- #1

rainstom07

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## Homework Statement

An electric field given by [tex]\vec E = 4.0 \hat i - 3.0(y^2+2) \hat j[/tex] 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

[tex]\phi = \int \vec E \cdot d \vec A[/tex]

## 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 [tex]\vec E = 4.0 \hat i - 3.0([2.0]^2+2) \hat j = 4.0 \hat i - 18.0 \hat j[/tex].

then this leads to, [tex]\phi = \int \vec E \cdot d \vec A = \vec E \int d \vec A = \vec E \cdot \vec A[/tex]

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

[tex]\phi = -18.0 * (2^2) = -72 \frac{Nm^2}{C}[/tex]

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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):

[tex]\int_S \int \vec F \cdot \vec N dS[/tex]

The unit vector normal to the surface:

[tex]\hat N dS = \frac{\bigtriangledown g(x,y,z)}{||\bigtriangledown g(x,y,z)||} dS = \bigtriangledown g(x,y,z) dA[/tex]

**4. The attempt at a solution without the shortcut**

The surface is probably given by [tex]z = g(x, y) = y - 2[/tex]

[tex]\hat N dS= dA \hat j[/tex]

[tex]\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[/tex]

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

Thanks in advance!

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