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Capacitors:

Using direct current I apply a voltage, $~V_o~$, to a capacitor of capacitance $~C~$. It acquire a charge of $~Q_o~$. I remove the charging source and I connect both sides of the capacitor to another capacitor. The second capacitor is identical but initially uncharged. In all of this, I am using low resistance cables to make the connections. After a short time, I find that the charges have equalized between the two capacitors, and each has a charge equal to $~\frac{Q_o}{2}~$. Furthermore, I find that the voltage is equal to $~\frac{V_o}{2}~$.

I know that the energy stored a capacitor is $~\frac{CV^2}{2}~$. I also know that the first capacitor has an energy of $~\frac{CV_O}{2}~$, and the energy stored in the two capacitors after charge equalization is $$2~ \left(C ~\frac{\left(\frac{V_o}{2}\right)^2}{2}\right) = \frac{CV_o^2}{4}~.$$

Unless my calculations are wrong, half the energy disappeared.

Where did it go?

nmasanta
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Angel Tran
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1 Answers1

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As discussed in the links given in comments, in a realistic circuit where the wires connecting the capacitors have a non-zero resistance, this half of the energy $\frac{1}{4}CV_0 ^2 $ is dissipated as heat in the wires. The exact value of the resistance doesn't change how much energy is dissipated, but determines how fast.

In an ideal circuit where the wires do have zero resistance, the loop formed by circuit will still have some self-inductance $L$. As soon as you make the connection, say at $t=0$, a current will start flowing. At the moment $t=t_0$ when both capacitors first have a voltage of $V_0/2$ across them, a current of $CV_0^2/2L^2$ will be flowing around the loop, with an energy of $\frac{1}{4}CV_0^2$ stored in the magnetic field associated with the inductance $L$.

At $t = 2t_0$, the current is instantaneously zero, the voltage across the first capacitor is zero and the magnitude of the voltage across the second inductor is $V_0$. The energy stored in the second capacitor is $\frac{1}{2}CV_0^2$.

The current then starts flowing in the other direction. At $t = 3t_0$, the situation is the same as at $t = 0$, except with the current flowing in the opposite direction. At $t = 4t_0 = 2\pi\sqrt{LC}$, the system has returned to its original state. Thus, there is a periodic energy transfer between the electric field in the capacitors and the magnetic field, which goes on for a long time. Eventually, half of the total energy would be radiated away in the form of electromagnetic energy until the capacitor voltages settle at $V_0/2$.

Puk
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