Two identical parallel-plate capacitors initially have capacitance C₀ = 1.200 × 10^-8 F and voltage V₀ = 40 V. Capacitor A is disconnected from its battery; B remains connected to an ideal battery at V₀. A dielectric of relative permittivity κ = 3 is then inserted fully into each capacitor, with unchanged plate geometry. Neglect fringing and leakage. Find final charge, voltage and stored energy for each, and explain why the energy changes differ.

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A: Q = 4.800 × 10^-7 C, V = 13.33 V, U = 3.200 × 10^-6 J. B: Q = 1.440 × 10^-6 C, V = 40 V, U = 2.880 × 10^-5 J. Full insertion changes each capacitance to κC₀ = 3.600 × 10^-8 F. Initially each charge is Q₀ = C₀V₀ = 4.800 × 10^-7 C and each energy is U₀ = ½C₀V₀² = 9.600 × 10^-6 J. Capacitor A is isolated, so charge cannot flow to or from its plates. Q_A = Q₀ and V_A = Q₀/(κC₀) = 13.33 V. Use the fixed-charge energy form U_A = Q₀²/(2κC₀) = U₀/κ = 3.200 × 10^-6 J. The decrease can supply mechanical work as the dielectric is drawn in. For B the ideal battery fixes voltage, not charge. Q_B = κC₀V₀ = 1.440 × 10^-6 C and U_B = ½κC₀V₀² = 2.880 × 10^-5 J. The battery transfers additional energy to B while charge flows, so increased field energy does not violate conservation. The constraint must be identified before choosing which version of the capacitor-energy formula to use.

A dielectric with two different electrical constraints · set 10

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