Two capacitors of capacitance $2C$ and $C$ are joined in parallel and charged to potential $V$. The battery is now removed and the capacitor $C$ is filled with a medium of dielectric constant $K$. The potential difference across each capacitor will be

$A$ parallel plate capacitor with air between the plates has a capacitance of $9 \ pF$. The separation between its plates is $d$. The space between the plates is now filled with two dielectrics. One of the dielectrics has a dielectric constant $k_1 = 3$ and thickness $d/3$,while the other one has a dielectric constant $k_2 = 6$ and thickness $2d/3$. The capacitance of the capacitor is now . . . . . . $pF$.

The separation between the plates of a parallel plate capacitor is $d$ and the area of each plate is $A$. When a slab of material of dielectric constant $k$ and thickness $t$ $(t < d)$ is introduced between the plates,its capacitance becomes

$A$ 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 capacitor is made of two circular plates separated by a distance $5 \ mm$ and with a dielectric of dielectric constant $2.2$ between them. When the electric field in the dielectric is $3 \times 10^4 \ Vm^{-1}$,the charge density of the positive plate will be close to:

The capacity of an air condenser is $2.0 \, \mu F$. If a medium is placed between its plates,the capacity becomes $12 \, \mu F$. The dielectric constant of the medium will be: