Derive an expression for the radial charge distribution of an E field

In summary, the conversation discusses the process of writing an expression for ##\rho (r)## using Gauss' law and the del equation. It is determined that the second and third terms of the del equation can be cancelled out due to the direction of the E field, leaving an expression of ##\nabla \cdot \vec{E}=\frac{1}{r^2}\frac{\partial}{\partial{r}}(r^2 E_r)##. The question asks for an expression of ##\rho (r)##, which can be achieved by isolating ##\rho## in the differential form of Gauss' law, resulting in ##\frac{\rho}{\epsilon_o}=\frac{2 E
  • #1
Jaccobtw
163
32
Homework Statement
Consider a spherical distribution of charge that creates a uniform, radial electric field described by: ##\vec{E}=E_o\hat{r}##

Use the differential form of Gauss's Law to derive an expression for the radial charge distribution, ##\rho##, that will create this field. You will need the divergence in spherical coordinates: ##\nabla \cdot \vec{V}=\frac{1}{r^2}\frac{\partial}{\partial{r}}(r^2 V_r)+\frac{1}{r\sin[{\theta}]}\frac{\partial}{\partial{\theta}}(\sin[{\theta}]V_{\theta})+\frac{1}{r\sin[{\theta}]}\frac{\partial}{\partial{\phi}}(V_{\phi})##
Enter your mathematical expression for ρ(r) in terms of ##\epsilon_o, E_o ##, and ##r##
Relevant Equations
$$\nabla \cdot \vec{V}=\frac{1}{r^2}\frac{\partial}{\partial{r}}(r^2 V_r)+\frac{1}{r\sin[{\theta}]}\frac{\partial}{\partial{\theta}}(\sin[{\theta}]V_{\theta})+\frac{1}{r\sin[{\theta}]}\frac{\partial}{\partial{\phi}}(V_{\phi}) $$
I know we're supposed to attempt a solution but I'm honestly super confused here. I think the second an third terms of the del equation can be cancelled out because there is only an E field in the r hat direction, so no e field in the theta and phi directions. That leaves us with ##\nabla \cdot \vec{E}=\frac{1}{r^2}\frac{\partial}{\partial{r}}(r^2 E_r)##. The question says to write an expression for ##\rho (r)##. I know gauss' law has the differential form of itself equaling ##\frac{\rho}{\epsilon_o}##. would you isolate ##\rho## to get an expression for ##\rho (r)##? Thanks for your help
 
Last edited:
Physics news on Phys.org
  • #2
Jaccobtw said:
I think the second an third terms of the del equation can be cancelled out because there is only an E field in the r hat direction, so no e field in the theta and phi directions.
Yes.

Jaccobtw said:
That leaves us with ##\nabla \cdot \vec{E}=\frac{1}{r^2}\frac{\partial}{\partial{r}}(r^2 V_r)##.
The symbol ##V_r## on the right side would be better written as ##E_r##.

Jaccobtw said:
The question says to write an expression for ##\rho (r)##. I know gauss' law has the differential form of itself equaling ##\frac{\rho}{\epsilon_o}##. would you isolate ##\rho## to get an expression for ##\rho (r)##?
Yes. See what you get.
 
  • Like
Likes topsquark and Jaccobtw
  • #3
TSny said:
Yes.The symbol ##V_r## on the right side would be better written as ##E_r##.Yes. See what you get.
Thank you. I got ##\frac{2 E_o \epsilon_o}{r}##
 
  • #4
Jaccobtw said:
Thank you. I got ##\frac{2 E_o \epsilon_o}{r}##
Looks good.
 
  • Like
Likes topsquark and Jaccobtw

FAQ: Derive an expression for the radial charge distribution of an E field

1. What is the purpose of deriving an expression for the radial charge distribution of an E field?

The purpose of deriving an expression for the radial charge distribution of an E field is to understand the behavior of electric charges in a given space. This expression helps us to calculate the electric field strength at any point in space due to a particular distribution of charges.

2. How is the radial charge distribution of an E field related to Coulomb's Law?

The radial charge distribution of an E field is directly related to Coulomb's Law, which states that the force between two point charges is directly proportional to the product of their charges and inversely proportional to the square of the distance between them. This means that the electric field strength at a point is dependent on the charge distribution and distance from the source charge.

3. What factors affect the radial charge distribution of an E field?

The radial charge distribution of an E field is affected by the magnitude and distribution of charges, as well as the distance from the source charge. Other factors that can affect the distribution include the presence of other charges or conductors in the vicinity, as well as the dielectric constant of the medium surrounding the charges.

4. How can the expression for the radial charge distribution of an E field be used in practical applications?

The expression for the radial charge distribution of an E field can be used in various practical applications, such as designing electrical circuits, analyzing the behavior of charged particles in accelerators, and calculating the electric field strength in different regions of a charged object. It is also useful in understanding the behavior of lightning and other atmospheric electrical phenomena.

5. Are there any limitations to the expression for the radial charge distribution of an E field?

Yes, there are limitations to the expression for the radial charge distribution of an E field. It assumes that the charges are point charges and that the medium surrounding them is homogeneous. In reality, charges have finite size and the medium may not be uniform, which can affect the accuracy of the calculated electric field strength. Additionally, the expression does not take into account the effects of relativity or quantum mechanics, which may be necessary in certain scenarios.

Back
Top