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

In summary, the problem asks to calculate the electric field at a point perpendicular to the midpoint of a thin wire with a length of 2.02 m and total charge of 5.04 uC. The initial attempt was to use integration, but Gauss's Law was suggested as an easier method. By treating the wire as an infinitely long cylinder and using charge divided by length for lambda, the electric field was calculated to be 2keλ/R. This was found to be the correct answer after a math error was corrected.
  • #1
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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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  • #2
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 [itex]\int_{\partial V}\mathbf{E}d\mathbf{A}=\frac{1}{\varepsilon_0} \int \rho dV[/itex]
becomes:
[itex]2\pi \rho E_\rho(\rho)\int dz=\frac{1}{\varepsilon_0} \lambda \int dz[/itex]​
with charge density [itex]\lambda[/itex] [C/m]. This is a rather dirty trick to see that
[itex]E_\rho(\rho)=\frac{\lambda}{2 \pi \varepsilon_0 \rho }\ . [/itex]​
This result is correct in the infinite-length-limit of a line charge. Here you can find the full calculation.
 

What is Gauss' Law?

Gauss' Law is a fundamental law in electromagnetism that relates the electric flux through a closed surface to the enclosed electric charge. It is named after the German mathematician and physicist Carl Friedrich Gauss.

How does Gauss' Law apply to a long, thin wire?

For a long, thin wire, Gauss' Law states that the electric field strength at a distance r from the wire is directly proportional to the amount of charge enclosed within a cylindrical surface of radius r centered at the wire. This means that the electric field strength decreases as the distance from the wire increases.

What is the significance of a long, thin wire in Gauss' Law?

A long, thin wire is used as an idealized model in Gauss' Law to simplify the calculations. This is because the electric field is assumed to be constant along the length of the wire and its direction is always radial, making the calculations easier.

How can Gauss' Law be applied in practical situations involving long, thin wires?

Gauss' Law can be used to determine the electric field strength at any point around a long, thin wire, which is useful in various practical situations. For example, it can be used to calculate the force on a charged particle placed near the wire or to find the voltage drop across the wire.

What are some limitations of Gauss' Law when applied to long, thin wires?

One limitation is that it assumes the wire to be infinitely long, which is not the case in real-world situations. Additionally, Gauss' Law only applies to static electric fields and does not take into account the effects of changing magnetic fields or moving charges.

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