Two parallel plate air capacitors of the same capacity $C$ are connected in series to a battery of emf $E$. Then one of the capacitors is completely filled with a dielectric material of constant $K$. The change in the effective capacitance of the series combination 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

Two identical parallel plate capacitors are placed in series and connected to a constant voltage source of $V_0 \, V$. If one of the capacitors is completely immersed in a liquid with dielectric constant $K$,the potential difference between the plates of the other capacitor will change to:

In a parallel plate air capacitor, the distance between plates is reduced to one-fourth and the space between them is filled with a dielectric medium of constant $2$. If the initial capacity of the capacitor is $4 \mu F$, then its new capacity is: (in $\mu F$)

The distance between the plates of a parallel plate capacitor is $8\,mm$ and the potential difference $(P.D.)$ is $120\,V$. If a $6\,mm$ thick slab of dielectric constant $K = 6$ is introduced between its plates,then:

$A$ parallel plate capacitor with width $4\,cm$,length $8\,cm$ and separation between the plates of $4\,mm$ is connected to a battery of $20\,V$. $A$ dielectric slab of dielectric constant $5$ having length $1\,cm$,width $4\,cm$ and thickness $4\,mm$ is inserted between the plates of the parallel plate capacitor. The electrostatic energy of this system will be ......... $\epsilon_{0}\,J$. (Where $\epsilon_{0}$ is the permittivity of free space)