Electric flux charge through a cube

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Homework Help Overview

The problem involves calculating the charge inside a cube based on the electric flux through its faces. The cube measures 4.50m on each side, and the electric flux is given as 130 Nm²/C.

Discussion Character

  • Exploratory, Assumption checking

Approaches and Questions Raised

  • Participants discuss the relationship between electric flux and charge, with one participant initially attempting to solve for the charge using the provided flux value without considering the total flux through all faces of the cube.

Discussion Status

The discussion has progressed with participants clarifying the need to account for the total flux through all six faces of the cube. Guidance has been provided regarding the correct approach to calculating the enclosed charge based on the total flux.

Contextual Notes

Participants are navigating the implications of applying Gauss's law and the properties of electric flux in relation to enclosed charge. There is an emphasis on understanding the concept of net flux through a Gaussian surface.

ZEli
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Homework Statement


The electric flux through each face of a 4.50m x 4.50m x 4.50m cube is 130Nm^2/C .
How much charge is inside the cube?


Homework Equations


Electric Flux = E*A = Qin/ε


The Attempt at a Solution


I know electric flux is given, 130Nm^2/C and I'm trying to solve for Qin. So, I plug those numbers into the equation and get
130Nm^2/C = Qin/ε
I end up with Qin = 1.1505 * 10^-9 C, but when I type this into the answer, it doesn't work!
I don't see what I'm doing wrong!:(
 
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You're looking for the total flux through the faces of the cube. If the flux was only through one face, you'd be correct, but there are six faces of the cube that have the same flux through it.
 
Ohhhhh, I see. So since there are 6 faces of the cube, I would multiply the given flux by 6 and then multiply by ε.

Got the right answer! Thanks!
 
Yep. Your net flux through an entire object is just the sum of the net fluxes through the component surfaces. This holds true for any Gaussian surface.
 
As it turns out, in fact, that combination of properties (that the net flux through a surface is dependent only on the enclosed charge and that any Gaussian surface that encloses exactly the same amount of charge will exhibit exactly the same flux) will prove incredibly useful in your future studies! It has some very helpful applications in electrostatics.
 

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