Partly filling a capacitor with dielectric1 Filling a capacitor with a dielectric II: Partial filling © Frits F.M. de Mul

Partly filling a capacitor with dielectric2 Partial filling a capacitor (1) Available: Flat capacitor : = surface area A, = distance of plates d, = no fill (  r . A d Assume: initially, plates are charged with +Q, -Q. +Q -Q Question: What will happen with Q,  E, D,  V and C upon PARTIAL filling the capacitor with dielectric material (with  r 

Partly filling a capacitor with dielectric3 Partial filling a capacitor (2) A. Series B. Parallel I. Free II. Connected to battery Options:

Partly filling a capacitor with dielectric4 Partial filling a capacitor (3) A A B B I I II Suppose: when empty: Q 0, V etc. fillingvolume filling for 1/3 of volume with  r = 5 Suppose: when empty: Q 0, V etc. fillingvolume filling for 1/3 of volume with  r = 5 Assumption : no “edge effects” : d << plate dimensions Assumption : no “edge effects” : d << plate dimensions Consequences : no E-field leakage Consequences : no E-field leakage planar symmetry everywhere Gauss’ Law is applicable straight field lines (E and D) planar symmetry everywhere Gauss’ Law is applicable straight field lines (E and D)

Partly filling a capacitor with dielectric5 Relations A d +Q -Q-Q QfQf D E Material constants: D =    r  E Material constants: D =    r  E VV

Partly filling a capacitor with dielectric6 Options: main features A.I and B.I: total charge unchanged A A B B I I II 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 when filled: Q’, V’, fillingvolume filling for 1/3 of volume (depth or surface area) with  r = 5 Suppose: when empty: Q 0, V when filled: Q’, V’, fillingvolume filling for 1/3 of volume (depth or surface area) with  r = 5

Partly filling a capacitor with dielectric7 A.I. Horizontal fill, not connected QfQf D E VV d b, d t : bottom and top layer Fill: d b = d 0 /3 with  r = 5 d t = 2d 0 /3 with  r = 1 d b, d t : bottom and top layer Fill: d b = d 0 /3 with  r = 5 d t = 2d 0 /3 with  r = 1 dbdb dtdt Q f,t ’ -Q f,b ’ Qf’Qf’ D’ E’ d’  V’ C’ o = old t = top b = bottom } total Start

Partly filling a capacitor with dielectric8 A.II. Horizontal fill, connected QfQf D E VV Fill: d b = d 0 /3 with  r = 5 d t = 2d 0 /3 with  r = 1 Fill: d b = d 0 /3 with  r = 5 d t = 2d 0 /3 with  r = 1 Qf’Qf’ D’ E’ d’  V’ C’ dbdb dtdt Q f,t -Q f,b V0V0  V’ E’D’Qf’Qf’ o = old t = top b = bottom } total Consider ratios: <0

Partly filling a capacitor with dielectric9 B.I. Vertical fill, not connected QfQf D E VV Fill: A r = A 0 /3 with  r = 5 A l = 2A 0 /3 with  r = 1 Fill: A r = A 0 /3 with  r = 5 A l = 2A 0 /3 with  r = 1 Qf’Qf’ D’ E’ C’ o = old l = left r = right } total Q f,l ’ -Q f,r ’ Q f,r ’ -Q f,l ’ AlAl ArAr Consider ratios:  V’ A’ Qf’Qf’ Qf’Qf’ D’  V’

Partly filling a capacitor with dielectric10 B.II. Vertical fill, connected QfQf D E VV Fill: A r = A 0 /3 with  r = 5 A l = 2A 0 /3 with  r = 1 Fill: A r = A 0 /3 with  r = 5 A l = 2A 0 /3 with  r = 1 Qf’Qf’ D’ E’ C’ o = old l = left r = right } total Consider ratios:  V’ A’ Qf’Qf’ Q f,l ’ -Q f,r ’ Q f,r ’ -Q f,l ’ AlAl ArAr V0V0

Partly filling a capacitor with dielectric11 Options: overview B B B.I and B.II: potentials parallel C : 15/11 x Q f : unchanged  V : 11/15 x Q f : unchanged  V : 11/15 x  V : unchanged Q f : 15/11 x  V : unchanged Q f : 15/11 x C : 7/3 x Q f : unchanged  V : 3/7 x Q f : unchanged  V : 3/7 x  V : unchanged Q f : 7/3 x  V : unchanged Q f : 7/3 x A A A.I and A.II: potentials in series I I II

Partly filling a capacitor with dielectric12 With combination rules B B B.I and B.II: parallel A A A.I and A.II: series I I II Filling with  r =5 in 1/3 of volume Filling with  r =5 in 1/3 of volume

Partly filling a capacitor with dielectric13Finally...Finally... B B B.I and B.II: parallel A A A.I and A.II: series I I II the end D =    r  E QfQf D E VV