Dielectric inserted with and without the battery: auditing the energy
A 20 cm x 20 cm parallel-plate capacitor with a 2.0 mm air gap is charged to 200 V from a regulated supply rated 200 V, 1 mA. A student then slides a 2.0 mm thick glass slab (dielectric constant 5.0) fully between the plates, doing it twice: in run A the supply stays connected throughout, in run B the supply is disconnected first. From the meter readings the stored energy has gone up fivefold in run A and down fivefold in run B. A classmate objects that run A violates conservation of energy, since the field energy grew with no visible source. The student also notices that the slab is pulled in strongly and that she has to hold it back to keep it half out, and she wonders whether the 1 mA supply can even deliver the extra charge quickly enough for the 200 V to stay constant.
1. Compute C0, the charge and the stored energy for the initial state and for both runs. Confirm the factors of five, and check whether the supply can maintain 200 V during the insertion.
Read the worked answer
C0 = eps0 A/d = 8.85e-12 x (0.20 x 0.20)/2.0e-3 = 8.85e-12 x 20 = 1.77e-10 F = 177 pF.
Initial: Q0 = C0 V = 1.77e-10 x 200 = 3.54e-8 C; U0 = 0.5 C0 V^2 = 0.5 x 1.77e-10 x 4.0e4 = 3.54e-6 J = 3.5 microjoule.
Run A (battery connected, V fixed): C = kappa C0 = 885 pF; V = 200 V; Q = 1.77e-7 C; U = 0.5 C V^2 = 17.7 microjoule = 5 U0. Check.
Run B (isolated, Q fixed): C = 885 pF; V = Q0/C = 3.54e-8/8.85e-10 = 40.0 V; U = Q0^2/(2C) = 0.5 x 3.54e-8 x 40.0 = 7.1e-7 J = U0/5. Check.
Supply check: the extra charge needed is 1.77e-7 - 3.54e-8 = 1.42e-7 C. Even if the slab takes only 0.1 s to insert, the average current is 1.4 microampere, roughly a thousandth of the 1 mA rating. The supply easily holds 200 V, so run A really is a constant-voltage process.
2. Resolve the classmate's objection with a full energy audit for run A, and compute the force pulling the slab in.
Read the worked answer
Work done by the battery when the extra charge is pushed at constant voltage:
W_batt = V x delta_Q = 200 x 1.42e-7 = 2.83e-5 J.
Change in stored energy: delta_U = 17.7 - 3.5 = 1.42e-5 J.
Balance: W_batt = delta_U + W_mech, so W_mech = 1.42e-5 J - exactly half the battery's work. That mechanical energy appears as kinetic energy of the slab (and ultimately heat when it is stopped); the battery is the source, so nothing is violated. The general result for a battery-connected capacitor is that a fixed-voltage change costs twice the change in stored energy.
Force: with the slab inserted a distance x and edge fringing neglected,
dC/dx = eps0 b (kappa - 1)/d = 8.85e-12 x 0.20 x 4/2.0e-3 = 3.54e-9 F/m.
F = 0.5 V^2 dC/dx = 0.5 x 4.0e4 x 3.54e-9 = 7.1e-5 N.
Check: F x 0.20 m = 1.4e-5 J, matching W_mech. In run B there is no battery, and the same 1.4e-5 J of mechanical energy must come out of the field, consistent with U falling by a factor 5.
3. Predict what happens when the slab is inserted only half-way (covering half the plate area) in the disconnected case, and explain why the pulling force does not depend on how far the slab is in.
Read the worked answer
Treat the capacitor as two parallel capacitors sharing the same voltage: one with air (area A/2) and one with glass (area A/2).
C = eps0 (A/2)/d + kappa eps0 (A/2)/d = C0 (1 + kappa)/2 = 3 C0 = 531 pF.
Isolated, so Q stays 3.54e-8 C:
V = Q/C = 3.54e-8/5.31e-10 = 66.7 V, and U = 0.5 Q V = 1.18 microjoule.
(Full insertion would give 40 V and 0.71 microjoule; half insertion is intermediate, as it must be.)
Why F is independent of x: dC/dx = eps0 b (kappa - 1)/d contains no x, so F = 0.5 V^2 dC/dx is constant while the slab is between the plates. Physically, each additional strip of slab that slides in adds the same increment of capacitance; the fringing field at the slab edge is what pulls, and that field is set by the voltage and geometry, not by the slab's position. The force is therefore uniform, giving uniform acceleration if the slab is released, and it vanishes only when the slab is fully out and the fringing field is lost. In practice fringing near the far edge makes the force drop as the slab nears full insertion.