In a parallel plate capacitor, the radius of each circular plate is $12\,cm$ and the distance between the plates is $5\,mm$. A glass slab of $3\,mm$ thickness and radius $12\,cm$ with a dielectric constant of $6$ is placed between the plates. The capacitance of the capacitor will be:

An air capacitor has capacitance $C_1$. The space between the two plates of the capacitor is filled with two dielectrics as shown in the figure. The new capacitance of the capacitor is $C_2$. The ratio $\frac{C_1}{C_2}$ is ($d=$ distance between two plates of the capacitor,$K_1$ and $K_2$ are dielectric constants of the two dielectrics respectively).

$A$ capacitor is charged and then the battery is disconnected. If a dielectric slab is inserted between the plates,choose the correct statement.

$A$ parallel plate capacitor has a dielectric slab of dielectric constant $K$ between its plates that covers $1/3$ of the area of its plates,as shown in the figure. The total capacitance of the capacitor is $C$ while that of the portion with dielectric in between is $C_1$. When the capacitor is charged,the plate area covered by the dielectric gets charge $Q_1$ and the rest of the area gets charge $Q_2$. Choose the correct option/options,ignoring edge effects.
$(A)$ $\frac{E_1}{E_2}=1$ $(B)$ $\frac{E_1}{E_2}=\frac{1}{K}$ $(C)$ $\frac{Q_1}{Q_2}=\frac{K}{2}$ $(D)$ $\frac{C}{C_1}=\frac{2+K}{K}$

$A$ parallel plate capacitor has a plate separation of $0.01\, mm$ and uses a dielectric (whose dielectric strength is $19\, kV/mm$) as an insulator. The maximum potential difference that can be applied to the terminals of the capacitor is......$V$

$A$ parallel plate capacitor with air between the plates has a capacitance of $15 \, pF$. The separation between the plates is doubled and the space between them is filled with a medium of dielectric constant $3.5$. Then the capacitance becomes $\frac{x}{4} \, pF$. The value of $x$ is $............$