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For a parallel plate capacitor, the capacitance C with a dielectric is C = kε0A/d, where k is the dielectric constant of the material. For air, kair ≈ 1. For oil, koil > 1.The electric field E in a capacitor is related to the charge Q, capacitance C, and plate separation d by E = V/d. Also, Q = CV, so V = Q/C. Thus, E = Q/(Cd).Since the capacitors are identical (same A, d) and carry the same charge Q:Eair = Q / (Cair * d) = Q / ((kairε0A) * d/d) = Q / (kairε0A)Eoil = Q / (Coil * d) = Q / ((koilε0A) * d/d) = Q / (koilε0A)The ratio Eair / Eoil = (Q / (kairε0A)) / (Q / (koilε0A)) = koil / kair.Since kair ≈ 1 and koil > 1 for oil, the ratio Eair / Eoil = koil, which will be greater than 1. Therefore, the ratio is between 1 and infinity.