Electric Flux thru the top face of a cube

[rainstom07]

Homework Statement

An electric field given by $$\vec E = 4.0 \hat i - 3.0(y^2+2) \hat j$$ pierces a 2.0 meter by 2.0 meter by 2.0 meter cube. What is the electric flux thru the top face?

Homework Equations

$$\phi = \int \vec E \cdot d \vec A$$

The Attempt at a Solution

I'm aware that you can take a shortcut to solving this problem by considering that the electric field at y = 2.0 is a constant $$\vec E = 4.0 \hat i - 3.0([2.0]^2+2) \hat j = 4.0 \hat i - 18.0 \hat j$$.

then this leads to, $$\phi = \int \vec E \cdot d \vec A = \vec E \int d \vec A = \vec E \cdot \vec A$$

And since $$\vec A$$ has no x-component, we only need to consider the y component. The area of the top face is just 2 times 2.

$$\phi = -18.0 * (2^2) = -72 \frac{Nm^2}{C}$$

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But how do I solve this same problem without taking the shortcut?

2. Homework Equations ?
Evaluating a flux integral (surface is oriented upward):
$$\int_S \int \vec F \cdot \vec N dS$$

The unit vector normal to the surface:
$$\hat N dS = \frac{\bigtriangledown g(x,y,z)}{||\bigtriangledown g(x,y,z)||} dS = \bigtriangledown g(x,y,z) dA$$

4. The attempt at a solution without the shortcut
The surface is probably given by $$z = g(x, y) = y - 2$$
$$\hat N dS= dA \hat j$$
$$\phi = \int_S \int \vec E \cdot \vec N dA = \int_R \int [4.0 \hat i - 3.0(y^2+2) \hat j] \cdot [dA \hat j] = \int_R \int -3.0(y^2+2)dA = -3.0 \int_0^2 \int_0^2 (y^2+2) dxdy = -40.0$$

:( this is obviously the wrong answer, any clue on what I'm doing wrong?

Thanks in advance!

 

[anathema]

Thank you for your post. I am a scientist and I would be happy to help you solve this problem. First, let's review the concept of electric flux. Electric flux is a measure of the flow of electric field through a given surface. It is calculated by taking the dot product of the electric field and the surface area vector, integrated over the surface. Mathematically, it can be written as:

\phi = \int_S \int \vec E \cdot \vec dA

where \vec E is the electric field and \vec dA is the surface area vector.

Now, let's apply this concept to the problem at hand. The electric field given in the problem is \vec E = 4.0 \hat i - 3.0(y^2+2) \hat j. The surface we are interested in is the top face of the cube, which can be described by z = 2.0. To calculate the surface area vector, we need to find the gradient of this surface, which is:

\bigtriangledown g(x,y,z) = 0 \hat i + 0 \hat j + 1 \hat k

Therefore, the surface area vector is \vec dA = 1 \hat k dA.

Now, we can rewrite the electric flux integral as:

\phi = \int_S \int \vec E \cdot \vec dA = \int_R \int [4.0 \hat i - 3.0(y^2+2) \hat j] \cdot [1 \hat k dA] = \int_R \int -3.0(y^2+2)dA = -3.0 \int_0^2 \int_0^2 (y^2+2) dxdy = -12.0

I believe the mistake in your attempt was in the calculation of the surface area vector. You used \hat j instead of \hat k, which resulted in a different answer. I hope this helps clarify the concept and solve the problem without using the shortcut. Let me know if you have any further questions.