- #1
Bashyboy
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Homework Statement
A large, flat, horizontal sheet of charge has a charge per unit area of 3.15 µC/m2. Find the electric field just above the middle of the sheet.
Homework Equations
[itex]dq = \sigma dA[/itex]
[itex]\oint \vec{E} \cdot d\vec{A} = \frac{q_enc}{\epsilon_0}[/itex]
[itex]\epsilon_0 = \frac{1}{4 \pi k_e}[/itex]
The Attempt at a Solution
I figured that if I could generate a Gaussian surface--specifically, a cube--,then I could solve this problem. Since two sides of the cube won't experience electric field lines propagating through, we won't have to consider those two sides; but the side of the cube above the sheet, and the side of the cube below the sheet, will experience electric flux. Furthermore, since the electric field lines are either emanating upwards or downwards, the amount of electric field lines will be the same through each of the two sides will be the same.
Hence, [itex]\vec{E}\oint d\vec{A} = \frac{q_enc}{\epsilon_0}[/itex].
In order for me to substitute in[itex]dq[/itex] for [itex]q[/itex], they need to be equal. I know they are equal, but I am having trouble justifying why the two are equal.
By doing this substitution, and solving for the electric field [itex]\vec{E}= \frac{\sigma dA}{\oint d\vec{A} \cdot \epsilon_0}[/itex].
Again, I know that that [itex]dA[/itex] and [itex]\oint d\vec{A}[/itex] cancel out, but I just can't qualify it.
Finally, since we are speaking about a point immediately above the center of the sheet, if we were to place a test particle at this position, it would only endure the electric field lines emanating from the top face of the sheet. Therefore, the electric field line at this point would be [itex]\frac{||\vec{E}||}{2}[/itex]
By using this method, I was able to solve the problem; but, as you can see, there are portions of this method I can't really justify. If someone could help me with using correct reasoning and terminology, I'd really appreciate it.