A slab of dielectric constant $$\mathrm{K}$$ has the same cross-sectional area as the plates of a parallel plate capacitor and thickness $$\frac{3}{4} \mathrm{~d}$$, where $$\mathrm{d}$$ is the separation of the plates. The capacitance of the capacitor when the slab is inserted between the plates will be :

(Given $$\mathrm{C}_{0}$$ = capacitance of capacitor with air as medium between plates.)

Two capacitors, each having capacitance $$40 \,\mu \mathrm{F}$$ are connected in series. The space between one of the capacitors is filled with dielectric material of dielectric constant $$\mathrm{K}$$ such that the equivalence capacitance of the system became $$24 \,\mu \mathrm{F}$$. The value of $$\mathrm{K}$$ will be :

A source of potential difference $$V$$ is connected to the combination of two identical capacitors as shown in the figure. When key '$$K$$' is closed, the total energy stored across the combination is $$E_{1}$$. Now key '$$K$$' is opened and dielectric of dielectric constant 5 is introduced between the plates of the capacitors. The total energy stored across the combination is now $$E_{2}$$. The ratio $$E_{1} / E_{2}$$ will be :

The total charge on the system of capacitors $$C_{1}=1 \mu \mathrm{F}, C_{2}=2 \mu \mathrm{F}, \mathrm{C}_{3}=4 \mu \mathrm{F}$$ and $$\mathrm{C}_{4}=3 \mu \mathrm{F}$$ connected in parallel is :

(Assume a battery of $$20 \mathrm{~V}$$ is connected to the combination)