A parallel plate capacitor with air between the plates has a capacitance of $9\mu \mathrm{F}$. The separation between its plates is $d$. The space between the plates is now filled with two dielectric. One of the dielectrics has dielectric constant $K_1=3$ and thickness $\frac{d}{3}$ while the other one has dielectric constant $K_2=6$ and thickness $\frac{2d}{3}.$ Capacitance of the capacitor is now

Four plates of equal area $A$ are separated by equal distance $d$ and are arranged as shown in the figure Fig. $4.105 .$ The equivalent capacitance is
 

Two parallel plates of area $A$ are separated by two different dielectric as shown in figure, Fig.$4.106$ . The net capacitance is

 A parallel plate capacitor with air as the dielectric has capacitance $C$ A slab of dielectric constant $K$ and having the same thickness as the separation between the plates is introduced so as to fill one-fourth of the capacitor as shown in figure. The new capacitance will be

Each capacitor shown in is$2\mu \mathrm{F}$. Then the equivalent capacitance between points $A$ and $B$ is

The work done is placing a charge of $8\times {10}^{-18}$ coulomb on a condenser of capacity $100\mu \mathrm{F}$ is