A frictionless dielectric plate $S$ is kept on a frictionless table $T$. A charged parallel plate capacitance $C$ (of which the plates are frictionless) is kept near it. The plate $S$ is between the plates. When the plate $S$ is left between the plates
It will remain stationary on the table
It is pulled by the capacitor and will pass on the other end
It is pulled between the plates and will remain there
All the above statements are false
A parallel plate capacitor has plate of length $'l',$ width $'w'$ and separation of plates is $'d'.$ It is connected to a battery of emf $V$. A dielectric slab of the same thickness '$d$' and of dielectric constant $k =4$ is being inserted between the plates of the capacitor. At what length of the slab inside plates, will be energy stored in the capacitor be two times the initial energy stored$?$
While a capacitor remains connected to a battery and dielectric slab is applied between the plates, then
The outer sphere of a spherical air capacitor is earthed. For increasing its capacitance
Between the plates of a parallel plate capacitor a dielectric plate is introduced just to fill the space between the plates. The capacitor is charged and later disconnected from the battery. The dielectric plate is slowly drawn out of the capacitor parallel to the plates. The plot of the potential difference across the plates and the length of the dielectric plate drawn out is
A parallel plate capacitor of area ' $A$ ' plate separation ' $d$ ' is filled with two dielectrics as shown. What is the capacitance of the arrangement?