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:

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:

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 has a capacitance of $5 \ \mu F$. When a glass plate is inserted between the plates of the capacitor,its potential becomes $1/8$ of its original value. What is the value of the dielectric constant?

The dielectric slab is introduced between the plates of a parallel plate charged capacitor. Which one of the following quantities will not change?

$A$ parallel plate capacitor having capacitance $12 \, pF$ is charged by a battery to a potential difference of $10 \, V$ between its plates. The charging battery is now disconnected and a porcelain slab of dielectric constant $6.5$ is slipped between the plates. The work done by the capacitor on the slab is.......$pJ$