Integral Form of Gauss' Law at Center of Finite Wire

In summary: So, if you cannot find the field at the central point of the wire, it's most likely because you are not considering a radial field.
  • #1
Nate Wellington
6
1
At the exact center of a finite wire (i.e. a distance, say $L/2$ from each end), why can I not apply Gauss's Law in integral form to find an EXACT solution for the electric field?

At the center of the wire, $E$ is entirely radial, so it seems like I should be able to draw an infinitesimally $\epsilon$ thick cylinder as my Gaussian surface, pull $E$ out of the integral (it should be a constant in the limit $\epsilon\to 0$), and get an exact expression for the electric field. Obviously, I do not.

Could someone explain in as mathematical a way as possible why this is the case?

BTW, I have similar questions regarding Ampere's Law. It seems like I should be able to apply that to finite wires as well. I know that the reason that doesn't work is that it only works for loops of wire, but again, just from the math, I cannot explain that either.

Thanks!
 
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  • #2
I am not an expert on this, but I will give your my thought. First, I think you need to specify that by finite wire, it means a finite line charge distribution(usually uniform, as the integral is easier to work with), not a current-carrying wire. Because there is hardly any electric field around a current-carrying wire, even though there is magnetic field around it. Then by exact center, do you mean at the central point of the wire? If so, then there is no electric field defined here, because we treat charges as concrete point without structure when we considering the problem. Mathematically, electric field vertical distance z away from the wire is given by E = kλ/z·[b/(z2+b2)1/2+a/(z2+a2)1/2] . You can see that E is not defined when r = 0.

You can learn more here: http://hyperphysics.phy-astr.gsu.edu/hbase/electric/elelin.html

As for Ampere's law, it works fine for a current carrying wire. ∮B·dS = μ0I. For magnetic field inside the wire, you just need to divide the current by relative surface area. So when you are at the exact center, I = 0 and B = 0.
 
  • #3
eifphysics said:
I am not an expert on this, but I will give your my thought. First, I think you need to specify that by finite wire, it means a finite line charge distribution(usually uniform, as the integral is easier to work with), not a current-carrying wire. Because there is hardly any electric field around a current-carrying wire, even though there is magnetic field around it. Then by exact center, do you mean at the central point of the wire? If so, then there is no electric field defined here, because we treat charges as concrete point without structure when we considering the problem. Mathematically, electric field vertical distance z away from the wire is given by E = kλ/z·[b/(z2+b2)1/2+a/(z2+a2)1/2] . You can see that E is not defined when r = 0.

You can learn more here: http://hyperphysics.phy-astr.gsu.edu/hbase/electric/elelin.html

As for Ampere's law, it works fine for a current carrying wire. ∮B·dS = μ0I. For magnetic field inside the wire, you just need to divide the current by relative surface area. So when you are at the exact center, I = 0 and B = 0.

For the Gauss's law case, I am talking about a wire with a uniform charge density lambda, as you say. And I am talking about finding the field a distance $z$ radially away from the midpoint of the wire. This is defined and easily calculable $\int_{-l/2}^{l/2} \lambda z dx/(x^2+z^2)^{3/2}$
 

Related to Integral Form of Gauss' Law at Center of Finite Wire

1. What is the integral form of Gauss' Law at the center of a finite wire?

The integral form of Gauss' Law at the center of a finite wire is a mathematical equation that relates the electric flux passing through a closed surface to the total charge enclosed within that surface. It is given by the equation ∮E·dA = Q/ε0, where E is the electric field, dA is the differential area element, Q is the total charge enclosed, and ε0 is the permittivity of free space.

2. How is the integral form of Gauss' Law at the center of a finite wire derived?

The integral form of Gauss' Law at the center of a finite wire can be derived from the differential form of Gauss' Law, which states that the divergence of the electric field at a point is equal to the charge density at that point divided by the permittivity of free space. The integral form is obtained by integrating the differential form over a closed surface and applying the Divergence Theorem.

3. What does the integral form of Gauss' Law at the center of a finite wire tell us?

The integral form of Gauss' Law at the center of a finite wire tells us that the electric flux passing through a closed surface is directly proportional to the total charge enclosed within that surface. This allows us to calculate the electric field at a point by knowing the charge distribution within a given region.

4. Does the integral form of Gauss' Law at the center of a finite wire apply to all types of charge distributions?

Yes, the integral form of Gauss' Law at the center of a finite wire applies to all types of charge distributions as long as the electric field is constant along the surface over which the integral is evaluated. It is a general law of electromagnetism and is not limited to any particular type of charge distribution.

5. How is the integral form of Gauss' Law at the center of a finite wire used in practical applications?

The integral form of Gauss' Law at the center of a finite wire is used in many practical applications, such as in the design of electric motors, generators, and capacitors. It is also used in the analysis of electric fields in transmission lines and in the calculation of the electric field inside a charged conductor. Additionally, it is an important tool in the study of electromagnetic waves and their propagation.

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