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

2. 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 

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

4. Partly filling a capacitor with dielectric4 Partial filling a capacitor (3) A A B B I I II Suppose: when empty: Q 0, V 0.....etc. fillingvolume filling for 1/3 of volume with  r = 5 Suppose: when empty: Q 0, V 0.....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)

5. 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

6. 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 0...... when filled: Q’, V’,...... fillingvolume filling for 1/3 of volume (depth or surface area) with  r = 5 Suppose: when empty: Q 0, V 0...... when filled: Q’, V’,...... fillingvolume filling for 1/3 of volume (depth or surface area) with  r = 5

7. 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

8. 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

9. 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’

10. 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

11. 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

12. 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

13. 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