Concept explainers
(II) Show that the capacitor in Example 24–12 with dielectric inserted can be considered as equivalent to three capacitors in series, and using this assumption show that the same value for the capacitance is obtained as was obtained in part (g) of the Example.
EXAMPLE 24–12 Dielectric partially fills capacitor. A parallel-plate capacitor has plates of area A = 250 cm2 and separation d = 2.00 mm. The capacitor is charged to a potential difference V0 = 150 V. Then the battery is disconnected (the charge Q on the plates then won’t change), and a dielectric sheet (K = 3.50) of the same area A but thickness ℓ = 1.00 mm is placed between the plates as shown in Fig. 24–18. Determine (a) the initial capacitance of the air-filled capacitor, (b) the charge on each plate before the dielectric is inserted, (c) the charge induced on each face of the dielectric after it is inserted, (d) the electric field in the space between each plate and the dielectric, (e) the electric field in the dielectric, (f) the potential difference between the plates after the dielectric is added, and (g) the capacitance after the dielectric is in place.
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Chapter 24 Solutions
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