Calculate the flux through a cube of size 1.0 m

[junglebobo]

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

 

[jtbell]

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?

 

[junglebobo]

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.

 

[vela]

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.

 

[Nickie]

To calculate the flux through the cube, we can use the equation \Phi = \int_S \bar{E} \cdot \hat{n} dA, where \bar{E} is the electric field, \hat{n} is the unit normal vector to the surface of the cube, and dA is the differential area element. In this case, the electric field is given by \bar{E} = \hat{i}xy^2, so we can rewrite the flux equation as \Phi = \int_S (\hat{i}xy^2) \cdot \hat{n} dA.

To find the charge enclosed, we can use Gauss's Law, which states that the flux through a closed surface is equal to the charge enclosed divided by the permittivity of free space ( \epsilon_0). In this case, the surface we are considering is the cube itself, so we can write the equation as \Phi = \frac{Q_{enc}}{\epsilon_0}.

To solve for the charge enclosed, we can rearrange the equation to Q_{enc} = \Phi \epsilon_0. Now, we can substitute the value of \Phi that we calculated earlier and the permittivity of free space, which is approximately \epsilon_0 = 8.85 \times 10^{-12} C^2/Nm^2.

So, the final equation to calculate the charge enclosed is Q_{enc} = \int_S (\hat{i}xy^2) \cdot \hat{n} dA \times 8.85 \times 10^{-12} C^2/Nm^2.

To solve this integral, we need to determine the limits of integration. Since the cube has sides of length 1.0 m, the limits for x, y, and z will be from 0 to 1.0 m. Therefore, the integral becomes Q_{enc} = \int_0^1 \int_0^1 \int_0^1 (\hat{i}xy^2) \cdot \hat{n} dxdydz \times 8.85 \times 10^{-12} C^2/Nm^2.

The unit normal vector \hat{n} for each face of the cube will be (1,0,0), (0,1,0), and (0,0,1) for the x, y,