$A$ parallel plate capacitor filled with a medium of dielectric constant $10$ is connected across a battery and is charged. The dielectric slab is replaced by another slab of dielectric constant $15$. Then the energy of the capacitor will ......................

The separation between the plates of a parallel plate capacitor is $d$ and the area of each plate is $A$. When a slab of material of dielectric constant $k$ and thickness $t$ $(t < d)$ is introduced between the plates,its capacitance becomes

$A$ parallel plate air capacitor is charged and then isolated. When a dielectric material is inserted between the plates of the capacitor,which of the following does not change?

An air capacitor has a capacitance of $1 \mu F$. Now the space between the two plates of the capacitor is filled with two dielectrics as shown in the figure. The capacitance of the capacitor is: [$d=$ distance between two plates of the capacitor,$K_1$ and $K_2$ are dielectric constants of the first and second dielectric respectively] (in $\mu F$)

$A$ capacitor with air as the dielectric is charged to a potential of $100\;V$. If the space between the plates is now filled with a dielectric of dielectric constant $K = 10$,the potential difference between the plates will be: (in $;V$)

When a dielectric slab is introduced between the plates of a parallel plate capacitor that is connected to a battery,the new charge on the plates is: