Potential of parallel plate capacitor

  • #1

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?


  • parallel plate 4.jpg
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Answers and Replies

  • #2
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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  • #3
Learning all this new info, it can be a lot and when things click I doubt myself! But thanks!

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