • Familiarity with electric fields and surface charge density
  • Knowledge of electrostatics concepts
  • Basic calculus for integration
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  • Study the application of Gauss's Law in various geometries
  • Learn about electric fields generated by different charge distributions
  • Explore the concept of electric potential and its relationship to electric fields
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sandy.bridge
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Homework Statement


[itex]\rho_{wire}=a[/itex], surface charge density [itex]\rho_S[/itex]

What is potential difference of a point a distance of [itex]b[/itex] measured from the centre of the infinitely long wire, and the surface of the wire?

Since all of the charge is dispersed on the surface of the conductor, there will exist no charge within the surface, and henceforth the electric field inside the conductor is zero. In addition, symmetry of the problem tells us that the electric field radiates outwards from the conductor.

Let the origin be centred in the wire, such that the z-axis runs parallel to the infinitely long wire. We can apply Gauss's Law utilizing a gaussian surface such that [itex]\rho_{gaussian}=g > a[/itex], and the height from the xy-plane will be [itex]l[/itex].

At this point, I just want to know if I am solving for the electric field properly. Thanks in advance.

We know that
[tex]\frac{1}{ε_o}\int _V \rho dV=\oint_S\vec{E}\cdot\vec{dS}[/tex]
[tex]\frac{\rho_S(2\pi al)}{ε_o}=\oint_S\vec{E}\cdot\vec{dS}=E\int_SdS=E(2\pi gl)\rightarrow E = \frac{\rho_Sa}{b}[/tex],
 
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The only mistake you made was to omit ε0 in the denominator.
 

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