Parallel-Plate Capacitor with a Leaky Dielectric
A parallel-plate capacitor has plates of area 0.050 m² separated by 2.0 mm, connected to a 12 V battery. A dielectric slab with relative permittivity 4.0 is inserted, filling the gap. After a few minutes the experimenter notices that the stored charge has fallen by 15% while the battery remains connected, and the dielectric feels slightly warm. A second identical capacitor without dielectric shows no such drop. The experimenter suspects the dielectric is slightly conducting.
1. Calculate the capacitance and stored charge before and after inserting the dielectric, and explain why the charge should be constant if the dielectric were ideal.
Read the worked answer
Vacuum capacitance: C₀ = ε₀A/d = (8.85 × 10⁻¹² × 0.050)/(2.0 × 10⁻³) = 2.21 × 10⁻¹⁰ F ≈ 0.22 nF.
With dielectric: C = κC₀ = 4.0 × 2.21 × 10⁻¹⁰ = 8.85 × 10⁻¹⁰ F ≈ 0.89 nF.
Charge: Q = CV = 8.85 × 10⁻¹⁰ × 12 = 1.06 × 10⁻⁸ C ≈ 10.6 nC.
Ideal case: With the battery connected, V is fixed at 12 V. An ideal dielectric polarises but does not conduct, so no charge can leave the plates. The extra charge comes from the battery supplying charge as C increases. Charge should be constant thereafter.
2. The observed 15% charge loss over 5.0 minutes suggests a leakage current. Estimate the effective resistance of the dielectric and explain the warming.
Read the worked answer
Lost charge: ΔQ = 0.15 × 10.6 nC = 1.59 nC over t = 300 s.
Current: I = ΔQ/t = 1.59 × 10⁻⁹/300 = 5.3 × 10⁻¹² A.
Resistance: R = V/I = 12/(5.3 × 10⁻¹²) ≈ 2.3 × 10¹² Ω. This is high but finite, consistent with a slightly conducting dielectric.
Warming: The leakage current dissipates power P = VI = 12 × 5.3 × 10⁻¹² ≈ 6.4 × 10⁻¹¹ W in the dielectric. Over 300 s this is about 1.9 × 10⁻⁸ J - tiny, so the warming is more likely from dielectric loss during the polarisation transient when the slab was inserted, not from steady leakage. The key judgement is that the steady leakage alone cannot explain perceptible warmth; the transient absorption is the better explanation.