The capacitance of an air capacitor is $15\,\mu F$ the separation between the parallel plates is $6\,mm$. A copper plate of $3\,mm$ thickness is introduced symmetrically between the plates. The capacitance now becomes………$\mu F$

A parallel plate capacitor of capacitance $C$ has spacing $d$ between two plates having area $A$. The region between the plates is filled with $N$ dielectric layers, parallel to its plates, each with thickness $\delta=\frac{ d }{ N }$. The dielectric constant of the $m ^{\text {th }}$ layer is $K _{ m }= K \left(1+\frac{ m }{ N }\right)$. For a very large $N \left(>10^3\right)$, the capacitance $C$ is $\alpha\left(\frac{ K \varepsilon_0 A }{ d \;ln 2}\right)$. The value of $\alpha$ will be. . . . . . . .

[ $\epsilon_0$ is the permittivity of free space]

The plates of a parallel plate capacitor with no dielectric are connected to a voltage source. Now a dielectric of dielectric constant $K $ is inserted to fill the whole space between the plates with voltage source remaining connected to the capacitor.

Two dielectric slab of dielectric constant $K_1$ and $K_2$ and of same thickness is inserted in parallel plats capacitor and $K_1 = 2K_2$ . Potential difference across slabs are $V_1$ and $V_2$ respectively then

With the rise in temperature, the dielectric constant $K$ of a liquid