In a parallel plate air capacitor, the distance between plates is reduced to one-fourth and the space between them is filled with a dielectric medium of constant $2$. If the initial capacity of the capacitor is $4 \mu F$, then its new capacity is: (in $\mu F$)

$A$ parallel plate capacitor is connected to the terminals of a battery. The distance between the plates is $6\,mm$. If a glass plate (dielectric constant $K = 9$) of thickness $4.5\,mm$ is introduced between them,then the capacity will become how many times the original capacity?

In a parallel plate capacitor,the separation between plates is $3x$. This separation is filled by two layers of dielectrics,in which one layer has thickness $x$ and dielectric constant $3k$,and the other layer has thickness $2x$ and dielectric constant $5k$. If the plates of the capacitor are connected to a battery,then the ratio of the potential difference across the dielectric layers is

Three parallel plate capacitors $C_1, C_2$ and $C_3$ each of capacitance $5 \mu F$ are connected as shown in the figure. The effective capacitance between points $A$ and $B$,when the space between the parallel plates of $C_1$ capacitor is filled with a dielectric medium having a dielectric constant of $4$,is (in $\mu F$)

$A$ parallel plate air capacitor has a capacitance $C$. When it is half filled with a dielectric of dielectric constant $5$,the percentage increase in the capacitance will be.....$\%$

Assertion : $A$ parallel plate capacitor is connected across a battery through a key. $A$ dielectric slab of dielectric constant $K$ is introduced between the plates. The energy stored becomes $K$ times.
Reason : The surface density of charge on the plate remains constant or unchanged.