$A$ parallel plate air-core capacitor is connected across a source of constant potential difference. When a dielectric plate is introduced between the two plates,then:

$A$ parallel plate capacitor of plate area $A$ and plate separation $d$ is charged to a potential difference $V$ and then the battery is disconnected. $A$ slab of dielectric constant $K$ is then inserted between the plates of the capacitor so as to fill the space between the plates. If $Q$,$E$,and $W$ denote respectively the magnitude of charge on each plate,the electric field between the plates (after the slab is inserted),and the work done on the system in the process of inserting the slab,then:

The capacity of a parallel plate capacitor is $10\,\mu F$ without a dielectric. If a dielectric of constant $K = 2$ is used to fill half the distance between the plates,the new capacitance in $\mu F$ is:

$A$ parallel plate capacitor with air between the plates has a capacitance of $15 \text{ pF}$. 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 $x/4 \text{ pF}$. The value of $x$ 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 the figure. The new capacitance will be

$A$ parallel plate capacitor with plate area $A$ and plate separation $d$ is filled with a dielectric material of dielectric constant $K = 4$. The thickness of the dielectric material is $x$,where $x < d$.
Let $C_1$ and $C_2$ be the capacitance of the system for $x = \frac{d}{3}$ and $x = \frac{2d}{3}$,respectively. If $C_1 = 2 \mu F$,the value of $C_2$ is $........... \mu F$.