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Filling a capacitor  with a dielectric: I. Complete filling © Frits F.M. de Mul
Filling a capacitor with dielectric 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  filling  the  capacitor  with  dielectric material  (with   r  A d Assume:   initially, plates are charged with  +Q , -Q.  +Q -Q
Analysis Case A  : yes Case B  : no Consequences : in A : voltage remains constant  in B : charges remain constant. A d +Q -Q Important :  Does the capacitor remain connected to the battery that loaded it?  V
Approach (1) Assumption  :  no “edge effects” :  d <<  plate dimensions   Consequences  : no  E -field leakage  planar symmetry everywhere : Gauss’ Law is applicable A d +Q -Q
Approach (2) A d +Q -Q Relevant variables  : Q ,   ,  E, D,    V  and  C Material constants  :    r  ,  A  and  d . Task  :  Establish the relations between the variables.
Approach (3) Relations between variables  : Q f A d +Q -Q D E Material constants: D =    r  E  V
Calculations (1) =>  D =   f   D Take Gauss pill box,  top/bottom area =  dA  ,  enclosed charge =   f ..dA dA
Calculations (2)   E Take integration path  from bottom to top,  length  d =>   V = E.d d
Calculations (3) D =   f  = Q f  /A D =     r  E  V = E.d C = Q f  /  V Case A  : capacitor connected to battery  V  remains constant !! Where to start????? Q f D E  V Q f  f  D  E   V  C   b = begin = end 1.   V  = const start 2.  E  = const 3.  D  :   r  x as large 4.    f   :   r  x as large 5.  Q f  :   r  x as large 6.  C   :   r  x as  large   7.   b   = (  r -1)   f, “old”
Calculations (4) Q f D E  V D =   f  = Q f  /A D =     r  E  V = E.d C = Q f  /  V Case B  : capacitor NOT connected to battery Q f  remains constant !! Q f  f  D  E   V  C   b Where to start????? 1.  Q f   = const start 2.   f   = const 3.  D   = const 4.   E   :   r  x as small 5.   V  :   r  x as small 6.  C   :   r  x as  large   7.   b   = (1-   r )  f, “old”
Conclusions ,[object Object],[object Object],[object Object],[object Object],[object Object],[object Object],[object Object],[object Object],[object Object],[object Object],the end

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Capacitor filling

  • 1. Filling a capacitor with a dielectric: I. Complete filling © Frits F.M. de Mul
  • 2. Filling a capacitor with dielectric 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 filling the capacitor with dielectric material (with  r  A d Assume: initially, plates are charged with +Q , -Q. +Q -Q
  • 3. Analysis Case A : yes Case B : no Consequences : in A : voltage remains constant in B : charges remain constant. A d +Q -Q Important : Does the capacitor remain connected to the battery that loaded it?  V
  • 4. Approach (1) Assumption : no “edge effects” : d << plate dimensions Consequences : no E -field leakage planar symmetry everywhere : Gauss’ Law is applicable A d +Q -Q
  • 5. Approach (2) A d +Q -Q Relevant variables : Q ,  , E, D,  V and C Material constants :    r , A and d . Task : Establish the relations between the variables.
  • 6. Approach (3) Relations between variables : Q f A d +Q -Q D E Material constants: D =    r  E  V
  • 7. Calculations (1) => D =  f   D Take Gauss pill box, top/bottom area = dA , enclosed charge =  f ..dA dA
  • 8. Calculations (2)   E Take integration path from bottom to top, length d =>  V = E.d d
  • 9. Calculations (3) D =  f = Q f /A D =    r E  V = E.d C = Q f /  V Case A : capacitor connected to battery  V remains constant !! Where to start????? Q f D E  V Q f  f  D E  V C  b = begin = end 1.  V = const start 2. E = const 3. D :  r x as large 4.  f :  r x as large 5. Q f :  r x as large 6. C :  r x as large 7.  b = (  r -1)  f, “old”
  • 10. Calculations (4) Q f D E  V D =  f = Q f /A D =    r E  V = E.d C = Q f /  V Case B : capacitor NOT connected to battery Q f remains constant !! Q f  f  D E  V C  b Where to start????? 1. Q f = const start 2.  f = const 3. D = const 4. E :  r x as small 5.  V :  r x as small 6. C :  r x as large 7.  b = (1-  r )  f, “old”
  • 11.