# Homework Help: Electric Flux thru the top face of a cube

1. Jan 26, 2013

### rainstom07

1. The problem statement, all variables and given/known data
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?

2. Relevant equations
$$\phi = \int \vec E \cdot d \vec A$$

3. 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. Relevant 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?