The equivalent capacitance is 20 \( \mu \)F by joining two capacitors of capacitances \( C_1 \) \( \mu \)F and \( C_2 \) \( \mu \)F in parallel. If they are joined in series, the equivalent capacitance is 4.8 \( \mu \)F. Find the ratio of the capacitances \( C_1 \) and \( C_2 \) ( \( C_1>C_2 \)).