Gauss Law and Flux: Calculate Charge Inside Box

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SUMMARY

The discussion focuses on applying Gauss's Law to calculate the charge inside a closed box given varying electric field strengths across its surfaces. The electric fields are specified as E1 = 90 V/m on the left, E2 = 400 V/m on the right, E3 = 120 V/m on the top, E4 = 175 V/m on the front and back, and E5 = 245 V/m on the bottom. The accurate permittivity of free space is provided as ε0 = 8.85e-12 C²/N·m². The solution involves calculating the net electric flux through the box, which is not zero due to differing electric field values, allowing for the determination of the enclosed charge.

PREREQUISITES
  • Understanding of Gauss's Law
  • Familiarity with electric field concepts
  • Knowledge of calculating electric flux
  • Basic geometry for determining areas of surfaces
NEXT STEPS
  • Study the application of Gauss's Law in non-uniform electric fields
  • Learn how to calculate electric flux through various geometries
  • Explore the relationship between electric field strength and charge density
  • Review examples of electric field calculations involving angles and slopes
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Students studying electromagnetism, physics educators, and anyone interested in mastering the application of Gauss's Law in complex geometries.

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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 = 90 V/m, and all over the right, slanting, side of the box E2 = 400 V/m. On the top the average field is E3 = 120 V/m, on the front and back the average field is E4 = 175 V/m, and on the bottom the average field is E5 = 245 V/m.

22-104-HW_prism_sym.jpg


How much charge is inside the box? Use the accurate value ε0 = 8.85e-12 C2/N·m2.


The Attempt at a Solution


I have been working these homework problems no problem when it is just rectangles and cylinders and other shapes where Nhat and E both point in the same direction, however I thought that when it was not the same you had to use cosin of the angle made between Nhat and E. However it does not give me an angle. Am I going about this the wrong way?
 
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It doesn't give you the angle, but it gives enough information for you to find the cosine of the angle from the geometry of a right triangle.
 
I asked a friend after posting this. They said you use the area of the left side and multiply it by the electric field coming out of the slope as if it was just a rectangle and there was no slope at all. No need for any cosin or angles at all. I tried this and it gave me the right answer but I am still confused as to how it works?
 
The cosine of the angle is 4/12, multiply (4/12) times the area of the slant rectangle, 12×5, gives the area of the vertical rectangle.
 
ahh clever
 
I don't understand, isn't it just 0? The net flux through a closed surface is not 0 here?
 
flyingpig said:
I don't understand, isn't it just 0? The net flux through a closed surface is not 0 here?
It doesn't say anything about charge inside. Only gives E field at the various surfaces.
 
The next flux would be zero if the E field had the same value at E1 and E2 ,
but since they have different E field values at those points it is not zero.
 

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