Electrical Engineering ⇒ Topic : Special Cases of Parallel-Plate Capacitor

Special Cases of Parallel-Plate Capacitor

We have seen that capacitance of a capacitor depends upon plate area, thickness of dielectric and value of relative permittivity of the dielectric.

We consider two cases by way of illustration

(1) Fig. (a) shows that dielectric thickness is d but plate area is divided into two parts; area A₁ having air as the dielectric and area A₂ having dielectric of relative permittivity ε_r. The arrangement is equivalent to two capacitors in parallel. Their capacitances are

The total capacitance C of this parallel plate capacitor is

                             C = C₁ + C₂

                                        _{(A)                                                                                        (B)}

                                                                   _FIGURE 

(2) Fig. 6.10 shows that plate area is divided into two parts ; area A1 has dielectric (air) of thickness d and area A2 has a dielectric (80 of thickness t and the remaining thickness is occupied by air The an-angement is equivalent to two capacitors connected in parallel. Their capacitances are

The total capacitance C of this parallel plate capacitor is

                          C = C₁ + C₂