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From: tan@ihlpb.ATT.COM (Bill Tanenbaum)
Newsgroups: sci.electronics,sci.physics
Subject: Re: HV Cap Fun!
Keywords: capacitor,energy,paradox
Message-ID: <10570@ihlpb.ATT.COM>
Date: 1 Jun 89 22:31:48 GMT
References: <4924@m2c.M2C.ORG> <3806@mit-amt> <20772@quacky.mips.COM> <62169@yale-celray.yale.UUCP>
Organization: AT&T Bell Laboratories - Naperville, Illinois
Lines: 27

In article <62169@yale-celray.yale.UUCP<, 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)
-----------------
Volts are joules/coulomb, but they are also coulombs/farad.  The amount
of charge (coulombs) is conserved.  The capacitance (farads) has doubled.
So the voltage (volts) must be cut in half.  V=Q/C. This is what you missed.

The energy (joules), on the other hand, is not conserved, since the system
is not closed.  Since U = QV, and V gets cut in half while Q is constant,
U must be cut in half.  Where does it go?   Mostly to resistive losses
if ohmic resistance is present, otherwise to radiative losses.
-- 
Bill Tanenbaum - AT&T Bell Labs - Naperville IL  att!ihlpb!tan