Calculating the Electric Field of a Long, Thin Wire Using Gauss' Law

AI Thread Summary
The discussion focuses on calculating the electric field at a point near a long, thin wire using Gauss' Law. The initial approach involved integration, but it was suggested that using Gauss' Law simplifies the process, treating the wire as a cylinder. The correct formula derived is E = λ / (2πε0R), where λ is the linear charge density and R is the distance from the wire. The final calculation confirmed that the electric field depends solely on the distance to the wire, reinforcing the effectiveness of Gauss' Law for this scenario. The solution emphasizes the importance of assuming an infinitely long wire for accurate results.
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Homework Statement


Calculate the electric field at a point 2.79 cm perpendicular to the midpoint of a 2.02 m long thin wire carrying a total charge of 5.04 uC.

You could integrate BUT if the wire is very long compared to the distance from the wire to where you are calculating the electric field, then the electric field will be radial and Gauss's law might be easier.

Homework Equations



ER=λ/2piε0R

The Attempt at a Solution


First I tried doing it by integral (getting E=kexQ/(x2+R2)3/2) but then it gave me the hint to use Gauss' Law, so I treated it like a cylinder, using charge divided by length for λ and 0.0279 m for R. I came out with 5.76x107 N/C which I am told is wrong.

THEN I worked that to E=2keλ/R after having a "duh" moment but I still seem to be getting the wrong answer (2.58x105)

Edit: calculator/math error. I solved it.
 
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The situation is very easy if you use assume really an infinitely long wire. Then, the electric field can only depend on the distance to the wire, say ρ in cylindrical coordinates.
Gauss's law \int_{\partial V}\mathbf{E}d\mathbf{A}=\frac{1}{\varepsilon_0} \int \rho dV
becomes:
2\pi \rho E_\rho(\rho)\int dz=\frac{1}{\varepsilon_0} \lambda \int dz​
with charge density \lambda [C/m]. This is a rather dirty trick to see that
E_\rho(\rho)=\frac{\lambda}{2 \pi \varepsilon_0 \rho }\ .​
This result is correct in the infinite-length-limit of a line charge. Here you can find the full calculation.
 
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