If a capacitor of capacity $$900 ~\mu \mathrm{F}$$ is charged to $$100 \mathrm{~V}$$ and its total energy is transferred to a capacitor of capacity $$100 ~\mu \mathrm{F}$$, then its potential will be

Two dielectric slabs having dielectric constant '$$\mathrm{K}_1$$' and '$$\mathrm{K}_2$$' of thickness $$\frac{\mathrm{d}}{4}$$ and $$\frac{3 \mathrm{~d}}{4}$$ are inserted between the plates as shown in figure. The net capacitance between $$A$$ and $$B$$ is $$\left[\varepsilon_0\right.$$ is permittivity of free space]

A parallel plate capacitor has plate area '$$\mathrm{A}$$' and separation between plates is '$$d$$'. It is charged to a potential difference of $$\mathrm{V}_0$$ volt. The charging battery is then disconnected and plates are pulled apart three times the initial distance. The work done to increase the distance between the plates is $$\left(\varepsilon_0=\right.$$ permittivity of free space)

Two capacitors $$\mathrm{C}_1=3 \mu \mathrm{F}$$ and $$\mathrm{C}_2=2 \mu \mathrm{F}$$ are connected in series across d.c. source of $$100 \mathrm{~V}$$. The ratio of the potential across $$C_2$$ to $$C_1$$ is