Does distance inbetween the plates of a capacitor have an effect on its charge?

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SUMMARY

The distance between the plates of a parallel plate capacitor significantly affects its charge depending on whether it is connected to a battery or not. When connected to a battery, increasing the distance between the plates results in a constant voltage, which leads to a decrease in capacitance and an increase in charge. Conversely, when the capacitor is disconnected from the battery, the charge remains constant, while the voltage changes with the distance. This confirms that the relationship between charge, voltage, and capacitance is dependent on the connection status to the battery.

PREREQUISITES
  • Understanding of capacitor fundamentals, including charge (Q), voltage (V), and capacitance (C).
  • Familiarity with the formula for capacitance: C = (εA)/d.
  • Knowledge of the relationship between electric field (E) and voltage (V = Ed).
  • Basic principles of circuit theory regarding battery connections and charge flow.
NEXT STEPS
  • Research the effects of dielectric materials on capacitance and charge in capacitors.
  • Learn about the implications of capacitor discharge in circuits.
  • Explore the concept of energy stored in capacitors and its relation to charge and voltage.
  • Investigate the role of capacitance in AC circuits and its frequency dependence.
USEFUL FOR

Students of electrical engineering, physics enthusiasts, and professionals involved in circuit design and analysis who wish to deepen their understanding of capacitor behavior in various configurations.

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You have a parallel plate capacitor, each with opposite charge Q, connected to a battery of some voltage.

What I'm wondering, is does the distance inbetween these plates have an effect on their charge, while connected to the battery? What about if the distance was changed after being disconnected from the battery?

What I think, is if the distance was doubled (while connected to the battery) the V=Ed would double, the C=(elipson)(A)/d would half, therefore the Q=CV would stay the same.

And if the distance was doubled after disconnected from the battery, V stays fixed, so does C, and therefore Q does as well.

Are those assumptions correct?
 
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It's the other way around. If the battery stays connected V stays the same and therefore the charge changes. If you disconnect the battery the charge stays the same since there is no way for the charge to flow away. So V changes.
 

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