Two identical condensers $M$ and $N$ are connected in series with a battery. The space between the plates of $M$ is completely filled with a dielectric medium of dielectric constant $8$,and a copper plate of thickness $d/2$ is introduced between the plates of $N$ ($d$ is the distance between the plates). Then the potential differences across $M$ and $N$ are,respectively,in the ratio:

$A$ capacitor has a capacitance of $15\ \mu F$ and a plate separation of $2\ mm$. If a dielectric slab of dielectric constant $K = 2$ and thickness $t = 1\ mm$ is inserted between the plates,what will be the new capacitance (in $\mu F$)?

The dielectric slab is introduced between the plates of a parallel plate charged capacitor. Which one of the following quantities will not change?

$A$ copper slab of thickness $d/2$ is inserted between the plates of a parallel plate capacitor,where $d$ is the distance between its two plates. If the capacitance of the capacitor without the copper slab is $C$ and with the copper slab is $C'$,find the ratio $C'/C$.

Assertion : If the distance between parallel plates of a capacitor is halved and the dielectric constant is increased to three times its original value,then the capacitance becomes $6$ times.
Reason : The capacity of a capacitor does not depend upon the nature of the material between the plates.

$A$ capacitor has air as dielectric medium and two conducting plates of area $12 \,cm^2$ and they are $0.6 \,cm$ apart. When a slab of dielectric having area $12 \,cm^2$ and $0.6 \,cm$ thickness is inserted between the plates, one of the conducting plates has to be moved by $0.2 \,cm$ to keep the capacitance same as in previous case. The dielectric constant of the slab is : (Given $\epsilon_0 = 8.834 \times 10^{-12} \,F/m$)