Calculate the capacitance and free charge on the plates of a parallel-plate capacitor with a dielectric slab.

Here's a step-by-step solution to the problem: First, let's list the given values and convert them to SI units: Plate area, A = 90 cm^2 = 90 × (10^-2 m)^2 = 9 × 10^-3 m^2 Plate separation, d = 1.14 cm = 1.14 × 10^-2 m Potential difference, V = 60 V Dielectric slab thickness, t = 0.6 cm = 0.6 × 10^-2 m = 6 × 10^-3 m Dielectric constant, K = 2.7 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 in a vacuum (or air, which is a good approximation). C = (_0 A)/(d) Step 2: Substitute the given values into the formula. C = (8.85 × 10^-12 F/m) × (9 × 10^-3 m^2)1.14 × 10^-2 m Step 3: Calculate the capacitance. C = 7.965 × 10^-14 F·m1.14 × 10^-2 m C = 6.9868 × 10^-12 F The capacitance is approximately 6.99 pF. C = 6.99 × 10^-12 F b) What free charge appears on the plate. Step 1: Use the relationship between charge, capacitance, and potential difference. Q = CV Step 2: Substitute the calculated capacitance from part (a) and the given potential difference. Q = (6.9868 × 10^-12 F) × (60 V) Step 3: Calculate the charge. Q = 4.19208 × 10^-10 C The free charge on the plate is approximately 4.19 × 10^-10 C. Q = 4.19 × 10^-10 C c) If the battery was disconnected and a dielectric slab of thickness 0.6 cm was inserted between the plates. What is the electric field in the gap between the plates and the dielectric slab? When the battery is disconnected, the charge Q on the plates remains constant. The electric field will be different in the air gap and inside the dielectric slab. Step 1: Calculate the electric field in the air gap (the region without the dielectric). The electric field in the air gap is determined by the surface charge density = Q/A. E_air = ()/(_0) = (Q)/(A_0) Step 2: Substitute the charge from part (b) and the given values. E_air = 4.19208 × 10^-10 C(9 × 10^-3 m^2) × (8.85 × 10^-12 F/m) E_air =