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 $\mu F$)

Two identical parallel plate air capacitors are connected in series to a battery of emf $V$. If one of the capacitors is inserted in a liquid of dielectric constant $K$,then the potential difference across the other capacitor will become:

$A$ parallel plate capacitor has a capacity of $80 \times 10^{-6} \ F$ when air is present between the plates. The volume between the plates is then completely filled with a dielectric slab of dielectric constant $K = 20$. The capacitor is connected to a battery of $30 \ V$. The dielectric slab is then removed while the capacitor remains connected to the battery. Calculate the charge that passes through the wire.

$A$ parallel plate air capacitor has capacity $C$ farad,potential $V$ volt,and energy $E$ joule. When the gap between the plates is completely filled with a dielectric material of dielectric constant $K > 1$,what happens to the potential $V$ and energy $E$?

$A$ parallel plate capacitor having a separation between the plates $d$,plate area $A$,and material with dielectric constant $K$ has capacitance $C_0$. Now,one-third of the material is replaced by another material with dielectric constant $2K$,such that effectively there are two capacitors: one with area $\frac{1}{3}A$,dielectric constant $2K$,and another with area $\frac{2}{3}A$ and dielectric constant $K$. If the capacitance of this new capacitor is $C$,then $\frac{C}{C_0}$ is:

$A$ parallel plate capacitor has capacitance $C$, when there is vacuum within the parallel plates. $A$ sheet having thickness $t = d/3$ (where $d$ is the separation between the plates) and relative permittivity $K$ is introduced between the plates. The new capacitance of the system is: