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Filling a capacitor  with a dielectric II: Partial filling © Frits F.M. de Mul
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
Partial filling a capacitor (2) A. Series B. Parallel I. Free II. Connected to battery Options:
Partial filling a capacitor (3) ,[object Object],[object Object],[object Object],[object Object],Assumption  :  no “edge effects” :  d <<  plate dimensions   Consequences  : no  E -field leakage  ,[object Object],[object Object],[object Object],A B I II
Relations Q f A d +Q -Q D E Material constants: D =    r  E  V
Options: main features A.I and B.I : total charge unchanged   A.II and B.II : total potential unchanged A.I and A.II : potentials in series B.I and B.II : potentials parallel ,[object Object],[object Object],[object Object],[object Object],[object Object],[object Object],A B I II
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
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
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’
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
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
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
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

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Capacitor partial filling

  • 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:
  • 4.
  • 5. Relations Q f A d +Q -Q D E Material constants: D =    r  E  V
  • 6.
  • 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