For changing the capacitance of a given parallel plate capacitor, a dielectric material of dielectric constant K is used, which has the same area as the plates of the capacitor. The thickness of the dielectric slab is $${3 \over 4}$$d, where 'd' is the separation between the plates of parallel plate capacitor. The new capacitance (C') in terms of original capacitance (C₀) is given by the following relation :

Consider the combination of 2 capacitors C₁ and C₂ with C₂ > C₁, when connected in parallel, the equivalent capacitance is $${{15} \over 4}$$ times the equivalent capacitance of the same connected in series. Calculate the ratio of capacitors, $${{{C_2}} \over {{C_1}}}$$.

An electron with kinetic energy K₁ enters between parallel plates of a capacitor at an angle '$$\alpha$$' with the plates. It leaves the plates at angle '$$\beta$$' with kinetic energy K₂. Then the ratio of kinetic energies K₁ : K₂ will be :

Two equal capacitors are first connected in series and then in parallel. The ratio of the equivalent capacities in the two cases will be :