If the distance between parallel plates of a capacitor is halved and the dielectric constant is doubled,then the capacitance will become:

When a parallel plate capacitor is charged up to $95 \ V$,its capacitance is $C$. If a dielectric slab of thickness $2 \ mm$ is inserted between the plates and the distance between the plates is increased by $1.6 \ mm$ such that the same potential difference is maintained,the dielectric constant of the material (slab) is:

In a parallel plate air capacitor of plate separation $d$,a dielectric slab of thickness $t$ is introduced between the plates $(t < d)$. The capacitance becomes one-third of the original value. The dielectric constant of the slab will be

$A$ parallel plate capacitor has a capacitance $C$ and the distance between the plates is $d$. If the space between the plates is filled with a dielectric material of dielectric constant $K$ as shown in the figure,find the new capacitance of the capacitor.

$A$ capacitor is fully charged with a battery and then disconnected. $A$ dielectric is then inserted into the capacitor. How do the charges on the surface of the dielectric and the outer surface of the plates of the capacitor change, respectively?

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 $\mu F$)