Two identical capacitors have the same capacitance C. One of them is charged to potential V1 and the other to V2. The negative ends of the capacitors are connected together. When the positive ends are also connected, the decrease in energy of the combined system is:
Correct answer: C. Energy decrease is 0.5C(V12 + V22 - 2Vf2)
- A. Energy decrease is 0.5C(V1 - V2)2
- B. Energy decrease is C(V1 - V2)2
- C. Energy decrease is 0.5C(V12 + V22 - 2Vf2)
- D. Energy decrease is 2C(V12 - V22)
Explanation
When two capacitors are connected in parallel, the total charge is conserved, and they share a common final voltage, Vf. Initially, the energy stored in the capacitors is given by 0.5CV12 and 0.5CV22. After connection, the final energy is 0.5C(Vf)2 for each capacitor. The decrease in energy is the difference between the initial and final energies, leading to the formula 0.5C(V12 + V22 - 2Vf2). Option C correctly represents this energy decrease. Options A, B, and D are incorrect as they misrepresent the energy changes due to incorrect assumptions about the system's final state.
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Electrostatics describes forces between charges through Coulomb's law, electric fields and field intensity, then uses electric potential to measure energy per unit charge. Capacitors store charge and electrical energy, with capacitance depending on geometry and dielectric material, not simply on the amount of charge present.
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