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Filling a capacitor  with a dielectric II: Partial filling 息 Frits F.M. de Mul
Partial filling a capacitor (1) Available:   Flat capacitor  :  = surface area  A  ,  = distance of plates  d ,   = no fill (  r 緒縁   .  Question:   What will happen with  Q ,  鰹 E,  D,   V  and  C  upon  PARTIAL filling  the  capacitor  with  dielectric material  (with   r 常縁随随 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) Suppose : when empty:  Q 0  , V 0  .....etc. filling  for 1/3 of  volume with   r  = 5 Assumption  :  no edge effects :  d <<  plate dimensions   Consequences  : no  E -field leakage  planar symmetry everywhere Gauss Law is applicable  straight field lines ( E  and  D ) 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 Suppose : when empty:  Q 0  , V 0   ...... when filled:  Q, V , ...... filling  for 1/3 of  volume (depth or surface area) with   r  = 5  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 息 Frits F.M. de Mul
  • 2. Partial filling a capacitor (1) Available: Flat capacitor : = surface area A , = distance of plates d , = no fill ( r 緒縁 . Question: What will happen with Q , 鰹 E, D, V and C upon PARTIAL filling the capacitor with dielectric material (with r 常縁随随 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. Partial filling a capacitor (3) Suppose : when empty: Q 0 , V 0 .....etc. filling for 1/3 of volume with r = 5 Assumption : no edge effects : d << plate dimensions Consequences : no E -field leakage planar symmetry everywhere Gauss Law is applicable straight field lines ( E and D ) A B I II
  • 5. Relations Q f A d +Q -Q D E Material constants: D = r E V
  • 6. 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 Suppose : when empty: Q 0 , V 0 ...... when filled: Q, V , ...... filling for 1/3 of volume (depth or surface area) with r = 5 A B I II
  • 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