Partly filling a capacitor with dielectric 1

Filling a capacitor with a dielectricII: Partial filling

© Frits F.M. de Mul

Partly filling a capacitor with dielectric 2

Partial filling a capacitor (1)

Available: Flat capacitor : = surface area A , = distance of plates d , = no fill (εr = 1) .

A

d Assume: initially, plates are charged with +Q , -Q.

+Q

-Q

Question: What will happen with Q , σ, E, D, ∆V and C upon PARTIAL filling the capacitor with dielectric material (with εr > 1) ???

Partly filling a capacitor with dielectric 3

Partial filling a capacitor (2)

A. Series B. Parallel

I. Free

II. Connected to battery

Options:

Partly filling a capacitor with dielectric 4

Partial filling a capacitor (3)

A B

I

II

Suppose:• when empty: Q0 , V0 .....etc.• fillingfilling for 1/3 of volumevolume• with εr = 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)

Partly filling a capacitor with dielectric 5

RelationsA

d

+Q

-Q

Qf

Gauss:

D dA• = = ∫∫∫∫∫ Q dVf fVA

ρ

D

E Material constants:

D = ε0 εr E

Potential:

∆Vpath

= •∫ E dl

∆V

Capacitance:

C Q Vf= / ∆

AADQ ff .. σ==

dEV .=∆

Partly filling a capacitor with dielectric 6

Options: main featuresOptions: main features

A.I and B.I: total charge unchanged

A B

I

IIA.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: Q0 , V0 ......• when filled: Q’, V’ , ......• fillingfilling for 1/3 of volumevolume

(depth or surface area)• with εr = 5

Partly filling a capacitor with dielectric 7

A.I. Horizontal fill, not A.I. Horizontal fill, not connectedconnected

Qf

DE

∆V

db , dt : bottom and top layerFill: db = d0 /3 with εr = 5

dt = 2d0 /3 with εr = 1db

dt

Qf,t’

-Qf,b’

Qf’ D’ E’ d’

51

31

∆V’

151

32

32

1511

∆V’ C’

1115

o = oldt = topb = bottom} total

11

Start

Partly filling a capacitor with dielectric 8

A.II. Horizontal fill, connectedA.II. Horizontal fill, connectedQf

D

E

∆VFill: db = d0 /3 with εr = 5

dt = 2d0 /3 with εr = 1

Qf’ D’ E’ d’51

31

∆V’

151

1510

32 =3

2

∆V’

C’

1115

db

dt

Qf,t

-Qf,b

∆V0

∆V’

111

1110

11

E’113

1115

32

1110 : =

1115

1115

D’ Qf’

o = oldt = topb = bottom} total

Consider ratios:

11

1115

<01115

Partly filling a capacitor with dielectric 9

B.I. Vertical fill, not connectedB.I. Vertical fill, not connectedQf

D

E

∆VFill: Ar = A0 /3 with εr = 5

Al = 2A0 /3 with εr = 1

Qf’D’E’

73

51

715 : =

73

E’ C’

37

o = oldl = leftr = right } total

Qf,l’

-Qf,r’

Qf,r’

-Qf,l’

Al Ar

Consider ratios:

∆V’

11

32

31

A’

32

Qf’

757

235

Qf’

715

31

75 : =

73

32

72 : =

D’

×5

73

73

∆V’

Partly filling a capacitor with dielectric 10

B.II. Vertical fill, connectedB.II. Vertical fill, connectedQf

D

E

∆VFill: Ar = A0 /3 with εr = 5

Al = 2A0 /3 with εr = 1

Qf’D’E’

×5

C’

37

o = oldl = leftr = right } total

Consider ratios:

∆V’

11 3

2

31

A’

32

Qf’

37

35

Qf,l’

-Qf,r’

Qf,r’

-Qf,l’

Al Ar

∆V0

Partly filling a capacitor with dielectric 11

Options: overviewOptions: overview

BB.I and B.II: potentials parallel

C : 15/11 xQf : unchanged∆V : 11/15 x

∆V : unchangedQf : 15/11 x

C : 7/3 xQf : unchanged∆V : 3/7 x

∆V : unchangedQf : 7/3 x

AA.I and A.II: potentials in series

I II

Partly filling a capacitor with dielectric 12

With combination rulesWith combination rules

B B.I and B.II: parallel

A A.I and A.II: series

I II d

AC rεε0= Filling with εr=5

in 1/3 of volume

A

d

A

d

A

d

CCC bt .15

11

.

3/2

5.

3/11

'

1

000 εεε=+=+=

01115' CC =

d

A

d

A

d

ACCC rl

.

3

73/.5.3/2.' 000 εεε =+=+= 03

7' CC =

Partly filling a capacitor with dielectric 13

Finally...Finally...

B B.I and B.II: parallel

A

A.I and A.II: seriesI II

bt CCC

11

'

1 +=

rl CCC +=' the end

fQ∫∫ =• dAD

D = ε0 εr E∫ •=∆ dlEV

Qf

DE

∆V

VQC f ∆= / AADQ ff .. σ==

dEV .=∆