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Calculate Net Charge Contained by the Cube

  • Thread starter MedEx
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18
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1. 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?


2. Homework Equations
net flux= qenc/Eo (Gauss's Law)
Net flux= Ecos(theta)A
A= r^2
3. 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.
 

kuruman

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Please post the figure.
 
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The only figure is the one that is already in the question. Can you not see it?
 

kuruman

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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).
 
18
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physics cube a.gif
 

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TSny

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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}##.
 
18
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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
 

TSny

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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.
 
18
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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
 
3
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Can the answer be 4.7849e-10 C ?
 
3
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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.
 
18
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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.
 

TSny

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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
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.
 
18
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Yup that worked. Thanks to everyone c:
 
3
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$$\vec{\nabla} . \vec{E} = \rho/\varepsilon_{0}$$
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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