Consider a parallel plate capacitor of $10\,\mu F$ with air filled in the gap between the plates. Now,one half of the space between the plates is filled with a dielectric of dielectric constant $K = 4$,as shown in the figure. The capacity of the capacitor changes to.......$\mu F$.

While a capacitor remains connected to a battery and a dielectric slab is inserted between the plates,then:

Two capacitors $C$ and $2C$ are connected in parallel and charged to a potential $V$ by a battery. The battery is then removed. The capacitor $C$ is filled with a dielectric of constant $K$. What is the new potential difference across the capacitors?

$A$ voltmeter reads $4 \ V$ when connected to a parallel plate capacitor with air as a dielectric. When a dielectric slab is introduced between the plates for the same configuration,the voltmeter reads $2 \ V$. What is the dielectric constant of the material?

$A$ parallel plate capacitor is made of two circular plates separated by a distance $5 \ mm$ and with a dielectric of dielectric constant $2.2$ between them. When the electric field in the dielectric is $3 \times 10^4 \ Vm^{-1}$,the charge density of the positive plate will be close to:

$A$ 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 $K_m = K(1 + \frac{m}{N})$. For a very large $N (> 10^3)$,the capacitance $C$ is $\alpha \left( \frac{K \varepsilon_0 A}{d \ln 2} \right)$. The value of $\alpha$ will be. . . . . . . .
[$\varepsilon_0$ is the permittivity of free space]