Inserting a Dielectric With the Battery Still Connected
In a school lab, a parallel-plate capacitor is built from two square metal plates, each 20.0 cm by 20.0 cm, separated by a 2.00 mm air gap. A student connects the plates to a regulated 300 V dc supply and leaves the supply connected. A mica sheet with dielectric constant 5.0, thickness 2.00 mm, and dielectric strength 40 kV/mm is available. The student wants to increase the stored energy by a factor of five by sliding the sheet completely between the plates. A classmate warns that the capacitor will discharge through the mica or that the voltage will fall to 60 V. The instructor asks for a quantitative check before insertion. Ignore edge effects and assume the mica fills the gap exactly. The supply can deliver up to 2 mA and has no current-limiting problem. The meter used to monitor voltage has a resistance of 10 MΩ.
1. Calculate the air-gap capacitance, the stored energy before insertion, and the stored energy after the mica is inserted with the supply still connected. Does the energy increase by the factor the student expects? Include the charge supplied by the battery and an energy balance.
Read the worked answer
Plate area A = 0.200 × 0.200 = 0.0400 m² and d = 2.00×10⁻³ m.
C₀ = ε₀A/d = (8.85×10⁻¹²)(0.0400)/(2.00×10⁻³) = 1.77×10⁻¹⁰ F ≈ 177 pF.
U₀ = ½C₀V² = 0.5(1.77×10⁻¹⁰)(300)² = 7.97×10⁻⁶ J ≈ 8.0 µJ.
With mica and the battery still connected, V remains 300 V, while C = KC₀ = 5(177 pF) = 885 pF. Then U = ½CV² = 0.5(8.85×10⁻¹⁰)(300)² = 3.98×10⁻⁵ J ≈ 40 µJ. The energy rises by a factor of 5, so the student's expectation is correct for this connection.
The extra charge supplied by the battery is ΔQ = (C − C₀)V = 4(1.77×10⁻¹⁰)(300) = 2.12×10⁻⁷ C. Battery work is VΔQ = 6.37×10⁻⁵ J. The stored energy increases by only 3.18×10⁻⁵ J, exactly half the battery work; the other half is dissipated or appears as mechanical work during the constant-voltage insertion.
2. Now suppose the student disconnects the battery before inserting the mica. Find the charge, voltage, electric field in the mica, and stored energy after insertion. Explain why the classmate's 60 V warning is correct in this case but not in the connected case.
Read the worked answer
With the battery disconnected, charge is fixed at Q = C₀V₀ = (1.77×10⁻¹⁰)(300) = 5.31×10⁻⁸ C.
After insertion C = 5C₀, so V = Q/C = V₀/5 = 60 V. The field in the mica is E = V/d = 60/(2.00×10⁻³) = 3.0×10⁴ V/m = 30 V/mm = 0.030 kV/mm.
The stored energy becomes U = Q²/(2C) = U₀/5 = 7.97×10⁻⁶/5 = 1.59×10⁻⁶ J ≈ 1.6 µJ. Energy decreases because the dielectric is pulled into the capacitor; the lost electrical energy does mechanical work on the dielectric, with some possible heat from friction.
The 60 V warning applies only when the battery is removed, because then Q is constant and V = Q/C falls by the factor K. With the battery connected, the supply maintains 300 V by supplying additional charge, so the voltage does not fall to 60 V.
3. Assess dielectric breakdown risk and the claim that the capacitor will discharge through the mica. Use the field values from both cases and the given dielectric strength. State one practical reason the real result might differ from the ideal calculation.
Read the worked answer
With the battery connected, E = V/d = 300/(2.00×10⁻³) = 1.5×10⁵ V/m = 150 V/mm = 0.15 kV/mm. With the battery disconnected, E = 0.030 kV/mm. The mica dielectric strength is 40 kV/mm, so the connected case is below breakdown by a factor of about 267, and the disconnected case by about 1333. There is no bulk breakdown in either case.
Mica is an insulator, not a conductor, so the capacitor does not 'discharge through' it in the sense of a short circuit. There is only a small leakage current, set by the very high resistivity of mica and the 10 MΩ monitoring meter.
A practical difference can arise from air voids or poor contact between the mica and the plates. Air has a lower permittivity than mica, so the field in an air gap can be higher than the average field by roughly a factor of K. Even then, 0.15 kV/mm × 5 = 0.75 kV/mm remains below the roughly 3 kV/mm breakdown field of air at room conditions, but partial discharges in voids and dielectric loss can still affect a real capacitor.