What will be the capacity of a parallel-plate capacitor when the half of parallel space between the plates is filled by a material of dielectric constant ${\varepsilon _r}$ ? Assume that the capacity of the capacitor in air is $C$

A parallel plate capacitor of capacitance $5\,\mu F$ and plate separation $6\, cm$ is connected to a $1\, V$ battery and charged. A dielectric of dielectric constant $4$ and thickness $4\, cm$ is introduced between the plates of the capacitor. The additional charge that flows into the capacitor from the battery is........$\mu C$

What is dielectric ?

The capacitance of an air filled parallel plate capacitor is $10\,p F$. The separation between the plates is doubled and the space between the plates is then filled with wax giving the capacitance a new value of $40 \times {10^{ - 12}}farads$. The dielectric constant of wax is

Two capacitors, each having capacitance $40\,\mu F$ are connected in series. The space between one of the capacitors is filled with dielectric material of dielectric constant $K$ such that the equivalence capacitance of the system became $24\,\mu F$. The value of $K$ will be.

The distance between the plates of a parallel plate capacitor is $d$. A metal plate of thickness $d/2$ is placed between the plates. The capacitance would then be