An isolated `16 muF` parallel plate air capacitor has a potential difference of 100 V. dielectric slab having relative permittivity (i.e., dielectric constant) = 5 is introduced to fill the space between the two plates completely. Calculate.

(i) the new capacitance of the capacitor
(ii) the new potential differece between the two plates of capacitor.

A dielectric slab of relative premittivity (i.e. dielectric constant) 6 is introduced between the two plates of an 8 mu F air capacitor, in order to completely occupy the space between the two plates. Find the new capacitance of the capacitor.

The force between the plates of a parallel plate capacitor of capacitance C and distance of separation of the plates d with a potential difference V between the plates, is.

A parallel plate capacitor is charged by a battery, which is then disconnected. A dielectric slab is now introduced between the two plates to occupy the space completely. State the effect on the following: (i) the capacitance of the capacitor. (ii) potential difference between the plates. (iii) the energy stored in the capacitor.

In a parallel plate capacitor of capacitor 1muF two plates are given charges 2muC and 4muC the potential difference between the plates of the capacitor is

A paller- plate capacitor has plate area 100cm^2 and plate separation 1.0cm.A glass plate (dielectric constant 4.0)are inserted one over the other the fill the space between the plate of the capacitor.Find the new capacitance.

If a thin metal foil of same area is placed between the two plates of a parallel plate capacitor of capacitance C, then new capacitance will be

Figure shows two identical capacitors C_1 and C_2 each of 2muF capacitance, connected to a battery of 5 V. Initially switch 'S' is closed. After some time 'S' is left open and dielectric slavs of dielectric constant K = 5 are inserted to fill completely the space between the plates of the two capacitors. How will the (i) charge and (ii) potential difference between the plates of the capacitors be affected after the slabs are inserted?

The plates of a parallel plate capacitor with no dielectirc are connected to a volatage source. Now a dielectric of dielectric constant K is inserted to fill the whose space between the plates with voltage source remainign connected to the capacitor-

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.

A charge of 20 microCoulomb is placed on the positive plate of on isolated parallel - plate capacitor of capacitance 10 microFarad calculate the potential difference developed between the plates .