The capacitance of an air-filled parallel plate capacitor is $9 \ pF$. If the space between the plates is filled with two dielectric slabs of thickness $d/3$ with dielectric constant $K_1 = 3$ and thickness $2d/3$ with dielectric constant $K_2 = 6$,the new capacitance will be ...... $pF$.

$A$ parallel plate capacitor is first charged and then a dielectric slab is introduced between the plates. The quantity that remains unchanged is

$A$ capacitor is charged by using a battery which is then disconnected. $A$ dielectric slab is then inserted between the plates. What is the result?

The expression for the capacity of the capacitor formed by compound dielectric placed between the plates of a parallel plate capacitor as shown in the figure,will be (area of plate $= A$)

The capacitance of an air-filled parallel plate capacitor is $10 \, pF$. The separation between the plates is doubled and the space between the plates is then filled with wax,giving the capacitance a new value of $40 \times 10^{-12} \, F$. The dielectric constant of wax is:

Two identical parallel plate capacitors of capacitance $C$ each are connected in series with a battery of emf $E$ as shown below. If one of the capacitors is now filled with a dielectric of dielectric constant $k$,then the amount of charge which will flow through the battery is (neglect internal resistance of the battery).