Lecture 18 Capacitance and Conductance

Transcription

1 Lecture 18 Capacitance and Conductance Sections: 6.3, 6.4, 6.5 Homework: See homework file

2 Definition of Capacitance capacitance is a measure of the ability of the physical structure to accumulate electrical free charge under certain voltage C = Q V, F=C/V C = S N P D εe ds E dl, F Gauss law notice proportionality to permittivity S + Q P E ε r V E E Q N - - LECTURE 18 slide

3 Capacitance of Parallel-Plate Capacitor ρ = a D< s n a n 0 E = ρs a ε z C P s n a n ρ = a D> E ds S Ez A A = ε = ε C = ε N Ez d d E dl 0, F S LECTURE 18 slide 3

4 1. A capacitor whose insulator has relative permittivity ε r1 = 1 has capacitance C 1 = 1 μf. What is going to be its capacitance if the insulator is replaced by another one with ε r = 30? C =. If the two capacitors (C 1 and C ) are biased with voltage V = 1 kv, what would be their respective charges (Q 1 and Q )? Q 1 = Q = 3. What is the nature of the charges Q 1 and Q? (a) free charge deposited on electrode surface (b) bound charge deposited on the insulator surface at the electrode (c) total charge at the electrode-insulator surface LECTURE 18 slide 4

5 Capacitance and Stored Energy 1 general energy expression (Lecture 9) 1 We = ( ρvf V ) dv v there are two electrodes: one at a potential V 1 and the other at V charge is distributed on the surface of the electrodes 1 1 W V ds V Q (1) (1) e = 1 ρsf = 1 1 S 1 1 W V ds V Q () () e = ρsf = S the capacitor is assumed charge neutral as a whole both before and after voltage is applied (charge conservation) Q + Q = 0 Q = Q = Q 1 1 LECTURE 18 slide 5

6 Capacitance and Stored Energy total energy of the two electrodes (1) () 1 We = We + We = QV ( 1 V ) W = e 1 QV We = ( C V ) V = CV Q Q 1 Q We = Q = C C C = V W e V 1 C = Q W e LECTURE 18 slide 6

7 Capacitance and Stored Energy Example Derive the formula for the capacitance of a parallel-plate capacitor by making use of the relation between stored energy and capacitance. LECTURE 18 slide 7

8 Capacitance Example 1: Double-Layer Plate Capacitor voltage between plates V= Ed 1 1+ Ed why ε1 at the dielectric interface Dn1 = Dn E = E1 ε V V E1 = ρs = D1= ε1e1= d + d ( ε / ε ) ( d / ε ) + ( d / ε ) Q ρs C = = V V S 1 1 C = = d1 d ε S ε S C C 1 1 capacitors in series ρ s ρ s LECTURE 18 slide 8

9 Equivalence of Metallic Structures principle: placing a PEC sheet at an equipotential surface does not change the field distribution capacitance does not change follows from the uniqueness theorem: potential values at the boundary surfaces remain the same V = 10 V V = 6 V E V = 0 V the structure is effectively split into two capacitors in series V =10 LECTURE 18 slide V = 6 V V = 0 V 9

10 V S Capacitance Example : Spherical Capacitor d = Q ε 4πr = Q D s E E= b 1 Q 4πε r a r Q 1 1 Q ( b a) = E d L = = 4πε a b 4πε ab a Q ab C = = 4 πε, F V ( b a ) S E Q +Q a r ε b ab single sphere capacitance: Ca = lim C = lim 4πε = 4 πε a, F b b ( b a) LECTURE 18 slide 10

11 Technology Brief: Capacitive Sensors A capacitor functions as a sensor if the stimulus changes either its geometry (e.g., distance between electrodes) or its effective permittivity. FLUID GAUGE whf w( h hf) C = Cf + Ca = εf + εa d d C = kh + C f empty V out C = ( C + C) 0 C V if C C out 0 4C0 k = C empty w( ε ε ) f d = ε a a wh d [Ulaby&Ravaioli, Fundamentals of Applied Electromagnetics, 7 th ed.] LECTURE 18 slide 11

12 Technology Brief: Capacitive Sensors HUMIDITY SENSOR typically uses interdigital capacitor structure to maximize capacitance and sensitivity spacing between digits is about 0. μm effective permittivity between the digits changes with humidity [Ulaby&Ravaioli, Fundamentals of Applied Electromagnetics, 7 th ed.] LECTURE 18 slide 1

