Two capacitors,each having capacitance $40\,\mu F$,are connected in series. The space between the plates of one of the capacitors is filled with a dielectric material of dielectric constant $K$ such that the equivalent capacitance of the system becomes $24\,\mu F$. The value of $K$ is:

The plates of a parallel plate capacitor are charged up to $200 \ V$. $A$ dielectric slab of thickness $4 \ mm$ is inserted between its plates. Then,to maintain the same potential difference between the plates of the capacitor,the distance between the plates is increased by $3.2 \ mm$. The dielectric constant of the dielectric slab is

$A$ capacitor stores $60\,\mu C$ charge when connected across a battery. When the gap between the plates is filled with a dielectric,an additional charge of $120\,\mu C$ flows through the battery. The dielectric constant of the dielectric inserted is

$A$ capacitor is connected to a battery of voltage $V$. If a dielectric slab of dielectric constant $k$ is completely inserted between the plates,what will be the final charge on the capacitor? (Assume the initial charge is $q_{0}$)

$A$ parallel plate capacitor with plate area $A$ and distance of separation $d$ is filled with a dielectric. What is the capacity of the capacitor when permittivity of the dielectric varies as :
$\varepsilon(x) = \varepsilon_{0} + kx$,for $(0 < x \leq d/2)$
$\varepsilon(x) = \varepsilon_{0} + k(d-x)$,for $(d/2 \leq x \leq d)$

$A$ parallel plate capacitor is filled with two dielectrics as shown in the figure. The area of each plate is $A \; m^2$ and the separation between the plates is $t \; m$. The dielectric constants are $k_1$ and $k_2$ respectively. Its capacitance in farad will be: