Is this statement correct with the right reasoning?

  • #1
Assume you have a simple battery-bulb-capacitor circuit. If you replace the capacitor with one that has twice the area as the old capacitor, the bulb will stay on longer because there has to be more charge to accumulate on the plates of the capacitor to zero out the net electric field. Mathematically speaking in (Q/A)/2εo (equation of the fringe field of a capacitor) if A increases, Q has to also to get a net fringe equivalent to the electric field inside the wire. Since current stays the same and more charge has to accumulate, it takes longer.
 

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  • #2
davenn
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Assume you have a simple battery-bulb-capacitor circuit.
depending on how it is wired, the bulb isn't likely to work
Do you understand what a capacitor is and how it works ?

show a circuit for your idea so we can confirm what you are talking about



Dave
 
  • #3
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Since current stays the same and more charge has to accumulate, it takes longer.
I think what you wrote was generally good, but it is a little hard to claim "current stays the same" in a situation like this where the current changes over time. I would say instead that the time constant is RC, since C is double, then RC is also double, and everything takes twice as long.
 
  • #4
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Besides what was already said, the fringe field is not described by that simple formula, it is quite complicated and depends on position.
The formula may be OK for points very far from the edges of the capacitor, so exactly what the fringe field is not. :)
 
  • #5
sophiecentaur
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Assume you have a simple battery-bulb-capacitor circuit.
A "simple" circuit still needs to be described - even when there are only three components. A diagram usually makes sure that everyone understands your question.
 
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