If a dielectric is inserted into a charged capacitor (battery removed),then the quantity that remains constant is

The distance between the plates of a parallel plate capacitor is $d$. $A$ metal plate of thickness $d/2$ is introduced between the plates. The capacitance will then be

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?

$A$ parallel-plate capacitor with plate area $A$ has a separation $d$ between the plates. Two dielectric slabs of dielectric constants $K_{1}$ and $K_{2}$,each having an area of $A/2$ and thickness $d/2$,are inserted in the space between the plates as shown in the figure. The equivalent capacitance of the capacitor will be:

The space between the plates of a parallel plate capacitor is filled with a mica sheet of thickness $1 \times 10^{-3} \,m$ and a fiber sheet of thickness $0.5 \times 10^{-3} \,m$. The dielectric constants of mica and fiber are $8$ and $2.5$ respectively. If the fiber breaks down at an electric field of $6.4 \times 10^6 \,V/m$, then the maximum voltage that can be applied to the capacitor is: (in $\,V$)

$A$ 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: