Finding the flux of an electrix field

  • Thread starter Thread starter spaghetti3451
  • Start date Start date
  • Tags Tags
    Field Flux
Click For Summary
SUMMARY

The discussion focuses on calculating the electric flux through a shaded face of a cube when a charge q is positioned at one corner. The solution involves applying Gauss's Law, which states that the electric flux through a closed surface is proportional to the enclosed charge. By surrounding the charge with seven additional cubes, the total flux can be calculated as 24 times the flux through one face, simplifying the integration process. This method is referenced in Griffiths' "Introduction to Electrodynamics."

PREREQUISITES
  • Understanding of Gauss's Law in electrostatics
  • Familiarity with surface integrals in vector calculus
  • Knowledge of electric field concepts
  • Basic principles of electrostatics from Griffiths' "Introduction to Electrodynamics"
NEXT STEPS
  • Study the application of Gauss's Law in different geometries
  • Learn about surface integrals and their role in electromagnetism
  • Explore electric field calculations for point charges
  • Review examples from Griffiths' "Introduction to Electrodynamics" related to electric flux
USEFUL FOR

Students studying electromagnetism, particularly those tackling problems involving electric flux and Gauss's Law, as well as educators seeking to clarify these concepts in a classroom setting.

spaghetti3451
Messages
1,311
Reaction score
31

Homework Statement



A charge q sits in the back corner of a cube, as shown in the attachment. What is the flux of E through the shaded side?

Homework Equations



The Attempt at a Solution



I know that I need the surface integral of E over the shaded area, but the problem is with choosing the proper coordinates and the origin.

Please help out this poor guy!
 

Attachments

  • screen.jpg
    screen.jpg
    16 KB · Views: 496
Physics news on Phys.org
Gauss's law finds the charge ENCLOSED,

So one way to do this question is to enclose the charge by surrounding the charge by another 7 boxes.

Then you can compute the integral as the charge is now enclosed.

There will be 24 faces after surrounding the charge, so integral will produce a flux 24 times the flux through one face (seeing as you only need the flux through one face) and I think you can take it from there,

I've seen this question in griffiths electrodynamics before I think
 

Similar threads

Elec. flux through the top side of a cube with q at a corner
  • · Replies 4 ·
Replies
4
Views
2K
Electric field of a neutral conductor just at the surface
  • · Replies 4 ·
Replies
4
Views
2K
Undergrad Calculation of electric flux on trapezoidal surface
  • · Replies 11 ·
Replies
11
Views
2K
Electric Flux through the Face of a Cube
  • · Replies 17 ·
Replies
17
Views
9K
How Do You Apply the Divergence Theorem to a Vector Field in a Unit Cube?
  • · Replies 2 ·
Replies
2
Views
2K
Electrix flux through a hemisphere
  • · Replies 3 ·
Replies
3
Views
2K
Undergrad Magnetic field when the area of the plates of a capacitor increases
  • · Replies 7 ·
Replies
7
Views
2K
Undergrad Motional EMF in Faraday disc, co-rotating magnet axial mean flux
  • · Replies 5 ·
Replies
5
Views
655
Thermal conductance of an angular section of a disc
  • · Replies 4 ·
Replies
4
Views
2K
How Long Does It Take for a Falling Square Loop to Clear a Magnetic Field?
  • · Replies 16 ·
Replies
16
Views
3K