CALCULATING EFFECTIVE CAPACITANCE ON INSERTION OF DIELECTRIC IN THE SPACE BETWEEN TWO PLATES

Insertion Of Dielectric Between Plates

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).

Capacitance Of Plates With Dielectric

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 capacitance of a parallel plate capacitor with plate area A and separation d is C . The space between the plates in filled with two wedges of dielectric constants K_(1) and K_(2) respectively. Find the capacitance of resulting capacitor.