A parallel plate air capacitor of plate separation d is charged to a potential difference `triangle V `. A slab of thickness d and dielectric constant k is introduced between the plates while the battery remains connected.
What happens to the charge on the capacitor?

A parallel plate air capacitor of plate separation d is charged to a potential difference triangle V . A slab of thickness d and dielectric constant k is introduced between the plates while the battery remains connected. Find the ratio of energy stored in the capacitor after and before the introduction of the dielectric slab.

A parallel plate capacitor of plate area A and plate separation d is charged to a potential difference V and then the battery is disconnected. A slab of dielectric constant k is then inserted between the plates of the capacitor so as to fill the space between the plates. If Q, E and W denote respectively the charge on each plate, the electric field between the plates and the work done in the process of inserting the slab, then

A parallel plate capacitor has a capacitance of 100muF . The plates are at a distance d apart. A slab of thickness t(t = d) and dielectric constant 5 is introduced between the parallel plates. Then capacitance can be,

A parallel plate capacitor has plates A and B each of area 0.5m^2 and a plate separation of 0.2 m. A dielectric slab of thickness 0.1m and dielectric constant 2.5 is placed between the two plates as shown - (a) FInd the capacitance of the capacitor, if the surface charge densities of the two plates are 8.854xx10^(-10)Cm^(-2) . (b) Graphically represent the variation of electric field between the two plates from A to B.

A parallel plate capacitor of capacitance 90 pF is connected to a battery of emf 20V. If a dielectric material of dielectric constant k=(5)/(3) is inserted between the plates, the magnitude of the induced charge will be

A parallel plate air capacitor has a capacitance C and the distance between the plates is d.A metallic plate of thickness d/3 is introduced between the plates of the capacitor. The new capacitance will be—

A parallel plate air capacitor has capacity 'C' distance of separation between plates is ‘d’ and potential difference 'V' is applied between the plates. Force of attraction between the plates of the parallel plate air capacitor is

Two parallel plate capacitors of capacity C and 2C are connected in parallel and charged to a potential difference V. The battery is then disconnected and the space between the plates of the capacitor C is completely filled with a material of dielectic constant K. What is the final potential difference across the capacitors?

The conducting plates of a parallel plate capacitor are separated by 2 cm from each other. A dielectric slab (k =5) of thickness 1 cm is inserted between the two plates. The distance between the plates is now so changed that the capacitance of the capacitor remains the same. What will be the new distance between the plates?