Is Left Side of Gauss' Law equation not the total electric flux?

AI Thread Summary
The discussion centers on the application of Gauss' Law to determine the charge enclosed by a cubical box given a total electric flux of 4700 Nm^2/C. It clarifies that the left side of Gauss' Law represents the total electric flux, and the size of the box is irrelevant to the calculation. The correct approach is to use the equation Q_enclosed = electric flux * ε0, where ε0 is a constant. The confusion arises from misunderstanding how to express the answer in Coulombs. The key takeaway is that the calculation should yield a charge value, not just a multiplication of the flux by ε0.
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Homework Statement



The total electric flux from a cubical box 40.0 cm on a side is 4700 Nm^2/C. What charge is enclosed by the box?

Homework Equations



Gauss Law: surface integral of E-field-vector * dA-vector = Q_enclosed/epsilon naught

The Attempt at a Solution



The reason I am making this post is that I was under the impression that the left side of Gauss' Law is the total (or net) electric flux over the surface of the shape. Why am I even being given the sides of the box? 4700 Nm^2/C, being the total electric flux, is the left side of the equation in Gauss' Law!

Why is my answer of [4700 * epsilon naught] incorrect?

Thanks.
 
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DocZaius said:

Homework Statement



The total electric flux from a cubical box 40.0 cm on a side is 4700 Nm^2/C. What charge is enclosed by the box?

Homework Equations



Gauss Law: surface integral of E-field-vector * dA-vector = Q_enclosed/epsilon naught

The Attempt at a Solution



The reason I am making this post is that I was under the impression that the left side of Gauss' Law is the total (or net) electric flux over the surface of the shape. Why am I even being given the sides of the box? 4700 Nm^2/C, being the total electric flux, is the left side of the equation in Gauss' Law!

Why is my answer of [4700 * epsilon naught] incorrect?

Thanks.
That answer is right, but remember that ε0 is a number. You should be able to give your answer in Coulombs.

Also, you are right in saying you don't need the size of the box.
 
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