This physics problem tests your understanding of fundamental physical laws and their applications. The step-by-step solution below breaks down the problem using relevant equations and physical reasoning.
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Another one CØTBØJ — let's solve it. Here are the solutions for problem 1.12: First, convert all given values to SI units: Plate area, A = 120 cm^2 = 120 × (10^-2 m)^2 = 0.012 m^2 Plate separation, d = 1.8 cm = 1.8 × 10^-2 m = 0.018 m Potential difference, V = 52 V Dielectric slab thickness, t = 0.6 mm = 0.6 × 10^-3 m Dielectric constant, k = 3 Permittivity of free space, _0 = 8.85 × 10^-12 F/m a) the capacitance of the capacitor Step 1: Use the formula for the capacitance of a parallel plate capacitor. C = (_0 A)/(d) Step 2: Substitute the values and calculate the capacitance. C = (8.85 × 10^-12 F/m) × (0.012 m^2)0.018 m C = 1.062 × 10^-13 F·m0.018 m C = 5.9 × 10^-12 F The capacitance of the capacitor is 5.9 pF. b) what free charge appears on the plate Step 1: Use the relationship between charge, capacitance, and voltage. Q = CV Step 2: Substitute the capacitance from part (a) and the given voltage. Q = (5.9 × 10^-12 F) × (52 V) Q = 306.8 × 10^-12 C Q = 3.068 × 10^-10 C The free charge on the plate is 3.068 × 10^-10 C. c) If the battery was disconnected and a dielectric slab of thickness 0.6 mm was inserted between the plates, what is the electric field in the gap between the plates and the dielectric slab if k=3 Step 1: When the battery is disconnected, the charge Q on the plates remains constant. The electric field in the air gap (the region without the dielectric) is determined by this constant charge density. The electric field in the air gap is given by E_gap = (Q)/(A _0). This is equivalent to the initial electric field before the dielectric was inserted, E_initial = (V)/(d). Step 2: Calculate the electric field in the gap using the initial voltage and plate separation. E_gap = (V)/(d) E_gap = 52 V0.018 m E_gap ≈ 2888.89 V/m The electric field in the gap between the plates and the dielectric slab is 2.89 × 10^3 V/m. d) what is the electric field in the dielectric slab.