Confusion with parallel plates

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SUMMARY

The discussion centers on the behavior of a parallel plate capacitor, specifically how the potential difference (voltage) is influenced by plate separation distance. The formula V=(Q*d)/(A*epsilon) indicates that voltage varies with distance when charge (Q) is held constant. However, when a battery is connected, the voltage remains fixed, and the charge adjusts accordingly. Disconnecting the battery allows the voltage to change with plate separation, confirming that the presence of a battery significantly alters the capacitor's behavior.

PREREQUISITES
  • Understanding of capacitor fundamentals
  • Familiarity with the formula V=(Q*d)/(A*epsilon)
  • Knowledge of electric charge and potential difference
  • Basic concepts of circuit theory, particularly regarding batteries and capacitors
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  • Research the effects of charge conservation in capacitors
  • Explore the relationship between capacitance and plate separation
  • Learn about the role of dielectric materials in capacitors
  • Investigate the behavior of capacitors in AC circuits
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Students of electrical engineering, physics enthusiasts, and anyone seeking to deepen their understanding of capacitor behavior in circuits.

Airsteve0
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Hi, I was wondering if someone could help me understand how the potential of a parallel plate capacitor can be altered. I understand that the potential difference across two plates is equal to the potential of the battery that charges them, but can this potential be altered by changing the plate separation distance? The reason I ask is because of the formula:

V=(Q*d)/(A*epsilon)​

where A is the area of the plates, d is the separation distance, epsilon is a constant and Q is the charge on the plates. According to this equation the voltage should change with distance but I suspect I am thinking about this incorrectly. As, well would this be different if there was no battery? Thanks you to anyone who can help clarify this problem I am having.
 
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Airsteve0 said:
According to this equation the voltage should change with distance but I suspect I am thinking about this incorrectly.
The voltage would change with distance if the charge was held fixed. With a battery attached, the voltage is held fixed so the charge adjusts itself.
As, well would this be different if there was no battery?
Absolutely. Disconnect the battery and the charge cannot change. Now you'll see that the voltage does change with the distance between the plates.
 
Thank you very much, that clears things up.
 

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