Disconnected Capacitor with a Partly Filled Dielectric
In a school laboratory, a parallel-plate capacitor is made from two metal plates of area 2.0 × 10−2 m2 separated by 4.0 mm. With the plates connected to a 12 V battery, a student measures the charge on the plates as 5.3 × 10−10 C. She then disconnects the battery, slides a 2.0 mm thick polythene sheet of dielectric constant 2.3 completely between the plates, and finds that the voltage across the plates falls. A second student argues that because the battery is disconnected the charge cannot change, so the voltage must remain 12 V and the stored energy cannot change. The teacher asks the group to reconcile the observation with energy conservation and to decide whether the sheet is pulled in or has to be pushed in during insertion. The plate area is unaffected and the sheet fills exactly half the gap.
1. Calculate the voltage across the plates after the polythene sheet is inserted and the change in stored energy. Account for where the energy goes.
Read the worked answer
Initial capacitance: C0 = ε0A/d = (8.85 × 10−12)(2.0 × 10−2)/(4.0 × 10−3) = 4.43 × 10−11 F. The measured charge is Q = 5.3 × 10−10 C, so the initial energy is U0 = Q2/(2C0) = (5.3 × 10−10)2/(2 × 4.43 × 10−11) ≈ 3.2 × 10−9 J. After insertion the gap is two layers in series: 2.0 mm of air and 2.0 mm of dielectric. C_air = (8.85 × 10−12)(2.0 × 10−2)/(2.0 × 10−3) = 8.85 × 10−11 F; C_die = 2.3 × 8.85 × 10−11 = 2.04 × 10−10 F. The series equivalent is C = (8.85 × 10−11 × 2.04 × 10−10)/(8.85 × 10−11 + 2.04 × 10−10) ≈ 6.16 × 10−11 F. Since the battery is disconnected, Q stays 5.3 × 10−10 C, so V = Q/C ≈ 8.6 V. The final energy is U = Q2/(2C) ≈ 2.3 × 10−9 J, a decrease of about 0.9 × 10−9 J. The missing energy is not lost: the electric field does mechanical work pulling the dielectric into the gap; if the sheet is released it gains kinetic energy and eventually this becomes heat as it settles.
2. Is the sheet pulled into the gap or must it be pushed in? Justify your answer using the field or energy argument.
Read the worked answer
It is pulled in; no external push is needed once insertion has started. At fixed charge, the energy is U = Q2/(2C). Inserting the dielectric increases C, so U decreases. A system free to move tends toward lower electrostatic energy, so the force is in the direction that increases capacitance, namely inward. Equivalently, the fringe field at the edge polarises the dielectric and exerts a force on its induced surface charges, pulling it into the region of stronger field. The energy decrease is exactly the work done by this force. If the sheet were held fixed after insertion, an external holder would have to do negative work, but during free insertion the field does positive work on the sheet.
3. The second student says that because the charge is fixed, the electric field between the plates cannot change. Evaluate this statement region by region.
Read the worked answer
The statement is only partly true. With free charge fixed, the free surface charge density σ = Q/A is fixed, and in the gap the electric displacement D = σ (ignoring edge effects). In the air layer E_air = D/ε0, which is the same as the original field, about 3.0 × 103 V m−1. In the dielectric layer, however, E_die = D/(κ ε0) = E_air/2.3 ≈ 1.3 × 103 V m−1. The potential difference is the line integral of E across the whole gap; because part of the path now has a smaller field, the voltage falls even though Q is unchanged. The stored energy density is ½D·E, so it is unchanged in the air layer but lower in the dielectric layer, giving the overall decrease in energy.