Dielectric Insertion with the Battery Still Connected
In a II PUC physics laboratory, a student charges a parallel-plate capacitor with plate area 0.020 m² and plate separation 1.0 mm while it remains connected to a 200 V battery. With air between the plates, the electrometer reads 200 V and the student records the charge. The student then slides a 1.0 mm thick dielectric slab (dielectric constant 5.0, dielectric strength 20 kV/mm) completely into the gap without disconnecting the battery. The electrometer still reads 200 V, but the charge stored has increased. A classmate argues that the stored energy must have decreased because the capacitance increased while the voltage stayed fixed. The teacher asks the group to check the capacitor for electrical breakdown and to account carefully for the energy supplied by the battery during the insertion. The slab is inserted slowly, and the student notices that the slab is pulled inward by the plates unless it is held back.
1. Calculate the air capacitance and the initial charge on the plates. State what stays fixed and what changes when the slab is inserted with the battery still connected.
Read the worked answer
Using C₀ = ε₀A/d:
C₀ = (8.85×10⁻¹² F/m)(0.020 m²)/(1.0×10⁻³ m) = 1.77×10⁻¹⁰ F ≈ 0.18 nF.
Q₀ = C₀V = (1.77×10⁻¹⁰ F)(200 V) = 3.54×10⁻⁸ C ≈ 35 nC.
With the battery connected, V is fixed at 200 V. Inserting the dielectric increases C, so Q = CV must increase; Q is not fixed.
2. After the slab fills the gap, find the new capacitance, the new charge, and the electric field inside the dielectric. Is the dielectric likely to break down? Justify with a numerical comparison.
Read the worked answer
C = K C₀ = 5.0(1.77×10⁻¹⁰ F) = 8.85×10⁻¹⁰ F.
Q = CV = (8.85×10⁻¹⁰ F)(200 V) = 1.77×10⁻⁷ C.
The field in the dielectric is E = V/d = 200 V/(1.0×10⁻³ m) = 2.0×10⁵ V/m = 0.20 kV/mm.
The dielectric strength is 20 kV/mm, so the actual field is about 100 times smaller than the breakdown field. Breakdown is very unlikely.
3. The classmate says stored energy decreases. Evaluate this claim and explain where the extra energy comes from or goes, using battery work and energy stored.
Read the worked answer
U = ½CV². Initially U₀ = ½(1.77×10⁻¹⁰)(200)² = 3.54×10⁻⁶ J. Finally U = ½(8.85×10⁻¹⁰)(200)² = 1.77×10⁻⁵ J. Stored energy increases by 1.42×10⁻⁵ J, so the classmate is wrong.
The battery supplies additional charge ΔQ = 1.77×10⁻⁷ − 3.54×10⁻⁸ = 1.42×10⁻⁷ C at 200 V, doing work W_batt = VΔQ = 2.83×10⁻⁵ J. This is twice the increase in stored energy. The other half is mechanical work done by the field pulling the slab inward, or is dissipated if the slab is not allowed to move slowly. Energy bookkeeping: W_batt = ΔU + W_mechanical.