- #1
exitwound
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Homework Statement
Two large metal plates of [tex]1m^2[/tex] are separated by a distance [tex].05m^2[/tex] and carry equal and opposite charges on their inner surfaces. If E between the plates is 55 N/C, what is their charge? Neglect edge effects
Homework Equations
[tex]\int \vec{E}\cdot d\vec{s}[/tex]
[tex]\epsilon_0EA=Q[/tex]
The Attempt at a Solution
First of all, can I ask if the Electric Field between the plates is equal to [tex]\frac{2\sigma}{\epsilon_0}[/tex]? It might not be relevant to the problem, but I don't even know if my attempt at the calculations from this point on would be accurate if it's not. It looks like the effect of a particle from the positive plate would be [tex]\frac{\sigma}{\epsilon_0}[/tex] and the effect from the negative plate would be [tex]\frac{\sigma}{\epsilon_0}[/tex] as well. Therefore, the Electric field is the sum of the two fields, or [tex]\frac{2\sigma}{\epsilon_0}[/tex] in the direction of the negative plate.
If I slap the values in the second equation mentioned above, the correct answer of [tex]4.9x10^-10 C[/tex] comes out. However, I don't understand where that equation comes from in terms of this problem.
[tex]\epsilon_0EA=Q[/tex]
[tex](8.85x10^-12)(55)(1)=4.9x10^-10 C[/tex]
I'm completely lost on the reasoning once again or the application of Gauss's Law (the chapter this problem comes from) on the problem.