What is the difference among:
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linear charge density,
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surface charge density, and
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volume charge density?
[Physics] Linear charge density, surface charge density and volume charge density
chargedensityelectrostatics
Related Solutions
The following relation describes the total electric flux density in a medium:
In the previous equation, D is the total electric flux density, epsilon is the free space permittivity . E is the external applied electric field, and P is the polarization vector. The polarization vector represents the reaction of the medium to the externally applied electric field. In general, one can write:
The previous equation shows the “induced” or “effective” volume charge density in a medium as a response to the external electric field. Now let us consider what happens in the case of perfect dielectric (non-lossy) and perfect conductor.
In case of perfect dielectric, the induced volume charges can’t move. This means they are going to stay where they were induced. The charges stay in the volume. In the case of perfect conductor, the induced charges move freely, such that instead of being distributed in the volume they all go and accumulate at the surface from which the electric field is applied. The number of charges or more accurately the surface charge density is just enough to cancel the external electric field within the medium. This basically means there won’t be any other charges induced in the volume.
A lossy dielectric is a dielectric that has finite conductivity, which means induced charges can move but not as freely as they would in a perfect conductor. If two lossy dielectrics are in contact, there will be no surface charge between them because that surface charge would move to the boundary from which the electric field is applied.
To make it clearer, have a look at the attached figure. Figure a shows the dielectric case, where charges are induced in pairs but they can’t move. That is why they are distributed everywhere. Figure b shows the perfect conductor case, where all negative charges moved to surface creating surface charge density. The positive charges left behind were neutralized by negative charges coming from ground. That is not a concern in your question. Figure c shows the two lossy dielectrics, if surface charge accumulated on the surface separating between the dielectric, it will eventually move to the boundary at which the external field is applied.
The difference between a perfect conductor and a lossy dielectric is the time scale at which the induction of charge in the volume and the motion of the negative charges happen. In perfect conductors the whole thing is immediate, while in lossy dielectrics it takes some time described by what is called relaxation time. That is defined as:
Epsilon is the permittivty of material, sigma is the conductivity of the material. For perfect conductor sigma is infinite and the relaxation time is zero, which is why it is instantaneous.
I recommend you to have a look at chapter 5 in Sadiku’s book “Elements of Electromganetics”. It describes more details on the same topic
Hope that was useful
This equation will always give you a volume charge density. One way to see this is that surface charge density and volume charge density have different units - $\mathrm{C/m^2}$ and $\mathrm{C/m^3}$ respectively - and in order for the units to be consistent, $\rho$ has to be the latter. The fact that the equation is written with $\rho$ is a helpful reminder that it is a volume charge density.
Of course, keep in mind that the potential is not $kx^{4/3}$ everywhere. That function only describes the potential within a certain region. You also have to think about what's happening outside that region, and on the boundaries of the region.
If you try solving Poisson's equation $\nabla^2\varphi = -\rho/\epsilon_0$ in region where the potential is not so nicely behaved (as you have to do here, if you think about the boundaries), you might get a solution that involves a delta function. Just to pull an example out of thin air, something like
$$\rho(x, y, z) = \delta(x - L) e^{-y^2 - z^2}$$
That is the signature of a surface charge density being expressed as a volume charge density. $\sigma$ is the part other than the delta function; in general:
$$\rho(x, y, z) = \delta(x - a)\sigma(y, z)$$
so in this purely hypothetical example you could discern that $\sigma(y, z) = e^{-y^2 - z^2}$.
This is consistent with the statement that surface charge densities correspond to discontinuities in the electric field, because remember you can write Poisson's equation as
$$\vec\nabla\cdot\vec{E} = \frac{\rho}{\epsilon_0}$$
When $\vec{E}$ is discontinuous, its derivative is "infinite", and therefore $\rho$ needs to be represented as a product involving a delta function.
Best Answer
Simple as the name suggest. The charge per unit length, area and volume are respectively linear , surface and volume charge densities.