0
\$\begingroup\$

Suppose there are two identical capacitors of capacitance C and one of them is charged to a potential V and gets a charge q. The energy stored in this capacitor is \$1/2cv^2\$ or \$q^2/2c\$ . Now this capacitor is connected with the uncharged capacitor. Now the net energy of this system will be \$q^2/4c\$ . How is the other half of the energy lost ? Where does it go ?

\$\endgroup\$
  • 1
    \$\begingroup\$ it go to the uncharged capacitor ? Energy is never lost ! \$\endgroup\$ – Long Pham Apr 19 '18 at 14:25
  • 1
    \$\begingroup\$ Of course energy is lost in some form. We can say that from the above equations. I just want to know where the other half goes \$\endgroup\$ – 0xVikas Apr 19 '18 at 14:28
  • 1
    \$\begingroup\$ Each capacitor will only have half the charge. \$\endgroup\$ – Oldfart Apr 19 '18 at 14:43
  • \$\begingroup\$ The only energy loss is in the resistance of the wiring - which is not being shown in the equations in the OP. Meanwhile, the second equation applies to both capacitors, not just the second. So, like oldfart says, the energy is split among both capacitors. \$\endgroup\$ – SDsolar Apr 19 '18 at 16:01
0
\$\begingroup\$

To expand on Spehro's answer, let's assume a superconducting capacitor and wires. Then, when connection is made, charge will start flowing from the first cap to the second. When the voltage is equal, current will not, due to the inductance of the wires. Instead, the first cap will completely discharge, and the second will fully charge. At this point, the process will repeat in reverse. The result will be a sinusoidal oscillation with frequency $$f = \frac{1}{2{\pi LC}} $$ where L is the inductance of the entire system, including wires and internal capacitor structure.

This, of course, assumes no losses anywhere. If the wires are not superconductors, the oscillation frequency will be a bit lower, and the oscillation amplitude will (approximately) look like an inverse exponential. Even if there is no resistance, as Spehro notes, the circuit will radiate power as RF radiation (or whatever frequency is involved) and this will also serve to damp the oscillations.

When everything comes to rest, the two caps will have equal voltage and equal charge. Half the original energy will have been lost via the mechanisms noted.

\$\endgroup\$
2
\$\begingroup\$

Mostly heat, typically in a real situation. Consider a non-zero resistance between the capacitors (and internal to them) and analyze what happens after the instant of connection. There will be a spark and some EM emission.

Even if the capacitors literally had zero resistance (superconducting) the energy would slop around because of inductance and eventually be lost because it was induced in lossy surrounding materials or emitted as EM waves.

\$\endgroup\$

Not the answer you're looking for? Browse other questions tagged or ask your own question.