Gauss Law and Flux: Calculate Charge Inside Box

Click For Summary

Homework Help Overview

The problem involves applying Gauss's Law to determine the charge inside a closed box based on the electric field measurements at various surfaces. The subject area is electrostatics, specifically focusing on electric flux and charge calculations.

Discussion Character

  • Exploratory, Assumption checking, Conceptual clarification

Approaches and Questions Raised

  • Participants discuss the relationship between the electric field and the geometry of the box, questioning the need for angles in calculations. Some suggest using the area of surfaces directly, while others explore the implications of differing electric field values on net flux.

Discussion Status

The discussion is ongoing, with participants exploring different interpretations of the problem. Some guidance has been offered regarding the calculation methods, but there is no explicit consensus on the correct approach or understanding of the underlying principles.

Contextual Notes

Participants note the absence of angles in the problem statement and question how this affects the calculations. There is also uncertainty regarding the implications of the electric field values provided at the surfaces of the box.

Kreamer
Messages
21
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 = 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?
 
Physics news on Phys.org
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.
 

Similar threads

  • · Replies 12 ·
Replies
12
Views
1K
  • · Replies 26 ·
Replies
26
Views
3K
  • · Replies 10 ·
Replies
10
Views
2K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 22 ·
Replies
22
Views
3K
Replies
6
Views
2K
  • · Replies 10 ·
Replies
10
Views
2K
Replies
23
Views
5K