One capacitor is charged until its stored energy is 17.0 J . A second uncharged

Question

One capacitor is charged until its stored energy is 17.0J. A second uncharged capacitor is then connected to it inparallel.

1. If the chargedistributes equally, what is now the total energy stored in theelectric fields? 2. Where did theexcess energy go? ("radiated"= radiated as infrared light directlyfrom the capacitors, "heat" = heated wires, " work" = work to movecharge from one capacitor to the other, "potential" = lostpotential energy, "other" = there must be anotherexplanation!) (a) heat (b) other (c) work (d) radiated (e) potential One capacitor is charged until its stored energy is 17.0J. A second uncharged capacitor is then connected to it inparallel.

1. If the chargedistributes equally, what is now the total energy stored in theelectric fields? 2. Where did theexcess energy go? ("radiated"= radiated as infrared light directlyfrom the capacitors, "heat" = heated wires, " work" = work to movecharge from one capacitor to the other, "potential" = lostpotential energy, "other" = there must be anotherexplanation!) (a) heat (b) other (c) work (d) radiated (e) potential 2. Where did theexcess energy go? ("radiated"= radiated as infrared light directlyfrom the capacitors, "heat" = heated wires, " work" = work to movecharge from one capacitor to the other, "potential" = lostpotential energy, "other" = there must be anotherexplanation!) (a) heat (b) other (c) work (d) radiated (e) potential

Explanation / Answer

It should be: 1) Just divide by two, since it is split evenly. 17/2 = 8.5 J 2) It's the thermal energy in the wires. a) heat I just did this minutes ago on CAPA, so it should work for you.:)

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