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From: tomb@hplsla.HP.COM (Tom Bruhns)
Newsgroups: sci.electronics
Subject: Re: HV Cap Fun!
Message-ID: <5170036@hplsla.HP.COM>
Date: 1 Jun 89 16:50:06 GMT
References: <4924@m2c.M2C.ORG>
Organization: HP Lake Stevens, WA
Lines: 30

naiman-jeffrey@CS.YALE.EDU (Jeffrey Naiman) writes: 
>
>OK, here's a silly question. Why is the voltage across the two identical
>capacitors 5V? Volts are joules/coulomb, or more simply, energy per
>electron. Voltage drops when going through a resistor (sticking to DC current).
>Why should the energy an electron has drop when it goes to another
>capacitor?
>
>The formulas: U=1/2 C V^2=100
>	      U=1/2 Q V=100 , Q=20 (Q is halved for two capcitor case)
>	      U=2 x .5 x 10 x 10=100 for two capacitors
>	      U=1 x .5 x 20 x 10=100 for one capacitor
>
>I'm obviously missing something....
>
>- Jeff Naiman (naiman@yale.edu)
>----------
Charge is conserved; energy is converted from potential energy to
kinetic (heat and electromagnetic) when you connect the two caps like
that.  The electrons have potential energy when on the cap.  If they
fall through a potential, they convert that potential energy to 
kinetic energy.  That energy won't come back as an electrical potential
(voltage) if you let it get out as heat or radiation.

Interesting question:  Can you design a simple circuit to connect the
two caps and end up with 7.07 volts on each? (no other energy source)
(I'll take a ckt that even manages to get 7 volts :-)

Tom Bruhns
tomb%hplsla@hplabs.hp.com