Re: Where did the energe go?
Half the energy is lost in the process of physically moving the charges from one capacitor to the other.
Let's look at some math:
The energy it requires to move a charge
q (in other words, the work done by the electric field) is
\[W = q\Delta V\].
However, the value \[\Delta V\] is the voltage difference between the two capacitors and is not constant over time. Therefore, we must set up an integral over the charge transfer.
Let's call
dq the incremental charge element that will be moved over from C1 to C2. And let's define \[\Delta V\] to be \[V_1 - V_2\]. If we call
q the amount of charge moved so far, we can find \[\Delta V\] as a function of
q.
\[V_1(q) = \frac{Q_o - q}{C_1}\], where \[Q_o = C_1 V_1\] is the initial charge on
C1.
\[V_2(q) = \frac{q}{C_1}\]
\[\Delta V(q) = V_1 - V_2 = \frac{Q_o - 2q}{C_1}\]
Now we are ready to set up the integral:
\[W = \int\limits_0^{\frac{1}{2}V_1 C_1} \Delta V(q) dq\]
\[W = \int\limits_0^{\frac{1}{2}V_1 C_1} \frac{Q_o - 2q}{C_1} dq\]
\[W = \frac{C_1V_1^2}{4}\]
The limits of this integral are from 0 to \[\frac{1}{2}V_1 C_1\], because as you stated in the original post, the amount of charge moved over is \[\frac{1}{2}V_1 C_1\].
This is work done by the electric field and is equal to the energy lost.
Added after 2 minutes:
claudiocamera said:
The message was edited to the correct way. There was a divided by 2 error.
the correct is:
Energy = (C1*(V1/2)^2) + (C2*(V2/2)^2)
We have C1=C2 that since I can write:
Energy = (C1*(V1/2)^2) + (C1*(V2/2)^2) =>
Energy = (2*C1*(V1/2)^2) =>
Energy = (2*C1*(V1^2))/4 =>
Energy = (C1*(V1^2))/2 and we have the same result, therefore there is no energy loss.
Ok now ?
claudio, your initial equation is incorrect. You dropped a factor of 1/2 in your energy equations.