Effective Capacitance on insertion of dielectrics of different dielectric constant

calculate the effective capacitance of the connection if a dielectric material of dielectric constant K is filled in between the plates as shown in the adjoining figure (n).

The parallel plate of a capacitor have an area 0.2 m^(2) and are 10^(-2) m apart. The original potential difference between them is 3000V, and it decreases to 1000 when a sheet of dielectric is inserted between the plates. Compute (a) Original capacitance C_(0) (b) The original charge Q on each plate . (C) Capacitance C after insertion of the dielectric . (d) Dielectric constant K . (e) The original field E_(0) between the plates and . (f) The electric field E_(0) between the plates and . (f) The electric field E after insertion of the dielectric . (epsi_(0) = 8.85 xx 10^(-12) S.I unit).

Insertion Of Dielectric Between Plates

A parallel plate capacitor of area A , plate separation d and capacitance C is filled with three different dielectric materials having dielectric constant K_(1),K_(2) and K_(3) as shown in fig. If a single dielectric material is to be used to have the same effective capacitance as the above combination then its dielectric constant K is given by :

Effect Of Dielectric On Different Parameter

A capacitor of capacitance C is charged to potential difference V and then disconnected from the battery. The air dielectric of capacitor is replaced by another dielectric of dielectric constant K. The fractional decrease in energy of the capacitor is

A parallel plate air capacitor has a capacitance C. When it is half filled with a dielectric of dielectric constant 5, the percentage increase in the capacitance will be

Capacitance Of Plates With Dielectric