Potential of parallel plate capacitor

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SUMMARY

The discussion centers on the electric field (E) of a parallel plate capacitor, specifically the equation E = -Q/(ε0A). The negative sign indicates the direction of the electric field vector, which points from the positive plate to the negative plate. When the orientation of the capacitor is reversed, the electric field direction changes, resulting in a positive x-component. This clarification helps in understanding the relationship between charge distribution and electric field direction in capacitors.

PREREQUISITES
  • Understanding of electric fields and their directionality
  • Familiarity with the concept of charge (Q) in capacitors
  • Knowledge of permittivity of free space (ε0)
  • Basic principles of capacitor construction and function
NEXT STEPS
  • Study the derivation of the electric field equation for parallel plate capacitors
  • Explore the effects of capacitor orientation on electric field direction
  • Learn about the role of permittivity (ε0) in electric field calculations
  • Investigate the applications of capacitors in electrical circuits
USEFUL FOR

Students studying electromagnetism, physics educators, and anyone interested in understanding the principles of electric fields in capacitors.

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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