$$ \text { A } 2 \mu \mathrm{~F} \text { capacitor is charged as shown, } $$

When switch $S$ is shifted to position $2, \%$ of energy lost is

A network of four capacitors of capacity equal to $C_1=C, C_2=2 C, C_3=3 C$ and $C_4=4 C$ are connected to a battery, as shown in the figure. The ratio of the charges on $C_2$ and $C_4$ is $\_\_\_\_$

Capacity of a parallel plate capacitor in the absence of dielectric medium is $C_0$. A sheet of dielectric constant $k$ and thickness of one-fourth of the plate separation is inserted between the plates. If the new capacity is $C$, then $\frac{C}{C_0}$ is

A parallel plate capacitor with air between the plates has a capacitance of $$15 \mathrm{~pF}$$. the separation between the plates is $$d$$. The space between the plates is now filled with two dielectrics constant $$k_1=3$$ and thickness $$d / 3$$ while the other one has dielectric constant $$k_2=6$$ and thickness $$2 d / 3$$. Capacitance of the capacitor is now