1. Filling a capacitor with a dielectric II: Partial filling © Frits F.M. de Mul

2. Partial filling a capacitor (1) Available: Flat capacitor : = surface area A , = distance of plates d , = no fill (  r  . Question: What will happen with Q ,  E, D,  V and C upon PARTIAL filling the capacitor with dielectric material (with  r  A d Assume: initially, plates are charged with +Q , -Q. +Q -Q

3. Partial filling a capacitor (2) A. Series B. Parallel I. Free II. Connected to battery Options:

5. Relations Q f A d +Q -Q D E Material constants: D =    r  E  V

7. A.I. Horizontal fill, not connected d b , d t : bottom and top layer Fill: d b = d 0 /3 with  r = 5 d t = 2 d 0 /3 with  r = 1 Q f ’ D’ E’ d’  V’  V’ C’ Start Q f D E  V d b d t Q f,t ’ -Q f,b ’ o = old t = top b = bottom } total

8. A.II. Horizontal fill, connected Fill: d b = d 0 /3 with  r = 5 d t = 2 d 0 /3 with  r = 1 Q f ’ D’ E’ d’  V’  V’ C’  V’ E’ D’ Q f ’ Consider ratios: Q f D E  V d b d t Q f,t -Q f,b  V 0 o = old t = top b = bottom } total <0

9. B.I. Vertical fill, not connected Fill: A r = A 0 /3 with  r = 5 A l = 2 A 0 /3 with  r = 1 Q f ’ D’ E’ Consider ratios:  V’ A’ Q f D E  V E’ C’ o = old l = left r = right } total Q f,l ’ -Q f,r ’ Q f,r ’ -Q f,l ’ A l A r Q f ’ Q f ’ D’  V’

10. B.II. Vertical fill, connected Fill: A r = A 0 /3 with  r = 5 A l = 2 A 0 /3 with  r = 1 Q f ’ D’ E’ C’ Consider ratios:  V’ A’ Q f ’ Q f D E  V o = old l = left r = right } total Q f,l ’ -Q f,r ’ Q f,r ’ -Q f,l ’ A l A r  V 0

11. Options: overview C : 15/11 x Q f : unchanged  V : 11/15 x  V : unchanged Q f : 15/11 x C : 7/3 x Q f : unchanged  V : 3/7 x  V : unchanged Q f : 7/3 x B B.I and B.II : potentials parallel A A.I and A.II : potentials in series I II

12. With combination rules Filling with  r =5 in 1/3 of volume B B.I and B.II : parallel A A.I and A.II : series I II

13. Finally... the end D =    r  E B B.I and B.II : parallel A A.I and A.II : series I II Q f D E  V