3-dimensional charge density for a finite thin wire

[mjordan2nd]

Homework Statement

Express the 3D charge density $$\rho$$ for a thin wire with length Z and uniform linear charge density $$\lambda$$ along the z-axis in terms of a two-dimensional dirac-delta function.

Homework Equations

The three dimensional charge density is the total charge over a volume.

The Attempt at a Solution

I am not sure how to proceed with this question. Last night I attempted to calculate the electric field components, which I believe I did correctly, but was heavy in the algebra. I had intended to use Gauss law to calculate the charge density from the Electric field, hoping that a delta function would pop out somewhere. I talked to my professor today and he told me I was using the wrong approach, and that the solution to this problem is much simpler. Unfortunately I'm not sure I entirely understand how to use the dirac-delta function, and I feel stuck. I'm not sure how to start this problem. All I know is that

$$Q=\int^{Z}_{0}\lambda dz$$

This integral just evaluates to $$\lambda Z$$.

I have no idea how to proceed. Any help would be appreciated.

Thanks.

 

[kuruman]

Do you know what a Dirac delta function is/does? You may wish to check

http://en.wikipedia.org/wiki/Dirac_delta_function

 

[tomcue]

I can provide a response to this content by first clarifying the concept of 3-dimensional charge density. The 3-dimensional charge density refers to the amount of charge per unit volume in a three-dimensional space. It is a physical quantity that describes the distribution of charge within a given volume.

In this problem, we are dealing with a thin wire with a finite length of Z along the z-axis. The wire has a uniform linear charge density, \lambda, which means that the amount of charge per unit length is constant along the wire. Our goal is to express the 3-dimensional charge density, \rho, of this wire in terms of a two-dimensional dirac-delta function.

To do this, we can first consider the total charge of the wire, Q, which is given by the integral \int^{Z}_{0}\lambda dz. This integral evaluates to \lambda Z, as you correctly mentioned. Now, we can consider a small volume element, dV, within the wire. The 3-dimensional charge density, \rho, can be calculated as the total charge within this volume divided by the volume itself. Mathematically, this can be expressed as:

\rho = \frac{Q}{dV} = \frac{\lambda Z}{dV}

Now, we need to express the volume element, dV, in terms of a two-dimensional dirac-delta function. This can be done by considering the definition of the dirac-delta function, which is:

\delta(x) = \begin{cases} \infty, & x=0 \\ 0, & x \neq 0 \end{cases}

We can see that the dirac-delta function is zero everywhere except at x=0, where it is infinite. This means that the volume element, dV, can be expressed as a two-dimensional dirac-delta function along the z-axis, since the wire is thin and only has a finite length along the z-axis. Mathematically, this can be written as:

dV = \delta(z) dx dy dz

Substituting this into our equation for \rho, we get:

\rho = \frac{\lambda Z}{\delta(z) dx dy dz}

Simplifying this further, we get:

\rho = \lambda Z \delta(z)

Therefore, the 3-dimensional charge density, \rho, for a finite thin wire with a uniform linear