Dielectric slab on a disconnected capacitor
To demonstrate dielectric effects, a parallel-plate capacitor is built from two circular plates of diameter 12.0 cm separated by 4.0 mm. A student charges the capacitor from a 200 V supply, disconnects the supply, and reads 200 V on an electrometer. A dielectric slab of the same thickness is inserted so that it fills exactly half the plate area; the electrometer reading falls to 129 V. The slab is withdrawn, the 200 V supply is reconnected, and the slab is inserted again. The electrometer still reads 200 V, and the slab is pulled inward. The student claims that the dielectric constant is 200/129 = 1.55 and that the force is repulsive because the voltage is held constant. A partner objects that the electrometer has its own capacitance, so the disconnected reading cannot be trusted.
1. Calculate the dielectric constant from the readings and say precisely what the ratio 1.55 represents. Does the result depend on the plate area?
Read the worked answer
C0 = ε0A/d. A = π(0.060 m)^2 = 1.13×10^-2 m^2, d = 4.0×10^-3 m, so C0 ≈ 8.85×10^-12 × 1.13×10^-2 / 4.0×10^-3 = 2.50×10^-12 F. With the supply disconnected, Q is constant, so V1/V2 = C2/C1. The half-filled capacitor is two capacitors in parallel: C2 = ε0(A/2)/d + κε0(A/2)/d = C0(1+κ)/2. Therefore (1+κ)/2 = 200/129 = 1.550, giving κ = 2.10. The student's 1.55 is C2/C1, not κ. The plate area cancels from the ratio, so the result does not require the absolute area, although edge effects make the ideal formula approximate.
2. Explain the direction of the force in both the connected and disconnected cases. Does the battery do work when the slab is inserted while connected?
Read the worked answer
The slab is pulled in because C increases as more dielectric enters the gap. At constant V, F = +1/2 V^2 dC/dx, which is inward. The battery supplies charge ΔQ = VΔC, doing work V^2ΔC = 2ΔU; half of this raises the field energy U = 1/2CV^2 and half appears as mechanical work on the slab. At constant Q, U = Q^2/(2C) decreases as C increases; F = -dU/dx = Q^2/(2C^2) dC/dx, also inward. The decrease in field energy supplies the mechanical work. Thus the force is attractive in both cases; the student's 'repulsive because voltage is constant' is wrong.
3. Assess the electrometer objection quantitatively. What voltage ratio would be observed if the electrometer had Ce = 5.0 pF? What does the observed 1.55 ratio imply about Ce?
Read the worked answer
If Ce = 5.0 pF, the total capacitance would be C0 + Ce = 7.5 pF before insertion and C2 + Ce = 3.89 + 5.0 = 8.89 pF after insertion. The voltage ratio would be 8.89/7.5 = 1.185, so the reading would fall only to 200/1.185 = 169 V, not 129 V. The observed ratio 1.55 is close to the ideal 1.55, so Ce must be much smaller than 2.5 pF; solving 3.89 + Ce = 1.55(2.50 + Ce) gives Ce ≈ 0.03 pF, effectively negligible. The objection is not supported, although leakage resistance in the electrometer could cause a slow drift rather than an immediate error.