Figure 25-18 shows a parallel-plate capacitor of plate areaA and plate separation d. A potential differenceV0 is applied between the plates. While thebattery remains connected, a dielectric slab of thicknessb and dielectric constant κ is placed between the plates as shown. Assume A = 190 cm2, d= 1.30 cm, V0 = 88.5 V, b = 0.950 cm,and κ = 4.89. Calculate (a) the capacitance,(b) the charge (in C) on the capacitor plates,(c) the electric field in the gap, and (d) the electric field in the slab, after the slab is in place.