How much charge inside a Gaussian surface?

AI Thread Summary
The discussion centers on calculating the electric flux through a closed box with varying electric fields on its sides. The participants analyze the contributions of the electric fields on the left and right sides, noting that only these sides contribute to the net flux due to their orientation relative to the area vectors. There is confusion regarding the calculation of flux on the right side, particularly in determining the angle between the electric field and the area vector. After several attempts and corrections, the net flux is recalculated, leading to a charge value derived from the net flux using the equation Q = ε0 * net flux. The main issue appears to be accurately accounting for the angle and uniformity of the electric field across the surfaces.
physicslove22
Messages
27
Reaction score
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!
 
Physics news on Phys.org
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.
 
Here is the figure!
 

Attachments

  • Screen Shot 2015-05-03 at 2.01.54 PM.png
    Screen Shot 2015-05-03 at 2.01.54 PM.png
    3.5 KB · Views: 1,053
You need the sine of the inverse sine, so 1/3.
 
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.
 
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?
 
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?
 
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.
 
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.
 
Back
Top