How to calculate the charge density in an electric field

In summary, the conversation discusses the relationship between charge density, electric field, and potential in the context of a single point charge. It is explained that the charge density should be zero everywhere except at the origin, but a discrepancy arises when the Laplacian of the electric field is taken. It is then discovered that the incorrect equation for electric potential was used, which explains the non-zero charge density. The correct form of the potential is provided.
  • #1
eschavez6
2
0
I've been studying electric fields in class for some time and one thing is seemingly contradictory and really confuses me.

The charge density ρ is related to the electric field E and the permiativity ε and the potential [itex]\Phi[/itex] by the following equation

ρ/ε=∇[itex]\cdot[/itex]E=-∇2[itex]\Phi[/itex]

if we examine the electric field created by a single point charge of magnitude q located at the origin, then the electrostatic potential can be expressed as follows

[itex]\Phi[/itex]=[itex]\frac{q}{4πε(x^2+y^2+z^2)}[/itex]

now I would expect the charge density in this system to be zero everywhere except the origin but if we take the laplacean of this electric field, instead we get

ρ/ε=-∇2[itex]\Phi[/itex]=[itex]\frac{2q}{4πε(x^2+y^2+z^2)^2}[/itex]

which is clearly non-zero.

is there an explanation for this discrepancy? Have I violated some fundamental assumption?

thanks in advance
 
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  • #2
Well you didn't really post your work so it's hard to tell where you went wrong. The Coulomb potential is much more elegantly expressed (and easier to work with) in spherical coordinates. ##\varphi = \frac{Q}{4\pi \epsilon_0 }\frac{1}{r}## hence ## \nabla^{2}\varphi = \frac{Q}{4\pi\epsilon_{0}}\frac{1}{r^{2}}\partial_{r}(r^{2}\partial_{r}\frac{1}{r}) = 0## for ##r \neq 0##.
 
  • #3
The potential goes like 1/r and not 1/r^2.
Try the correct form of the potential. It works even in Cartesian coordinates even though is more work.
 
  • #4
thanks

it seems like i was using the wrong equation for electric potential. i had an r2 in the denominator instead of just r
 
  • #5


First of all, it is important to clarify that the charge density ρ is a measure of the amount of charge per unit volume in a given region of space. It is not the same as the charge itself, which in this case is represented by q. So, the charge density is a property of the electric field, not the charge itself.

Now, to calculate the charge density in an electric field, we can use the relationship ρ/ε=∇\cdotE, which is derived from Gauss's law. This equation tells us that the charge density is equal to the divergence of the electric field, which is a measure of how much the field is spreading out or converging in a given region.

In the case of a single point charge at the origin, the electric field is given by E=q/4πε(x^2+y^2+z^2)^2. This means that the electric field is strongest at the origin and decreases as we move away from it. However, the divergence of this field is not zero, as you correctly pointed out. This is because the electric field is not spherically symmetric around the point charge. In other words, the field is not spreading out or converging equally in all directions.

So, even though the charge density is zero at all points except the origin, the divergence of the electric field is not zero. This is not a violation of any fundamental assumption, but rather a consequence of the asymmetry of the electric field around a single point charge.

In summary, the charge density is a measure of the amount of charge per unit volume, while the electric field is a measure of the force experienced by a test charge at a given point. The discrepancy you observed is due to the fact that the electric field is not spherically symmetric in this case.
 

1. What is charge density in an electric field?

Charge density in an electric field refers to the amount of electric charge per unit volume in a given area. It is represented by the symbol ρ (rho) and is measured in units of Coulombs per cubic meter (C/m^3).

2. How do you calculate charge density in an electric field?

To calculate charge density in an electric field, you can use the formula ρ = Q/V, where Q is the total charge and V is the volume that the charge is spread over. Alternatively, you can use ρ = ε0E, where ε0 is the permittivity of free space and E is the electric field strength.

3. What is the relationship between charge density and electric field?

The charge density and electric field are directly related, meaning that as one increases, the other also increases. This relationship is described by the equation ρ = ε0E, where ε0 is a constant value and E is the electric field strength.

4. How does charge density affect the behavior of an electric field?

The behavior of an electric field is affected by the charge density in the area. In areas with higher charge density, the electric field will be stronger and the force between charged particles will be greater. In contrast, areas with lower charge density will have a weaker electric field and less force between charged particles.

5. What are some common applications of calculating charge density in an electric field?

Calculating charge density in an electric field is important in various fields such as physics, engineering, and chemistry. It is used in the design and analysis of electrical circuits, the study of electromagnetic fields, and the understanding of atomic and molecular structures. It is also essential in the development of technologies such as capacitors, batteries, and semiconductors.

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