Can a capacitor increase voltage beyond initial charge?

AI Thread Summary
Charging a capacitor and then separating its plates can indeed result in an increase in voltage beyond the initial charge, due to the decrease in capacitance while maintaining the same charge. This phenomenon is explained by the equation Q = CV, where a reduction in capacitance (C) leads to a higher voltage (V) if the charge (Q) remains constant. However, to achieve a voltage higher than the initial source, the voltage source must be disconnected before the plates are pulled apart. This process is counterintuitive but is rooted in the principles of work and energy in electrical circuits. Understanding this concept is essential for exploring advanced applications of capacitors.
Blackhawk4560
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Question: If you charge a capacitor circuit and then separate the capacitor, does voltage increase beyond the initial source?

It seams counter intuitive that if you plug a 9 Volt battery into two metal plates and pull the plates apart that you'll get massive voltage spikes, but that's what the equations seem to point to...

You'll see a diagram of this attached,
Thanks for your time!
 

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Blackhawk4560 said:
t seams counter intuitive that if you plug a 9 Volt battery into two metal plates and pull the plates apart
Why counter-intuitive ? You yourself already write you need to pull (i.e. do work) the plates apart.

The capacitance decreases, so, in accordance with Q = CV the voltage must increase if the charge remains the same.

By the way, the guy's name is 'van de Graaff'
 
Ohhhhhhhh I hadn't thought of that work being done... makes sense!

Can that lead to the voltage being HIGHER than the initial source? Or does it cap off at the voltage of the battery/van de GraAff?

Thanks again, blown away by that response time by the way!
 
Blackhawk4560 said:
Can that lead to the voltage being HIGHER than the initial source
Yes. Of course, you first have to disconnect the voltage source before starting to pull on the plates.
 
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