# Calculate the flux through a cube of size 1.0 m

• junglebobo
In summary, the conversation discusses the calculation of flux through a cube with an E field described by \bar{E} = \hat{i}xy^2. The concept of flux is defined as \Phi = EA, but the integration formula \Phi = \int {\vec E \cdot \hat n da} is mentioned if the E field does not have a constant magnitude and direction. The conversation ends with a suggestion to refer to the textbook for more information on the topic.
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

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.

Last edited by a moderator:
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.

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,

## 1. What is flux?

Flux is a measure of the amount of a quantity passing through a surface. It is usually represented by the symbol φ (phi) and is measured in units of quantity per unit area per unit time.

## 2. How is flux calculated?

To calculate flux, you need to first determine the quantity passing through the surface. This could be something like the flow of water, the amount of light, or the strength of a magnetic field. Then, you need to determine the surface area that the quantity is passing through. Finally, divide the quantity by the surface area to get the flux value.

## 3. What is the unit of flux?

The unit of flux depends on the quantity being measured. For example, if the quantity is electric field, the unit of flux would be volts per meter. If the quantity is light, the unit of flux would be lumens per square meter.

## 4. Why is the size of the cube important when calculating flux?

The size of the cube is important because it determines the surface area that the quantity is passing through. A larger cube will have a larger surface area, resulting in a higher flux value compared to a smaller cube.

## 5. Can flux be negative?

Yes, flux can be negative. This can happen if the quantity passing through the surface is decreasing over time or if it is flowing in the opposite direction of the surface normal. Negative flux values can also indicate a decrease in a field, such as a decrease in light or a decrease in magnetic field strength.

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