- #1
karanbir
- 7
- 0
i'm reposting this question as nobody answered it previously. please answer it this time.
in given type of arrangement we normally take it equivalent to two capacitors(plate 1 and surface 2) and (surface 2 and plate 3) in series. My question is that a capacitor is formed with two plates of equal and opposite charge but the induced charge on dielectric slab(surface 2) is less than the charge on plates 1 and 3 which is clear by the formula q(ind)=q(1-1/k), then how can we assume plate 1 and surface 2 as one capacitor?
in given type of arrangement we normally take it equivalent to two capacitors(plate 1 and surface 2) and (surface 2 and plate 3) in series. My question is that a capacitor is formed with two plates of equal and opposite charge but the induced charge on dielectric slab(surface 2) is less than the charge on plates 1 and 3 which is clear by the formula q(ind)=q(1-1/k), then how can we assume plate 1 and surface 2 as one capacitor?