A line of charge as a delta function

In summary, the charge density is in terms of delta function volumes, and is given by: \rho(\vec{r}') = \frac{1}{a}\lambda \delta(r' - a) \delta (\phi' - \phi_0).
  • #1
AxiomOfChoice
533
1
Can someone tell me how to express a line of charge of charge per unit length [tex]\lambda[/tex] as a delta function volume charge density in cylindrical coordinates?
 
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  • #2
Ok, here's my guess:

[tex]
\rho(\vec{r}') = \frac{1}{a}\lambda \delta(r' - a) \delta (\phi' - \phi_0).
[/tex]

Assuming that the line of charge is infinite, parallel to the [itex]z[/itex]-axis, a distance [itex]a[/itex] from the origin, and at an angle [itex]\phi = \phi_0[/tex] w.r.t. the [itex]x[/itex]-axis. Sound good?
 
  • #3
...anyone? Please?

Would this be the correct charge density in rectangular coordinates:

[tex]
\rho(\mathbf{r}') = \lambda \delta(x' - a) \delta (y' - a)
[/tex]
 
  • #4
AxiomOfChoice said:
Ok, here's my guess:

[tex]
\rho(\vec{r}') = \frac{1}{a}\lambda \delta(r' - a) \delta (\phi' - \phi_0).
[/tex]

Assuming that the line of charge is infinite, parallel to the [itex]z[/itex]-axis, a distance [itex]a[/itex] from the origin, and at an angle [itex]\phi = \phi_0[/tex] w.r.t. the [itex]x[/itex]-axis. Sound good?

This is correct. Why? There are two things: first the 2-D delta function is right, and second, the charge density is right. First the delta function. You are integrating over an area in the x-y plane, so
[tex]
dA = r\, d\phi \, dr.
[/tex]
You require your 2-D delta function [tex]\delta(\mathbf{r}-\mathbf{\rho})[/tex] to satisfy
[tex]
f(a,\phi_0) = \int \, dA \, \delta(\mathbf{r}-\mathbf{\rho}) \, f(r, \phi).
[/tex]
Above, [tex]\mathbf{\rho}[/tex] is your source location in the x-y plane, and [tex]\delta(\mathbf{r}-\mathbf{\rho})[/tex] is simply a symbol to represent the delta function that you are looking for. Clearly your choice above satisfies this:
[tex]
\int dr \, \int r\, d\phi \, \frac{1}{a} \delta(r - a) \delta (\phi - \phi_0) \, f(r, \phi)
= \int dr \, r\, \frac{1}{a} \delta(r - a) \, f(r, \phi_0)
= a \frac{1}{a} f(a,\phi_0) = f(a,\phi_0).
[/tex]
Note that your cartesian coordinate version you posted is also okay.

Now for the charge density. Beyond the delta function you basically just need to make sure you have the right units. your "r" delta function has units of one over length, so your charge density has teh right units.
 
Last edited:
  • #5


A line of charge can be represented as a delta function in cylindrical coordinates by using the formula for volume charge density, which is given by ρ = λδ(ρ)δ(φ)δ(z). Here, ρ represents the radial distance from the origin, φ represents the angle in the cylindrical coordinate system, and z represents the distance along the axis of the line of charge. By multiplying the charge per unit length, λ, with the delta function in the cylindrical coordinates, we can express the line of charge as a delta function volume charge density. This representation allows us to simplify calculations and analyze the electric field and potential due to the line of charge in cylindrical coordinates. I hope this helps.
 

1. What is a line of charge as a delta function?

A line of charge as a delta function is a theoretical concept used in physics to represent an infinitely thin, one-dimensional line of charge with a very high charge density. It is often used as a simplified model for calculating the electric field and potential of a charged object.

2. How is a line of charge as a delta function different from a regular line of charge?

A regular line of charge has a finite thickness and a non-uniform charge distribution, while a line of charge as a delta function has zero thickness and a uniform charge distribution. This makes calculations with a delta function line of charge much simpler, but it is not a physically realistic representation of a real line of charge.

3. Can a line of charge as a delta function exist in the real world?

No, a line of charge as a delta function is a theoretical concept and cannot exist in the real world. It is used as a simplified model for mathematical calculations, but it does not accurately represent any physical object.

4. How is the electric field of a line of charge as a delta function calculated?

The electric field of a line of charge as a delta function can be calculated using Coulomb's Law, which states that the electric field at a point is equal to the charge at that point divided by the square of the distance from the point to the charge. In the case of a delta function line of charge, the charge is considered to be concentrated at a single point with zero distance, resulting in an infinite electric field.

5. What are some real-life examples of situations where a line of charge as a delta function may be used?

A line of charge as a delta function may be used in theoretical calculations for objects such as a charged wire or an infinitely long charged rod. It can also be used to model the electric fields and potentials of point charges or charged particles moving along a straight path.

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