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Consider a capacitor of capacitance C initially charged with charge Q. A second initially uncharged capacitor,...

Consider a capacitor of capacitance C initially charged with charge Q. A second initially uncharged capacitor, also of capacitance C, is subsequently connected in parallel to the first capacitor. It is straight forward to show that charge state of the combined system, when it reaches equilibrium, is one where the charge divides between the capacitors, so that each capacitor now carries charge Q=2. However, it is interesting to note that the energy stored in the capacitors has changed between the initial configuration (1 capacitor) and the final configuration (2 capacitors in parallel). Energy must be conserved. Where did it go? Hint: consider the resistance of the wire used to connect the capacitors (easier), and, optionally, the inductance of this wire. Obtain a quantitative solution for the charge on the capacitors as a function of time after they are connected, using realistic estimates for actual parameter values. Assume C = 10 * 10^-6 F.

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