Capacitor with a Partly Filling Dielectric
A student charges a parallel-plate capacitor of plate area 0.020 m² and plate separation 2.0 mm to 200 V from a battery, then disconnects the battery. She inserts a glass slab of dielectric constant 6.0, thickness 1.8 mm and area 0.018 m², intending to fill the gap. An electrometer of resistance 10^11 Ω connected across the plates reads 45 V, not the 33 V she expected. The slab does not cover the plates completely and leaves an air gap. The laboratory is humid, and the electrometer reading drifts downward even before the slab is inserted. She has a micrometer, a second identical capacitor, and several thin glass sheets. She must decide whether the discrepancy comes from incomplete dielectric filling, leakage across the humid surfaces, or the electrometer, and report a defensible value of the effective dielectric constant.
1. Calculate the voltage expected if the glass slab completely filled the plate gap, and infer the effective dielectric constant from the observed 45 V. What does the comparison already rule out?
Read the worked answer
The air capacitance is C₀ = ε₀A/d = (8.85×10⁻¹²)(0.020)/(2.0×10⁻³) = 8.85×10⁻¹¹ F. The stored charge is Q = C₀V = (8.85×10⁻¹¹)(200) = 1.77×10⁻⁸ C. If the slab filled the whole gap, C = 6C₀ and V = Q/(6C₀) = 200/6 = 33.3 V. The observed 45 V gives C_eff = Q/V = 1.77×10⁻⁸/45 = 3.93×10⁻¹⁰ F = 4.44C₀, so the apparent dielectric constant is about 4.4. Since 1 < 4.4 < 6, dielectric is present but the effective capacitance is below the full-fill value. Incomplete coverage and the residual air gap put air in series and parallel with glass, lowering C. Leakage alone would lower V below the full-fill prediction; it cannot explain a voltage above 33 V if the initial charge is still used. However, because charge may already have leaked, 4.4 is only an apparent value.
2. The electrometer reading drifts downward even before the slab is inserted. How does that affect the value of k inferred from V, and how would you test whether the drift is surface leakage or the electrometer?
Read the worked answer
If charge leaks, the true charge at the later time is less than Q_initial. Using k = C_eff/C₀ = Q_initial/(V C₀) therefore overestimates k, because the numerator is too large. The drift also means the voltage is time-dependent, so a single reading cannot give a reliable k. To separate the causes, clean and dry the plates and insulators, then charge and disconnect with no slab and record V against time. A fast initial drop that depends on humidity points to surface leakage; a slow drop that remains after careful cleaning points to the electrometer input resistance or the capacitor itself. Check the electrometer with a known capacitor and estimate the RC time constant. A guard electrode or a dry enclosure would reduce surface leakage. The key is to measure V quickly and repeat with a control capacitor before inserting the slab.
3. Design a procedure to obtain a defensible effective dielectric constant for the actual slab, stating what you would measure and how you would reduce the two largest errors.
Read the worked answer
Measure the slab thickness at several points with the micrometer and confirm its area; record whether it covers the plates completely. Clean and dry the plates, slab and supports, and use a guard ring if available. Charge the capacitor, disconnect the battery, insert the slab quickly, and read the electrometer within a fraction of a second; repeat several times and also record the discharge curve with no slab. Compute k_eff from C_eff = Q/V using the charge measured immediately before insertion, and report the geometry separately: glass area, air-gap thickness and any uncovered area. The two largest errors are leakage, which makes Q uncertain, and incomplete filling, which makes the geometry uncertain. Reduce them by using stacked thin glass sheets to fill the 2.0 mm gap completely, by keeping the assembly dry, and by comparing with the second identical capacitor as a control.