The div in cartesian coordinates

AI Thread 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 'Minimum mass of a block'
Here we know that if block B is going to move up or just be at the verge of moving up ##Mg \sin \theta ## will act downwards and maximum static friction will act downwards ## \mu Mg \cos \theta ## Now what im confused by is how will we know " how quickly" block B reaches its maximum static friction value without any numbers, the suggested solution says that when block A is at its maximum extension, then block B will start to move up but with a certain set of values couldn't block A reach...
TL;DR Summary: Find Electric field due to charges between 2 parallel infinite planes using Gauss law at any point Here's the diagram. We have a uniform p (rho) density of charges between 2 infinite planes in the cartesian coordinates system. I used a cube of thickness a that spans from z=-a/2 to z=a/2 as a Gaussian surface, each side of the cube has area A. I know that the field depends only on z since there is translational invariance in x and y directions because the planes are...
Thread 'Calculation of Tensile Forces in Piston-Type Water-Lifting Devices at Elevated Locations'
Figure 1 Overall Structure Diagram Figure 2: Top view of the piston when it is cylindrical A circular opening is created at a height of 5 meters above the water surface. Inside this opening is a sleeve-type piston with a cross-sectional area of 1 square meter. The piston is pulled to the right at a constant speed. The pulling force is(Figure 2): F = ρshg = 1000 × 1 × 5 × 10 = 50,000 N. Figure 3: Modifying the structure to incorporate a fixed internal piston When I modify the piston...
Back
Top