Potential of parallel plate capacitor

AI Thread Summary
The discussion centers on the electric field (E) of a parallel plate capacitor, specifically addressing the sign convention based on the orientation of the plates. The electric field is defined as E = -Q/(ε0A), with the negative sign indicating that the field points from the positive plate to the negative plate, which is to the left if the positive axis is defined to the right. If the capacitor's orientation is reversed, the electric field would have a positive x-component. The participant expresses uncertainty about the concepts but acknowledges the complexity of the material. Understanding the directionality of electric fields is crucial in grasping capacitor behavior.
cookiemnstr510510
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Homework Statement


I have attached a problem from within the chapter of my book. I have a question regarding my highlighted part.

My question:
Is the reason they are saying E= -Q/(ε0A) because they defined the positive "s" axis to the right? and since E is pointing from right to left we say it is a negative electric field?
 

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Yes. The x-component of the electric field is negative, since the electric field vector is pointing to the left. If you turn the capacitor around, placing the negative plate to the right of the positive plate, you would have a positive x component of the electric field. Why did you get the doubt?
 
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Learning all this new info, it can be a lot and when things click I doubt myself! But thanks!
 

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