A 6.00 μF parallel-plate capacitor is charged to 12.0 V. A dielectric of constant κ = 3.00 completely fills the gap. Compare two cases: A, the battery is disconnected before insertion; B, the ideal 12.0 V battery remains connected. For each case find final capacitance, charge, voltage and stored energy. Determine the external work in case A for quasistatic insertion and explain the direction of the dielectric force.

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Both cases C_f = 18.0 μF. A: Q = 72.0 μC, V = 4.00 V, U = 144 μJ, W_ext = −288 μJ. B: Q = 216 μC, V = 12.0 V, U = 1.296 mJ. The dielectric is pulled inward. Full insertion multiplies capacitance by κ, so C_f = 3.00(6.00 μF) = 18.0 μF. Initial charge is C₀V₀ = 72.0 μC and initial energy is ½C₀V₀² = 432 μJ. In case A, isolated charge remains 72.0 μC. Thus V_f = Q/C_f = 4.00 V and U_f = Q²/(2C_f) = 144 μJ. For quasistatic insertion with no battery exchange, W_ext = ΔU = 144−432 = −288 μJ. The negative sign means the field does +288 μJ of mechanical work while pulling the dielectric inward. In case B, voltage remains 12.0 V. Then Q_f = C_fV = 216 μC and U_f = ½C_fV² = 1.296 mJ; the battery supplies additional charge and energy. Insertion increases capacitance. At fixed charge this lowers field energy; at fixed voltage the battery–capacitor system also favors insertion after battery work is included. In both physical cases the net electromagnetic force draws the dielectric farther between the plates.

Compare connected and isolated dielectric insertion

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