Dielectric Inserted After the Charging Battery Is Disconnected
A parallel-plate capacitor in a school lab has plates of area 0.020 m² separated by 2.0 mm of air. A student charges it to 120 V from a battery, then disconnects the battery and slides in a 2.0 mm thick plastic slab of dielectric constant 3.0 that exactly fills the gap. An electrometer now reads 40 V. A second student says the missing stored energy must have been converted entirely into heat in the plastic, but the slab is barely warm. The lab manual asks the group to account for the energy quantitatively and to predict what would change if the battery had stayed connected during insertion. The capacitor is far from other objects, edge effects are ignored, and ε0 = 8.85 × 10⁻¹² F m⁻¹.
1. Calculate the capacitance, charge and stored energy before insertion, and the capacitance, voltage and stored energy after insertion. Show that the measured 40 V is consistent with charge conservation.
Read the worked answer
Before insertion: C0 = ε0A/d = (8.85 × 10⁻¹²)(0.020)/(2.0 × 10⁻³) = 8.85 × 10⁻¹¹ F. Q = C0V = (8.85 × 10⁻¹¹)(120) = 1.06 × 10⁻⁸ C. U0 = ½C0V² = ½(8.85 × 10⁻¹¹)(120)² = 6.37 × 10⁻⁷ J.
After insertion the capacitor is isolated, so Q is unchanged. The dielectric multiplies capacitance by K: C = KC0 = 3(8.85 × 10⁻¹¹) = 2.66 × 10⁻¹⁰ F. V′ = Q/C = (1.06 × 10⁻⁸)/(2.66 × 10⁻¹⁰) = 40 V, matching the electrometer. U′ = ½Q²/C = U0/K = (6.37 × 10⁻⁷)/3 = 2.12 × 10⁻⁷ J. The voltage falling to one-third while charge stays fixed is exactly what K = 3 predicts.
2. Account for the decrease in stored energy. Is the second student's heat explanation acceptable? Estimate the energy involved and describe a test that would distinguish the mechanisms.
Read the worked answer
The energy decrease is ΔU = U0 − U′ = 6.37 × 10⁻⁷ − 2.12 × 10⁻⁷ = 4.25 × 10⁻⁷ J. In an isolated capacitor the field exerts a force pulling the dielectric into the gap. If the slab is released, that energy appears mainly as kinetic energy of the slab; if it is held, the holder does negative work while the field does positive work. Only if the motion is damped does the energy end up as internal energy. The slab being barely warm argues against direct resistive heating being the main channel. A distinguishing test is to release the slab from rest and measure its speed as it enters, or to attach a force sensor and integrate force over displacement. If the mechanical energy gained is close to 4.25 × 10⁻⁷ J, the heat explanation fails as the primary account. Repeating with a highly damped slab should then produce measurable warming.
3. If the battery had remained connected during insertion, determine the new charge, voltage and stored energy, and how much work the battery would supply. Comment on energy conservation.
Read the worked answer
With the battery connected, V stays 120 V while C triples: C = 2.66 × 10⁻¹⁰ F, Q′ = CV = (2.66 × 10⁻¹⁰)(120) = 3.19 × 10⁻⁸ C. U′ = ½CV² = ½(2.66 × 10⁻¹⁰)(120)² = 1.91 × 10⁻⁶ J, triple the original. The extra charge supplied by the battery is ΔQ = Q′ − Q = 2.12 × 10⁻⁸ C, so the battery does work W = VΔQ = (120)(2.12 × 10⁻⁸) = 2.55 × 10⁻⁶ J. The increase in stored energy is ΔU = 1.27 × 10⁻⁶ J. The remaining 1.27 × 10⁻⁶ J is mechanical work done by the field in pulling the slab in (or heat if the motion is damped). Thus W = ΔU + mechanical work, and energy is conserved; the battery supplies both the extra stored energy and the work of drawing the slab into the field.