Re: 50%

To: tesla-at-pupman-dot-com
Subject: Re: 50%
From: Steve Roys <sroys-at-umabnet.ab.umd.edu>
Date: Tue, 5 Nov 1996 09:47:03 -0500 (EST)

On Mon, 4 Nov 1996, Tesla List wrote:

> This capacitor problem is a tricky question and and covers many
> possibilities and is sometimes given by electrical/electronic instructors
> because it has to do with both capacitor theory and unfamiliar algebra.
> 
>    Your Q = CV = coulombs is correct
>    My   J = 1/2 CV^2  = joules (energy) is also correct
> 
[snip]

> There are several ways to solve this problem. However, the voltage across
> the two capacitors after reconnecting them must be found by the conservation
> of energy equation.

Why should it be the energy that's constant (not conserved) within the
described system?  I would think that the charge would be the constant. 
Both charge and energy are ultimately conserved, but in this case, the
charge has nothing to do other than to divide itself between the two
capacitors.  The energy can be lost from the system as heat, EM radiation,
etc...

> I will start with both capacitors equal to 1 farad and the voltage at 10
> volts.

Given your initial conditions, you have:
  Q0 = C0 * V0  = 1F * 10V = 10 Coulombs on the capacitor.
After shorting the two together, you say that you would have
  Q1 = C1 * V1 = 2F * 7.07V = 14.14 Coulombs divided between the two caps.

Where does the extra charge come from?

Steve Roys.