The capacity of an air condenser is $2.0 \, \mu F$. If a medium is placed between its plates,the capacity becomes $12 \, \mu F$. The dielectric constant of the medium will be:

$A$ parallel-plate capacitor of capacitance $40 \mu F$ is connected to a $100 V$ power supply. Now,the intermediate space between the plates is filled with a dielectric material of dielectric constant $K=2$. Due to the introduction of the dielectric material,the extra charge and the change in the electrostatic energy in the capacitor,respectively,are:

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.

$A$ parallel plate capacitor has a capacitance of $2\ \mu F$ with a separation of $0.4\ cm$ between the plates. If the distance between the plates is halved and the space is filled with a dielectric material of constant $K = 2.8$,the final capacitance of the capacitor will be .....$\mu F$.

$A$ slab of dielectric constant $K$ has the same cross-sectional area as the plates of a parallel plate capacitor and thickness $\frac{3}{4}d$,where $d$ is the separation of the plates. The capacitance of the capacitor when the slab is inserted between the plates will be (Given $C_{0} =$ capacitance of capacitor with air as medium between plates):

When the battery across the plates of a charged condenser is disconnected and a dielectric slab is introduced between its plates,then the energy stored: