Two capacitors $C$ and $2C$ are connected in parallel and charged to a potential $V$ by a battery. The battery is then removed. The capacitor $C$ is filled with a dielectric of constant $K$. What is the new potential difference across the capacitors?

$A$ parallel plate capacitor of capacitance $5\,\mu F$ and plate separation $6\, cm$ is connected to a $1\, V$ battery and charged. $A$ dielectric of dielectric constant $4$ and thickness $4\, cm$ is introduced between the plates of the capacitor. The additional charge that flows into the capacitor from the battery is........$\mu C$

An air capacitor has a capacitance of $1 \mu F$. Now,the space between the two plates of the capacitor is filled with two dielectrics as shown in the figure. The capacitance of the capacitor is ($d=$ distance between two plates,$K_1=8$ and $K_2=4$ are the dielectric constants of the two dielectrics respectively).

$A$ parallel plate capacitor has a capacitance of $10 \ \mu F$ with air between its plates. Now,half of the space between the two plates is filled with a dielectric material of dielectric constant $K = 4$ as shown in the figure. Find the capacitance of the capacitor in $\mu F$.

$A$ parallel plate air capacitor has capacitance $C_p$. It is equally filled with parallel layers of materials of dielectric constants $K_1$ and $K_2$. Now its capacity becomes $C_K$. The ratio $C_P$ to $C_K$ is

$A$ parallel plate air capacitor, with plate separation $d$, has a capacitance of $9 \text{ pF}$. The space between the plates is now filled with two dielectrics, the first having $K_1=3$ and thickness $d_1=d/3$, while the second has $K_2=6$ and thickness $d_2=2d/3$. The capacitance of the new capacitor is: (in $\text{ pF}$)