Parallel plate capacitor (voltage and charge relative to distance)

AI Thread Summary
In a parallel plate capacitor without a battery, moving the plates farther apart increases the potential (V) due to the work done against the electric field. The relationship between voltage and distance is influenced by the electric field (E), which remains approximately constant as long as the separation distance is small relative to the plate dimensions. The confusion arises from misapplying the inverse square law, which is not applicable in this context. As the distance increases beyond a certain point, the electric field does decrease, but within practical limits, it remains relatively stable. Understanding this distinction clarifies why potential increases with distance in a parallel plate capacitor.
swarm
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Hi all.

I'm struggling to understand something about PP capacitors. Let's say there is no battery connected to a PP capacitor.

Why does the potential (V) INCREASE when the plates are moved farther apart?

- we know that: dv = -Ed (in a constant E)
however as d increases, E decreases with the inverse square of d, falling off exponetially while d falls of linearly, so since E is getting smaller faster than d is getting larger, souldn't dv decrease as d increases?

- From a non-math perspective I understand it, work must be done to pull the plates apart, thus potential energy is increased, so dv would increase. My math just doesn't agree though.

Can someone point out my error? Many thanks in advance, this has been driving me nuts for a couple days now!
 
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swarm said:
Why does the potential (V) INCREASE when the plates are moved farther apart?

- we know that: dv = -Ed (in a constant E)
however as d increases, E decreases with the inverse square of d, falling off exponetially while d falls of linearly, so since E is getting smaller faster than d is getting larger, souldn't dv decrease as d increases?
As long as the separation distance remains small with respect to the dimensions of the plates, the electric field remains constant. It does not drop off exponentially or as an inverse square with distance (as would happen with a point charge).
 
Thanks Doc Al, I think my problem was that I was using unrealistically high values for 'd' (1 meter or more), it makes sense now.
 
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