Find the net outward flux through the surface and the charge density

In summary, the conversation discusses taking the divergence of the theta component of the vector field D and finding the charge density to determine the flux through a spherical surface. There is a disagreement over the numerical coefficient and hints are given for finding the answer. There is also a discussion about the limits of integration and a potential difficulty due to the singularity of the field at the origin.
  • #1
vboyn12
16
2
Homework Statement
Find the net outward flux through the surface and the charge density. (Help please)
Relevant Equations
Relevant equations in a photo below. I try to solve it but the result is 0 which makes me confused
1.jpg
2.jpg
 
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  • #2
4.jpg
 
  • #3
When taking the divergence, note that the ##\theta## component of ##\mathbf D## has a numerical coefficient of 10, not 20.

After you find the charge density, you might be able to see whether or not a zero answer for the flux through the spherical surface makes sense.

What do you get for part (b)?
 
  • #4
Sorry. I now see where the factor of 20 comes from in evaluating the ##\theta## component of the divergence. Your work looks OK to me.
 
  • #5
TSny said:
When taking the divergence, note that the ##\theta## component of ##\mathbf D## has a numerical coefficient of 10, not 20.

After you find the charge density, you might be able to see whether or not a zero answer for the flux through the spherical surface makes sense.

What do you get for part (b)?
I think it must be 20 because when taking partial derivative of D(theta component)*sin(theta) respect to theta we can obtain derivative of sin(theta)^2=2sin(theta)cos(theta). This is the first time I post thread so excuse me about the math formulas
 
  • #6
TSny said:
Sorry. I now see where the factor of 20 comes from in evaluating the ##\theta## component of the divergence. Your work looks OK to me

TSny said:
Sorry. I now see where the factor of 20 comes from in evaluating the ##\theta## component of the divergence. Your work looks OK to me.
thank you. Can you give me some hints to do part (b), please?
 
  • #7
vboyn12 said:
I think it must be 20 because when taking partial derivative of D(theta component)*sin(theta) respect to theta we can obtain derivative of sin(theta)^2=2sin(theta)cos(theta). This is the first time I post thread so excuse me about the math formulas
Yes, you are right. Your work looks correct.
 
  • #8
vboyn12 said:
thank you. Can you give me some hints to do part (b), please?
All you need is a minor modification of your work for part (a). Would any of the limits of integration change?
 
  • #9
TSny said:
All you need is a minor modification of your work for part (a). Would any of the limits of integration change?
theta from 0 to pi/2, is it correct?
 
  • #10
vboyn12 said:
theta from 0 to pi/2, is it correct?
Yes. Do you know if the hemisphere is meant to include a flat base?
 
  • #11
TSny said:
Yes. Do you know if the hemisphere is meant to include a flat base?
So we have to take a double integral of the flat base with limits r from 0 to 1 and phi from 0 to 2pi, i guest
 
  • #12
vboyn12 said:
So we have to take a double integral of the flat base with limits r from 0 to 1 and phi from 0 to 2pi, i guest
Yes. However, there could be a difficulty here due to the fact that the field blows up as ##1/r^3## for ##r## going to zero. So, maybe they don't want you to include the base. I don't know.

This singularity of the field also means that the divergence of ##\mathbf D## is not actually defined at ##r = 0##. Your result ##\nabla \cdot \mathbf D = 0## is valid for all points except at the origin.
 
  • #13
TSny said:
Yes. However, there could be a difficulty here due to the fact that the field blows up as ##1/r^3## for ##r## going to zero. So, maybe they don't want you to include the base. I don't know.

This singularity of the field also means that the divergence of ##\mathbf D## is not actually defined at ##r = 0##. Your result ##\nabla \cdot \mathbf D = 0## is valid for all points except at the origin.
thank you a lot.
 
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  • #14
vboyn12 said:
thank you a lot.
Can you take some time to have a look at another topic of mine?
 

Related to Find the net outward flux through the surface and the charge density

1. What is the concept of net outward flux?

The net outward flux is a measure of the flow of an electric field through a closed surface. It is the sum of all the electric field vectors passing through the surface in the outward direction.

2. How do you calculate the net outward flux through a surface?

The net outward flux can be calculated by taking the dot product of the electric field vector and the surface area vector at each point on the surface and then summing these values over the entire surface.

3. What factors affect the net outward flux through a surface?

The net outward flux is affected by the strength of the electric field, the orientation of the surface, and the shape and size of the surface. It is also affected by the presence of any charges inside the surface.

4. What is the relationship between net outward flux and charge density?

The net outward flux is directly proportional to the charge density inside the surface. The higher the charge density, the greater the net outward flux through the surface will be.

5. How is the net outward flux related to Gauss's Law?

Gauss's Law states that the net outward flux through a closed surface is equal to the total charge enclosed by the surface divided by the permittivity of free space. This relationship allows us to use the net outward flux to calculate the charge density inside a surface.

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