The document discusses the effects of filling a capacitor with a dielectric material on various electrical parameters such as charge, electric field, voltage, and capacitance. It analyzes two scenarios: one where the capacitor remains connected to a battery, resulting in constant voltage, and another where it is disconnected, resulting in constant charge. The relationships between relevant variables are established, with calculations supporting the outcomes in both cases.

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. Conclusions if connected to battery  V = const E = const D,  f , Q f :  r x as large C :  r x as large if NOT connected Q f = const  f , D = const E ,  V  :  r x as small C :  r x as large the end