The div in cartesian coordinates

Click For Summary
The discussion revolves around understanding the divergence of a vector field \( \mathbf{F} \) in the context of surface integrals and flux calculations. The integral of \( \mathbf{F} \) over a surface can be expressed as the sum of integrals over smaller surfaces, leading to a relationship with the divergence of \( \mathbf{F} \) over a volume. The user seeks clarification on why the book averages the values of \( \mathbf{F}_z \) at the center of the upper and lower plates when calculating flux, rather than using the values directly at the surfaces. There is confusion regarding how to derive the same flux for different surfaces and the approach to obtaining a closed surface. The discussion highlights the complexities of applying divergence in practical examples and the need for a clearer method.
kirito
Messages
77
Reaction score
9
Homework Statement
trying to understand how to derive it ,
Relevant Equations
gauss's theorem
I am currently studying a section from \textit{Electricity and Magnetism} by Purcell, pages 81 and 82, and need some clarification on the following concept. Here’s what I understand so far:

1. The integral of a function $ \mathbf{F} $ over a surface \( S \) is equal to the sum of the integrals of $ \mathbf{F} $ over smaller surfaces \( S_i \):

$$
\int_S \mathbf{F} \cdot d\mathbf{A} = \sum_i \int_{S_i} \mathbf{F} \cdot d\mathbf{A}_i
$$

2. This can be rewritten as:

$$
\sum_i \int_{S_i} \mathbf{F} \cdot d\mathbf{A}_i = \sum_i \int_{S_i} \frac{V_i}{V} \mathbf{F} \cdot d\mathbf{A}
$$

3. This is equal to the integral of the divergence of $ \mathbf{F} $ over a volume \( V \):

$$
\int_V \nabla \cdot \mathbf{F} \, dV
$$

Now, I want to find the divergence of $ \mathbf{F} $ in a book example, specifically the flux through the upper and lower plates in the \( z \)-direction.

In the example, I know that the function $ \mathbf{F} $ changes only in the \( z \)-direction and the area of each surface is \( dx \, dy \). The direction is \( \hat{z} \). Using the second expression above, I have:

$$
\mathbf{F}_z(x,y,z+\Delta z) \, dx \, dy - \mathbf{F}_z(x,y,z) \, dx \, dy = \left( \frac{\partial \mathbf{F}_z}{\partial z} \right) \Delta z \, dx \, dy
$$

However, in the derivation in the book, they look at the average of $ \mathbf{F}_z $ on the top and bottom plates and take the net contribution by considering the difference between them.

Why are they looking at the value of $ \mathbf{F}_z $ at the center of each plate \( \left(x + \frac{dx}{2}, y + \frac{dy}{2}, z \right) \) and at \( \left(x + \frac{dx}{2}, y + \frac{dy}{2}, z + dz \right) \)? I was only following the definition $$ \mathbf{F} \cdot d\mathbf{A}_1 + \mathbf{F} \cdot d\mathbf{A}_2 $$ and so on.
 
Physics news on Phys.org
question.png

in addition I can t see how c has the same flux as a and be I tried to rearrange it to get a closed surface yet got stuck seems like there is a simpler way to approach this
 
kirito said:
View attachment 348625
in addition I can t see how c has the same flux as a and b I tried to rearrange it to get a closed surface yet got stuck seems like there is a simpler way to approach this
 
Thread 'Correct statement about size of wire to produce larger extension'
The answer is (B) but I don't really understand why. Based on formula of Young Modulus: $$x=\frac{FL}{AE}$$ The second wire made of the same material so it means they have same Young Modulus. Larger extension means larger value of ##x## so to get larger value of ##x## we can increase ##F## and ##L## and decrease ##A## I am not sure whether there is change in ##F## for first and second wire so I will just assume ##F## does not change. It leaves (B) and (C) as possible options so why is (C)...

Similar threads

Replies
6
Views
3K
Replies
3
Views
2K
  • · Replies 7 ·
Replies
7
Views
2K
Replies
19
Views
3K
  • · Replies 9 ·
Replies
9
Views
2K
  • · Replies 18 ·
Replies
18
Views
2K
Replies
1
Views
1K
  • · Replies 3 ·
Replies
3
Views
1K
Replies
3
Views
2K
  • · Replies 1 ·
Replies
1
Views
468