Electric Flux Problem: Can't Calculate Answer - Please Help!

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Homework Help Overview

The problem involves calculating the electric flux through one face of a cube that contains a point charge at its center. The subject area is electrostatics, specifically focusing on electric flux and Gauss's law.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • The original poster attempts to calculate the electric flux but struggles with the method. Some participants suggest considering the total flux through the entire surface of the cube and invoking symmetry to simplify the problem.

Discussion Status

The discussion includes hints and attempts to clarify the relationship between total flux and the flux through one face of the cube. There is an acknowledgment of symmetry in the setup, but no explicit consensus or resolution has been reached.

Contextual Notes

Participants are working under the constraints of the problem as posed, with specific values given for the charge and dimensions of the cube. There is an indication of assumptions regarding the direction of the electric field and area elements.

crazynut52
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I've been thinking about this problem for a while now, and I can't get it to work out. A 9.6X10^-6 C point charge is at the center of a cube with sides of length .5m What is the electric Flux through one of the six faces of the cube? I know the answer is 1.81X10^5 but I can't figure out how to get there. Please Help!
 
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Hint 1: What would the total flux be through the entire cubic surface?

Hint 2: Invoke symmetry!
 
[tex]\oint {E \cdot dA} = \frac{q}{\epsilon_0}[/tex]

Let's say B = the area of one side then 6B = A if A is the total surface area so...

[tex]\oint {E \cdot da} = \frac{q}{\epsilon_0} = 6b \times E[/tex]

the flux through B is therefore what? (I Realize this isn't entirely correct because I am assuming E and dA are parallel but the answer remains the same through symmetry)
 
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sure is easy when you figure it out... thanks for the help
 

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