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:

$A$ parallel plate air capacitor has a capacitance $C$. When it is half filled as shown in the figure with a dielectric constant $K = 5$, the percentage increase in the capacitance is . . . . . . .

In a parallel plate capacitor,two dielectric slabs of thicknesses $t_1$ and $t_2$ and dielectric constants $K_1$ and $K_2$ respectively are inserted. What is the capacitance of this capacitor?

$A$ capacitor of $10 \mu F$ capacitance,whose plates are separated by $10 \text{ mm}$ through air and each plate has an area of $4 \text{ cm}^2$,is now filled equally with two dielectric media of $K_1=2$ and $K_2=3$ respectively,as shown in the figure. If the new force between the plates is $8 \text{ N}$,the supply voltage is . . . . . . $V$.

If the distance between the plates of a parallel plate capacitor is halved and the dielectric constant of the dielectric is doubled,then its capacity will

$A$ parallel plate capacitor with air between the plates has a capacitance of $4 \ pF$. If the distance between the plates is reduced by half and the space between them is filled with a substance of dielectric constant $K = 6$,then the value of capacitance will be . . . . . . . (in $pF$)