Dielectric Insertion: Who Supplies the Extra Energy?
In a school laboratory, a parallel-plate capacitor is made from two metal plates of area 0.020 m² held 2.0 mm apart by insulating supports. It is charged from a 100 V battery and the battery is then disconnected. A student slides a 2.0 mm thick polythene sheet (dielectric constant 4.0, dielectric strength 20 kV/mm) between the plates so that it completely fills the gap. A classmate predicts that the voltage will rise to 400 V because 'the dielectric adds charge.' The teacher asks the group to measure the new voltage with an electrometer and to account for the stored energy before and after insertion. A second group repeats the experiment but keeps the battery connected throughout. Both groups must explain why the energy change has opposite signs in the two cases and where the energy comes from or goes to. The plate separation is fixed.
1. With the battery disconnected before the slab is inserted, calculate the capacitance, charge, field in the dielectric and stored energy before and after insertion. Is the classmate's prediction of 400 V correct?
Read the worked answer
Take ε₀ = 8.85 × 10⁻¹² F/m.
Before insertion
- C₀ = ε₀A/d = (8.85 × 10⁻¹²)(0.020)/(2.0 × 10⁻³) = 8.85 × 10⁻¹¹ F ≈ 88.5 pF.
- Q = C₀V₀ = (8.85 × 10⁻¹¹)(100) = 8.85 × 10⁻⁹ C. Because the battery is disconnected, this free charge cannot change.
- E₀ = V₀/d = 100/(2.0 × 10⁻³) = 5.0 × 10⁴ V/m = 50 kV/m.
- U₀ = ½C₀V₀² = ½(8.85 × 10⁻¹¹)(100)² = 4.43 × 10⁻⁷ J.
After insertion
- C = KC₀ = 4 × 88.5 pF = 354 pF.
- V = Q/C = (8.85 × 10⁻⁹)/(3.54 × 10⁻¹⁰) = 25 V, not 400 V.
- E = V/d = 25/(2.0 × 10⁻³) = 1.25 × 10⁴ V/m = 12.5 kV/m.
- U = Q²/2C = ½(8.85 × 10⁻⁹)²/(3.54 × 10⁻¹⁰) = 1.11 × 10⁻⁷ J.
The classmate is wrong. The dielectric does not add free charge; it polarises, and the bound surface charge produces a field opposing the original field. With Q fixed, the potential difference and the field both fall by the factor K. The stored energy falls by 3.32 × 10⁻⁷ J; this appears mainly as mechanical work done by the field in pulling the slab into the gap (plus a little heat if there is friction). Energy is conserved.
2. Repeat the comparison for the group that keeps the 100 V battery connected. Account for the battery work and state where the extra energy goes.
Read the worked answer
Here V is held at 100 V, so the charge must change.
- C = 354 pF, Q = CV = (3.54 × 10⁻¹⁰)(100) = 3.54 × 10⁻⁸ C.
- Increase in charge: ΔQ = 3.54 × 10⁻⁸ − 8.85 × 10⁻⁹ = 2.66 × 10⁻⁸ C.
- U = ½CV² = ½(3.54 × 10⁻¹⁰)(100)² = 1.77 × 10⁻⁶ J, an increase of ΔU = 1.33 × 10⁻⁶ J.
- Work done by the battery: W = VΔQ = 100 × 2.66 × 10⁻⁸ = 2.66 × 10⁻⁶ J.
Only half of the battery work appears as extra stored energy. The remaining 1.33 × 10⁻⁶ J is the mechanical work done in pulling the slab into the capacitor, exactly as in the isolated case. For a linear dielectric, W_battery = 2ΔU and W_mechanical = ΔU. So the sign of the energy change is opposite in the two experiments, but in both cases the field pulls the slab in; with the battery connected the source resupplies the charge that the polarisation would otherwise reduce.
3. In a third trial the battery is again disconnected and a slab only 1.0 mm thick is inserted, leaving a 1.0 mm air gap. Find the new voltage and the field in the air gap and in the dielectric. Which region limits the safe operating voltage?
Read the worked answer
The air gap and dielectric are now in series, so the effective separation is d_eff = d_air + d_d/K = 1.0 mm + (1.0 mm)/4 = 1.25 mm.
- C = ε₀A/d_eff = (8.85 × 10⁻¹²)(0.020)/(1.25 × 10⁻³) = 1.42 × 10⁻¹⁰ F = 142 pF.
- V = Q/C = (8.85 × 10⁻⁹)/(1.42 × 10⁻¹⁰) = 62.5 V.
The free charge is unchanged, so the electric displacement D = Q/A = (8.85 × 10⁻⁹)/(0.020) = 4.43 × 10⁻⁷ C/m² is the same in both layers.
- In air: E_air = D/ε₀ = 4.43 × 10⁻⁷/(8.85 × 10⁻¹²) = 5.0 × 10⁴ V/m = 50 kV/m (unchanged from the original field).
- In the dielectric: E_d = D/(Kε₀) = 12.5 kV/m.
Check: V = E_air(1.0 mm) + E_d(1.0 mm) = 50 V + 12.5 V = 62.5 V.
The air gap is the weak point. It still experiences the full original field of 50 kV/m, while the dielectric sees only one quarter of it. The dielectric strength of 20 kV/mm is 20,000 kV/m, far above 50 kV/m, so this particular sheet is safe, but if the charging voltage were raised, breakdown would start in the air gap first. Partial dielectric filling does not protect the remaining air gap.