Verifying Gauss' Law: Homework Statement Solutions

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The discussion revolves around verifying Gauss' Law using a given flux density. The charge density was determined to be 5x, leading to an enclosed charge of 5/2 C after integration. However, there is confusion regarding the net flux calculation, with an initial result of 7/2, prompting doubts about the accuracy of both calculations. Clarifications were provided on integrating the flux density over the cube's surfaces, emphasizing the importance of considering the vector nature of the surface element. The conversation highlights the value of collaborative learning in mastering complex physics concepts.
ryukyu
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Homework Statement


For the given flux density: \vec{D}=(2y2+z)\widehat{x}+(4xy)\widehat{y}^+(xz)\widehat{z}
a)Determine the charge density.
b)Find the total charge enclosed if the surface is 0<x<1, 0<y<1, 0<z<1 (unit cube)
c)Confirm Gauss’s law by finding the net flux through the surface of the volume.

Homework Equations



The Attempt at a Solution



I used divergence to find the \rhov=5x

To find Qenc I integrated \int\int\int5xdxdydz and came up with
Qenc=5/2 C

The last step I know is to verify that \oint\vec{D}dS=Qenc.

From what I gather since the divergence only has an x-component we will integrate only the x-component over the dxdydz, but this gives me 7/2. I'm guessing both are incorrect, but obviously at least one of them is.
 
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ryukyu said:
1.

The last step I know is to verify that \oint\vec{D}dS=Qenc.

From what I gather since the divergence only has an x-component we will integrate only the x-component over the dxdydz, but this gives me 7/2. I'm guessing both are incorrect, but obviously at least one of them is.


The divergence is a scalar, and you have to integrate the flux density for all sides of the cube. Remember that the surface element dS is a vector normal to the surface.


ehild
 
Thanks for the response and the insight.

So for the top of said cube I would integrate the z-hat coefficient by dydx?
the bottom by z-hat (-dydx)
the right by y-hat (dxdz)
left by y-hat(-dxdz)
front by x-hat(dydz)
and back x-hat(-dydz)...
 
It looks OK.

ehild
 
I apologize for not saying thanks again. I do find this site a valuable resource in attempting to learn this material instead of just blindly using equations and hoping that things work out.
 
You are welcome.

ehild
 
The book claims the answer is that all the magnitudes are the same because "the gravitational force on the penguin is the same". I'm having trouble understanding this. I thought the buoyant force was equal to the weight of the fluid displaced. Weight depends on mass which depends on density. Therefore, due to the differing densities the buoyant force will be different in each case? Is this incorrect?

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