An air-filled parallel plate capacitor has a capacity of $2 \ pF$. If the separation between the plates is doubled and the interspace between the plates is filled with a dielectric material,the capacity increases to $6 \ pF$. The dielectric constant of the material is:

$A$ parallel plate capacitor of area $A$ and plate separation $d$ is filled with two dielectrics as shown. What is the equivalent capacitance of the arrangement?

Initially,the circuit is in a steady state. Now,one of the capacitors is filled with a dielectric of dielectric constant $K = 2$. Find the heat loss in the circuit due to the insertion of the dielectric.

$A$ parallel plate capacitor with air as the dielectric has capacitance $C$. $A$ slab of dielectric constant $K$ and having the same thickness as the separation between the plates is introduced so as to fill one-fourth of the capacitor as shown in the figure. The new capacitance will be

$A$ parallel plate capacitor is charged and then isolated. If the separation between the plates is increased,which one of the following statements is $NOT$ correct?

$A$ slab of material of dielectric constant $K$ has the same area as the plates of a parallel-plate capacitor but has a thickness $(3/4)d$,where $d$ is the separation of the plates. How is the capacitance changed when the slab is inserted between the plates?