-1

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
  • 293
Angel Tran
  • 9
  • 1
    Welcome New contributor Angel Tran! I've downvoted your question because of the "does not show any research effort" reason. Have you taken the time to search this site for the related questions and answers? The 'two capacitor missing energy' problem is so well known, that I have to conclude that you've made no effort to research this. For example, there is this Wikipedia article devoted to the problem: Two capacitor paradox – Alfred Centauri Aug 22 '19 at 00:23
  • This question has been asked many times. It is sometimes referred to as the two capacitor paradox. Look at this Wikipedia article https://en.wikipedia.org/wiki/Two_capacitor_paradox – Bob D Aug 22 '19 at 00:25
  • Geesh Alfred. Yes I have researched and I am still lost. I am new to this site because I was recommended to helpful it is, but guess not if rude commenters like you show up. I know that a capacitor has only a finite charge Q and as the charge of one capacitor is flowing to the other, the capacitor giving the charge is not replenished. But I feel like I am missing something. – Angel Tran Aug 22 '19 at 01:16
  • This is all new to me and when I tried looking it up, I didn't get much good results. I never knew it was called the two capacitor paradox so thank you to the kind commenters for pointing it out to me – Angel Tran Aug 22 '19 at 01:17
  • Angel Tran, you might find the following link helpful: How do I ask a good question?: "Have you thoroughly searched for an answer before asking your question? Sharing your research helps everyone. Tell us what you found and why it didn’t meet your needs. This demonstrates that you’ve taken the time to try to help yourself, it saves us from reiterating obvious answers," – Alfred Centauri Aug 22 '19 at 12:35

1 Answers1

0

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
  • 13,526
  • 1
  • 22
  • 42