- #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
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 formulasTSny 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)?
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?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.
Yes, you are right. Your work looks correct.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
All you need is a minor modification of your work for part (a). Would any of the limits of integration change?vboyn12 said:thank you. Can you give me some hints to do part (b), please?
theta from 0 to pi/2, is it correct?TSny said:All you need is a minor modification of your work for part (a). Would any of the limits of integration change?
Yes. Do you know if the hemisphere is meant to include a flat base?vboyn12 said:theta from 0 to pi/2, is it correct?
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 guestTSny said:Yes. Do you know if the hemisphere is meant to include a flat base?
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.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
thank you a lot.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.
Can you take some time to have a look at another topic of mine?vboyn12 said:thank you a lot.
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.
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.
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.
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.
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.