How much charge inside a Gaussian surface?

In summary: I didn't check that.In summary, the electric field on the closed box is measured to be horizontal and to the right, with different values on each side. The flux on the left side is -0.16, while the flux on the right side is 0.8. The net flux is 0.64. However, there may be an error in the conversion from net flux to charge.
  • #1
physicslove22
27
0

Homework Statement


The electric field has been measured to be horizontal and to the right everywhere on the closed box shown in the figure. All over the left side of the box E1 = 80 V/m, and all over the right, slanting, side of the box E2 = 400 V/m. On the top the average field is E3 = 260 V/m, on the front and back the average field is E4 = 200 V/m, and on the bottom the average field is E5 = 230 V/m.

Homework Equations


Flux = |E|*|A|*cos (theta)
ε0 = 8.85 ✕ 10-12 C2/N·m2

The Attempt at a Solution


The flux on the sides of the box = 0 because the electric field is perpendicular to the area vector. The flux on the left side of the box is (80)(.002)(cos 180) = -0.16 The flux on the right side of the box is (400)(.006)(cos (inverse sine of 4/12)) = 2.26
Then net flux is -0.16 + 2.26 = 2.1
2.1 = Q / (8.85x10^-12), Q = 1.86x10^-11

I can't figure out where I went wrong with this!
 
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  • #2
physicslove22 said:

Homework Statement


The electric field has been measured to be horizontal and to the right everywhere on the closed box shown in the figure.
There is no figure.
 
  • #3
Here is the figure!
 

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  • #4
You need the sine of the inverse sine, so 1/3.
 
  • #5
The electric field is uniform, so I believe you have the correct equation.

Notice that ##\vec E_3, \vec E_4##, and ##\vec E_5## cause no electric flux because each area vector ##d \vec A_{3, 4, 5}## is perpendicular to the electric field.

Only ##\vec E_1## and ##\vec E_2## are going to cause any flux.

I believe you have calculated the electric flux caused by ##\vec E_1## properly.

As for the electric flux caused by ##\vec E_2##, you need to separate the area vector for that side of the surface into it's ##x## and ##y## components ##d \vec A_x## and ##d \vec A_y##. Only the ##x## component should allow you to calculate a flux because the angle between it and ##\vec E_2## is zero.
 
  • #6
I think I may have found my mistake: wouldn't Flux2 = (.12) (.05) (400) cos (90 - (inverse sine (.04/.12)) = .01424 ? Then net flux would be -0.1458?
 
  • #7
Ok so I've checked over this several times, and I plugged it into my calculator wrong, so Flux2 = 0.8. However, it is still wrong! Does anyone know if there's some trick to it that I'm missing?
 
  • #8
I believe it is because the electric flied lines through that face are not uniform. You can't apply ##\Phi_{\vec E_2} = E_2 A \text{cos}(\theta)## directly because finding the angle between ##d \vec A = \vec n \space dA## and ##\vec E_2## is difficult.
 
  • #9
However, it is still wrong
I can't see what's wrong, you don't show your calculation nor the result.

Zondrina said:
I believe it is because the electric flied lines through that face are not uniform. You can't apply ##\Phi_{\vec E_2} = E_2 A \text{cos}(\theta)## directly because finding the angle between ##d \vec A = \vec n \space dA## and ##\vec E_2## is difficult.
It is given that the field is to the right and has the same value everywhere on the right side of the thing. I don't see why the 0.8 of love is incorrect.
 
  • #10
The answer is -0.16+400*0.006*1/3=0.64.
See my previous post.
 
  • #11
my2cts said:
The answer is -0.16+400*0.006*1/3=0.64.
See my previous post.
Yes, the -0.16 from love's post #1 and the 0.8 from love's post #7 combined. So if the answer is wrong it must lie in the conversion from net flux to charge.
 

1. What is a Gaussian surface?

A Gaussian surface is an abstract concept used in physics to simplify calculations of electric and magnetic fields. It is a hypothetical closed surface that is chosen to enclose a specific region of space.

2. How is the charge inside a Gaussian surface determined?

The charge inside a Gaussian surface is determined by integrating the electric field over the surface. This is known as Gauss's law, which states that the electric flux through any closed surface is equal to the total enclosed charge divided by the permittivity of the surrounding medium.

3. Can a Gaussian surface have a net charge?

No, a Gaussian surface must always have a net charge of zero. This is because the electric flux through the surface is directly proportional to the enclosed charge, and if there is a net charge, the electric flux would not be equal to zero.

4. How does the shape of a Gaussian surface affect the calculation of charge inside?

The shape of a Gaussian surface does not affect the calculation of charge inside. This is because the electric field is a conservative vector field, meaning that the path of integration does not affect the result. As long as the surface encloses the same amount of charge, the calculation will be the same regardless of the shape of the surface.

5. Why is a Gaussian surface often chosen to be a symmetrical shape?

A Gaussian surface is often chosen to be a symmetrical shape because it simplifies the calculation of the electric field. In symmetrical situations, the electric field has the same magnitude and direction at every point on the surface, making the integration process easier. Additionally, symmetrical shapes often have a higher degree of symmetry, making the integration even simpler.

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