Calculate the flux through a cube of size 1.0 m

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To calculate the electric flux through a cube in an electric field described by E = xy², one must understand that the flux is given by the integral Φ = ∫ E · n da, where n is the unit normal vector to the surface. The formula Φ = EA applies only when the electric field is uniform and perpendicular to the surface, which is not the case here. The discussion emphasizes the need to grasp the underlying concepts and notation before attempting to solve the problem. Participants suggest reviewing textbook examples to build foundational knowledge. Understanding these principles is crucial for successfully calculating the flux and determining the enclosed charge.
junglebobo
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An E field exists in a region of space and it can be described by:
\bar{E} = \hat{i}xy^2

Calculate the flux through a cube of size 1.0m, with one end extending into the positive x,y and z directions.

Find the charge enclosed.

I have no idea how to start this? can someone point me in the right direction please?

I know that:
\Phi = EA

But what does it mean that the E field exists in a region of space?

All help is appreciated.

And I am not looking for a solution, but all help will be appreciated.

Cheers
 
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junglebobo said:
I know that:
\Phi = EA

This applies only if ##\vec E## has the same magnitude everywhere on the surface, and is perpendicular to the surface everywhere. If these are not true, you have to integrate:

$$\Phi = \int {\vec E \cdot \hat n da}$$

Does this ring a bell?
 
Sorry, not really.
What values do I put in for for \hat{i}?
And what's the \hat{n}? And do I pull E out of the integration as a constant? Then integrate da just to a?
I'm sorry, I just got into this class two weeks late, and this is due tomorrow. But I'm going to spend this week catching up.
 
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You really just need to sit down with your textbook and learn the basics. After all, if you don't even understand what the notation means, how can you expect to solve the problem? The examples in your textbook should answer most of your questions.
 
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