Calculate Net Charge Contained by the Cube

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The discussion focuses on calculating the net charge contained within a cube using Gauss's Law, given a specific electric field. Initial calculations of net flux from the front and back faces of the cube led to varying results, with participants correcting each other's numerical errors. The correct total net flux was determined to be 54.0672, resulting in a net charge of approximately 4.7849e-10 C. Additionally, an alternative method involving the divergence of the electric field was discussed, confirming the same charge value. The conversation emphasizes the importance of accuracy in calculations and significant figures.
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Homework Statement


The figure shows a closed Gaussian surface in the shape of a cube of edge length 3.20 m. It lies in a region where the electric field is given by E=(1.65x+2.86)i + 4.39j + 5.47k N/C, with x in meters. What is the net charge contained by the cube?
fig23_53.gif


Homework Equations


net flux= qenc/Eo (Gauss's Law)
Net flux= Ecos(theta)A
A= r^2

The Attempt at a Solution


The top, bottom and sides don't matter because they're constant, so I should only need to look at the front and back.
For the front side I had net flux = 2.86(cos (0deg))(10.24)=29.2864
For the back side I had net flux = (1.65*3.2+2.86)(cos 180deg)(10.24) = -83.3536
So the total net flux would be -54.0672.
Multiply this number by 8.55e-12 and I got the net charge to be -4.62275e-10. Any help would be appreciated.
 
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Please post the figure.
 
The only figure is the one that is already in the question. Can you not see it?
 
MedEx said:
The only figure is the one that is already in the question. Can you not see it?
I cannot. Please use the UPLOAD button (lower right).
 
physics cube a.gif
 

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What is the value of ##x## on the back face (including sign)?

##\epsilon_0## is ##8.85 \times 10^{-12} \frac{C^2}{N m^2}##, not ##8.55 \times 10^{-12} \frac{C^2}{N m^2}##.
 
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okay so x is heading backwards into the negative direction. so the back face should be (1.65*-3.2+2.86)(cos 180deg)(10.24) = 83.3536
the total net flux is then 83.3536+29.2864 = 112.64
112.64*8.85e-12= 9.96864e-10C for my net charge?
That doesn't seem like it works either
 
MedEx said:
the back face should be (1.65*-3.2+2.86)(cos 180deg)(10.24) = 83.3536
This looks set up OK, but I get a different numerical result here.
 
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Did you get 24.781? I might have just took the original 83 number and flipped the sign.
alright 24.7808+29.2864=54.0672
54.0672*8.85e-12 is 4.78495e-10C
 
  • #10
Can the answer be 4.7849e-10 C ?
 
  • #11
coils said:
Can the answer be 4.7849e-10 C ?

Well I thought it could, but now I'm nervous.
 
  • #12
MedEx said:
Well I thought it could, but now I'm nervous.

Well, my approach to the solution was just take the divergence of the electric field and integrate over the volume.
 
  • #13
coils said:
Well, my approach to the solution was just take the divergence of the electric field and integrate over the volume.
What do you get when you try to do it that way? I'm really not very good at calc so that wouldn't be my go to.
 
  • #14
MedEx said:
Did you get 24.781? I might have just took the original 83 number and flipped the sign.
alright 24.7808+29.2864=54.0672
54.0672*8.85e-12 is 4.78495e-10C
This looks good to me. @coils method is nice, if you are familiar with the divergence equation for the electric field.

You might consider how many significant figures you should keep in your answer.
 
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  • #15
Yup that worked. Thanks to everyone c:
 
  • #16
$$\vec{\nabla} . \vec{E} = \rho/\varepsilon_{0}$$
Now for the calculation of the LHS, we have;
$$\begin{split}\vec{\nabla} .\vec{E} &= \left(\frac{\partial}{\partial x}\hat{x} + \frac{\partial}{\partial y}\hat{y} + \frac{\partial}{\partial z}\hat{z}\right) . \left(\left(1.65x + 2.86\right)\hat{x} + 4.39\hat{y} + 5.47\hat{z}\right)\\
&= \frac{\partial \left(1.65x + 2.86\right)}{\partial x} + \frac{\partial \left(4.39\right)}{\partial y} + \frac{\partial \left(5.47\right)}{\partial z}\\
&= 1.65
\end{split}$$

Multiply with the ##\varepsilon_{0}## and integrate over the volume of the cube as,
$$\begin{split}
Q &= \int dxdydz\ \left(1.65\varepsilon_{0}\right)\\
&= 1.65\varepsilon_{0}\left(2.30\right)^{3}\\
&= 4.7849\times 10^{-10} C
\end{split}$$
 
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