In the given figure,there are two capacitors in series. $A$ rigid conducting slab of thickness $b$ can slide vertically. Find the equivalent capacitance of the system.

$A$ parallel plate capacitor is charged to a potential difference of $100\,V$ and disconnected from the source of $emf$. $A$ slab of dielectric is then inserted between the plates. Which of the following three quantities change?
$(i)$ The potential difference
$(ii)$ The capacitance
$(iii)$ The charge on the plates

$A$ combination of parallel plate capacitors is maintained at a certain potential difference. When a $3 \, mm$ thick slab is introduced between all the plates,in order to maintain the same potential difference,the distance between the plates is increased by $2.4 \, mm$. Find the dielectric constant of the slab.

Half of the space between the plates of a parallel plate capacitor is filled with a medium of dielectric constant $K$ parallel to the plates. If the initial capacitance is $C$,then the new capacitance will be:

Two square-shaped metal plates of side $1 \,m$, kept $0.01 \,m$ apart in air, form a parallel plate capacitor. It is connected to a battery of $500 \,V$. The plates of the capacitor are then immersed in an insulating oil by lowering the plates vertically with a speed of $0.001 \,m/s$. If the dielectric constant of the oil is $11$, then the current drawn from the battery during this process is:

Two parallel plate capacitors are connected in series and then connected to a $100 \ V$ battery. $A$ dielectric slab of dielectric constant $K = 4.0$ is inserted between the plates of the second capacitor. What will be the potential difference across each capacitor respectively?