Two ideal capacitors (i.e., purely capacitance—no resistance or inductance), each with a capacitance of(adsbygoogle = window.adsbygoogle || []).push({}); C,are connected together through ideal wires (zero resistance), and an ideal switch (i.e., whenopen,the switch offers infinite resistance; whenclosed,it offers zero resistance), which initially (say, fort<0)is open.

A charge Q is placed on one of the capacitors; the other has zero charge. Since the energy stored in an ideal capacitor is given by

Energy = (1/2) (1/capacitance) X (charge)^2

we see that the energy stored in the two capacitors is

E(1st Capacitor, t<0) = (1/2) (1/C) X Q^2

and

E(2nd Capacitor, t<0) = 0.

Hence the total energy stored in the circuit while the switch is open is

E(total, t<0) = (1/2) (1/C) X Q^2.

https://www.physicsforums.com/attachment.php?attachmentid=5476&stc=1&d=1131215321

Suppose (say, att=0)the switch is then closed. The charge Q will flow from the 1st (charged) capacitor onto the 2nd (uncharged) capacitor until it’s evenly distributed between the two capacitors (because each has the same capacity to store charge). Hence, each will end up storing a charge of Q/2.

The energy stored in the two capacitors after the switch is closed is, therefore,

E(1st Capacitor, t>0) = (1/8) (1/C) X Q^2

and

E(2nd Capacitor, t>0) = (1/8) (1/C) X Q^2.

Hence the total energy stored in the circuit after the switch is closed is

E(total, t>0) = (1/4) (1/C) X Q^2 .

https://www.physicsforums.com/attachment.php?attachmentid=5477&stc=1&d=1131215312

Clearly, half of the initial (before the switch is closed) energy islostafter the switch is closed.

Where did it go?

(The answer is not at all obvious.)

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# Where did it go?

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