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dipole knight
dipole knight is offline
#1
Nov8-12, 10:52 AM
P: 3
Hi,

on page 63 of David J. Griffiths' "Introduction to Electrodynamics" he calculates the electric field at a point z above a line charge (with a finite length L) using the electric field in integral form.
[itex]E_z = \frac{1}{4 \pi \epsilon_0} \int_{0}^{L} \frac{2 \lambda z}{\sqrt{(z^2 + x^2)^3}} dx = \frac{1}{4 \pi \epsilon_0} \frac{2 \lambda L}{z \sqrt{z^2 + L^2}}[/itex]
whereas [itex] \lambda = \frac{Q}{L}[/itex]
Basically it's a two-dimensional system with a horizontal x-axis and a vertical z-axis, the charges go from -L to L on the x-axis and we look at the electric field a distance z above the line (i.e. on the z-axis).

That's all fine and dandy but I have some serious troubles trying to reproduce that same result with the corresponding Maxwell equation in differential form and the divergence theorem.
This is what I got so far:

[itex]
\vec{\nabla} \vec{E} = \frac{\rho}{\epsilon_0} \\
\int \vec{\nabla} \vec{E} dV = \int \frac{\rho}{\epsilon_0} dV \\
\int \vec{E} d\vec{A} = \int \frac{\lambda}{\epsilon_0} dl \\
[/itex]
whereas I used [itex] \rho dV \propto \lambda dl [/itex]

For the left side I use cylindrical coordinates and get:
[itex]
\vec{E} \hat{x} = \frac{1}{2 \pi \epsilon_0 x z} \int {\lambda} dl
[/itex].

Since λ is constant I can pull it out of the integral and when I integrate I get as a final result:
[itex]\vec{E} \hat{x} = \frac{2 \lambda L}{4 \pi \epsilon_0 x z}[/itex]

Now this is a completely different result than what I get when I use the formula for the electric field in the integral form. One of the problems that this happens is that the integral form actually has the vector difference between the position of the charge and the point at which you want to calculate the electric field, i.e. [itex] \int \lambda \frac{\vec{r}-\vec{r'}}{\| \vec{r} - \vec{r'}\|^3} dl [/itex] whereas this is not the case in the Maxwell equation.

What am I doing wrong? Why cannot I reproduce the same result?
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