If q_f is the free charge on the capacitor plates and q_b is the bound charge on the dielectric slab of dielectric constant k placed between the capacitor plates, then bound charge q_b an be expressed as :

A parallel plate capacitor with plate area 'A' and distance of separation 'd' is filled with a dielectric. What is the capacity of the capacitor when permittivity of the dielectric varies as :

$$\varepsilon (x) = {\varepsilon _0} + kx$$, for $$\left( {0 < x \le {d \over 2}} \right)$$

$$\varepsilon (x) = {\varepsilon _0} + k(d - x)$$, for $$\left( {{d \over 2} \le x \le d} \right)$$

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, $${{{C_2}} \over {{C_1}}}$$.