Question on zero electric flux through a closed surface.

Click For Summary
SUMMARY

The discussion centers on the application of Gauss's law in the context of electric flux through a closed surface, specifically a cube with no enclosed charge. It is established that when an electric field is applied parallel to two faces of the cube, the total electric flux (\Phi) is zero, as per Gauss's law, due to the cancellation of flux from opposing faces. The user raises a critical question regarding scenarios where the electric field strength differs across the two faces, suggesting potential nonzero flux, which contradicts Gauss's law. This highlights the importance of uniform electric fields in applying Gauss's law correctly.

PREREQUISITES
  • Understanding of Gauss's law in electrostatics
  • Familiarity with electric flux concepts
  • Basic knowledge of electric fields and their properties
  • Ability to analyze geometric shapes in physics, such as cubes
NEXT STEPS
  • Study the implications of non-uniform electric fields on electric flux
  • Learn about applications of Gauss's law in different geometries
  • Explore the relationship between electric field strength and surface area in flux calculations
  • Investigate real-world scenarios where Gauss's law is applied in electrostatics
USEFUL FOR

Students of physics, educators teaching electrostatics, and anyone interested in the principles of electric fields and flux calculations.

student14
Messages
4
Reaction score
0
I'm relearning basic electricity concepts and I can't find an answer to a situation I've thought up.

Imagine a cube with no enclosed charge and an electric field through it parallel to two of its faces. Gauss's law says that the flux should be zero because there is no enclosed charge.

Every example of this that I have seen goes something like this: The four perpendicular faces have zero flux due to the sides being perpendicular to the field, and the flux from the other two sides cancel each other out. I.e if E is the strength of the electric field, A is the area of the sides of the cube, and \Phi is the total flux, then

\Phi = EA + (-EA) = 0.

But what if the strength of the electric field is not the same at both sides? Then there won't be any cancelling out and there will be nonzero flux, which is contrary to Gauss's law?
 
Last edited:
Physics news on Phys.org
Sorry, I see that this thread should have been posted in the coursework section. I shall post this there and this thread can be deleted.
 

Similar threads

Undergrad Why is the net flux through a closed surface equal to zero?
  • · Replies 25 ·
Replies
25
Views
7K
Undergrad Calculation of electric flux on trapezoidal surface
  • · Replies 11 ·
Replies
11
Views
2K
Undergrad Magnetic field when the area of the plates of a capacitor increases
  • · Replies 7 ·
Replies
7
Views
2K
Undergrad What is Electric Flux and How is it Calculated?
  • · Replies 30 ·
2
Replies
30
Views
3K
Undergrad Another way of stating Gauss' law?
  • · Replies 83 ·
3
Replies
83
Views
5K
Graduate Magnetic Flux through a Closed Surface and Maxwell's laws
  • · Replies 14 ·
Replies
14
Views
3K
High School Electric Flux - confused about integrating over a tilted area
  • · Replies 8 ·
Replies
8
Views
2K
Undergrad Motional EMF in Faraday disc, co-rotating magnet axial mean flux
  • · Replies 5 ·
Replies
5
Views
977
Graduate What is Electrical Field in a Closed surface with no charge
  • · Replies 3 ·
Replies
3
Views
3K
Undergrad Question regarding the use of Electric flux and Field Lines
  • · Replies 25 ·
Replies
25
Views
3K