A parallel plate capacitor `(C=50 muF. d=4mm)` is charged to `200V` and then charging battery is removed. Now a dielectric slab `(K=4)` of thickness `2mm` is placed between the plates. Find new potential difference across capacitor.

After charging a capacitor the battery is removed. Now by placing a dielectric slab between the plates

A capacitor with plate separation d is charged to V volts. The battry is disconnected and a dielectric slab of thickness (d)/(2) and dlelectric constant '2' is inserted between the plates. The potential difference across its terminals becomes

Two identical parallel plate capacitors are connected in series to a battery of 100V . A dielectric slab of dielectric constant 4.0 is inserted between the plates of second capacitor. The potential difference across the capacitors will now be respectively

The space between plates of a parallel plate capacitor is filled by a dielectric and it is charged and then battery is removed. Now dielectric slab is slowly drawn out of the capacitor parallel to the plates. The variation of the potential of capacitor with respect to the length of the dielectric plate drawn out is

A capacitor is charged by a cell of emf E the charging battery is then removed. If an identical capacitor is now inserted in the circuit in parallel with the previous capacitor, the potential difference across the new capacitor is

The plates of a parallel plate capacitor are charged up to 200 V. A dielectric slab of thickness 4 mm is inserted between its plates. Then, to maintain the same potential difference between the plates of the capacitor, the distance between the plates increased by 3.2 mm. The dielectric constant of the dielectric slab is

Separation between the plates of parallel plate capacitor is 5 mm. this capacitor, having air as the dielectric medium between the plates, is charged to a potential difference 25 V using a battery. The battery is then disconnected and a dielectric slab of thickness 3mm and dielectric constant K=10 is placed between the plates as shown. potential difference between the plates after the dielectric slab has been introduced is-

The parallel combination of two air filled parallel plate capacitor of capacitance C and nC is connected to a battery of voltage, V. When the capacitors are fulley charged, the battery is removed and after that a dielectric material of dielectric constant K is placed between the two plates of the first capacitors. the new potential difference of the combined system is

Two identical parallel plate capacitors are connected in series and then joined in series with a battery of 100 V . A slab of dielectric constant K=3 is inserted between the plates of the first capacitor. Then, the potential difference across the capacitor will be, respectively.

A parallel plate capacitor is charged by a battery. After sometime the battery is disconnected and a dielectric slab with its thickness equal to the plate separation is inserted between the plates. How will (i) the capacitance of the capacitor, (ii) potential difference between the plates and (iii) the energy stored in the capacitor be affected ? Justify your answer in each case.