$A$ thin metal plate $P$ is inserted halfway between the plates of a parallel plate capacitor of capacitance $C$ in such a way that it is parallel to the two plates. The capacitance now becomes

Between the plates of a parallel plate capacitor, a dielectric plate is introduced to completely fill the space between the plates. The capacitor is charged and then disconnected from the battery. The dielectric plate is slowly drawn out of the capacitor parallel to the plates. The plot of the potential difference $V$ across the plates versus the length $x$ of the dielectric plate drawn out is:

$A$ parallel plate capacitor is charged by connecting a $2 \ V$ battery across it. It is then disconnected from the battery and a glass slab is introduced between the plates. Which of the following pairs of quantities decrease?

$A$ parallel-plate capacitor of area $A,$ plate separation $d$ and capacitance $C$ is filled with four dielectric materials having dielectric constants $K_1, K_2, K_3$ and $K_4$ as shown in the figure. If a single dielectric material is to be used to have the same capacitance $C$ in this capacitor,then its dielectric constant $K$ is given by

$A$ parallel plate capacitor with air medium between the plates has a capacitance of $10 \mu F$. The area of the capacitor is divided into two equal halves and filled with two media (as shown in the figure) having dielectric constants $K_1=2$ and $K_2=4$. The capacitance of the system will be (in $\mu F$)

The capacitance of a parallel plate capacitor with air as the medium is $3 \mu F$. With the introduction of a dielectric medium between the plates,the capacitance becomes $15 \mu F$. The permittivity of the medium in $SI$ units is: [Given: $\epsilon_{0} = 8.85 \times 10^{-12} \text{ SI units}$]