13 Technology Brief: Capacitive Sensors 3 PRESSURE SENSOR [Ulaby&Ravaioli, Fundamentals of Applied Electromagnetics, 7 th ed.] LECTURE 18 slide 13

14 Technology Brief: Capacitive Sensors 4 NONCONTACT SENSOR When external object is placed in the proximity of capacitor, it changes effective permittivity, thereby changing capacitance. [Ulaby&Ravaioli, Fundamentals of Applied Electromagnetics, 7 th ed.] LECTURE 18 slide 14

15 Capacitance per Unit Length: Parallel-Plate Line C A wl = ε = ε h h, F C w C = = ε, F/m l h h w l LECTURE 18 slide 15

16 Capacitance per Unit Length: Coaxial Cable apply Gauss law to find E field cross-section ρ 1 b l Q Eρ =, V/m a περ = περ l find voltage from E b Q b V0 = Ed ln, V a ρ ρ = πεl a find capacitance from voltage Q πεl C = =, F V ln( b / a ) 0 find PUL capacitance C πε C = =, F/m l ln( b/ a) V 0 + E V = 0 l S ρ LECTURE 18 slide 16

17 Capacitance per Unit Length: Twin-Lead Cable 1 (optional) Step 1: Find equation of equipotential lines of two line charges at x = s and x = s. ρl s ρl s at observation point P: V+ = ln, V = ln πε ρ πε ρ1 ρl ρ1 V( P) = V+ + V = ln πε ρ y h h P r ρ l s V = 0 0 ρ 1 s equipotentials ρ r x +ρ l LECTURE 18 slide 17

18 Capacitance per Unit Length: Twin-Lead Cable (optional) at an equipotential line V = V c we have ρ1 πεvc ρ1 πεvc ln K exp ρ = = = ρl ρ ρ l K at P(x,y) K ρ ρ 1 = = ( s+ x) + y ( s x) + y squaring and re-arranging we obtain the equation of a circle + 1 Ks K x s + y = K 1 K 1 x-coordinate of center h r radius LECTURE 18 slide 18

19 Capacitance per Unit Length: Twin-Lead Cable 3 (optional) the equation of the equipotential line V = V c + 1 Ks K x s + y = K 1 K 1 is a circle of radius Ks r = K 1 h h K = + and a center on the x-axis at a distance 1 r r from the origin K + 1 h = s s = h r K 1 Step : Construct an equivalent problem of wires of finite radius r by placing the wires so that their surfaces coincide with the equipotential lines of the ideal line charges (a distance h from origin). LECTURE 18 slide 19

20 Capacitance per Unit Length: Twin-Lead Cable 4 ρ potential at the positive wire l V1 = ln K πε ρl potential at the negative wire V = ln K πε potential difference between wires ρ = = l C V 1 h h ρ l h h V1 = V1 V = ln + 1 πε r r πε, F/m ln + 1 r r K C πε, h h r ln r LECTURE 18 slide 0

21 Analogy between Capacitance and Conductance Q = = S C V N E d P εe ds L, F I S G = = V N E d P σ E ds L,S for a given geometry the expressions for capacitance and conductance are identical except for the material constant Examples: 1) coaxial capacitor/resistor πεl πσ l ln( b/ a) C =, F G =, S; R=, Ω ln( b/ a) ln( b/ a) πσl ) homework: conductance G of a parallel-plate resistor LECTURE 18 slide 1

22 Conductance per Unit Length 1 3) coaxial cable with lossy (dissipative) insulator (σ d 0) πε πσ d C =, F/m G =, S/m ln( b/ a) ln( b/ a) note that a cable in general suffers loss not only due to the conducting wires (described by Rꞌ) but also due to its non-ideal insulator (current flows through the insulator, described by Gꞌ) the loss in the metallic leads of a coaxial cable Rꞌ was obtained in Lecture 14: R = +, /m σ Ω mπ a c b no skin effect taken into account! I metal NOTE: G 1/ R I metal I diel LECTURE 18 slide

23 Conductance per Unit Length 4) twin-lead cable with lossy insulator (σ d 0) πε πσ d C =, F/m G =, S/m h h h h ln + 1 ln + 1 r r r r the loss in the metallic wires of a twin-lead cable was obtained in Lecture 14 1 R =, Ω/m σ A m G 1/ R 5) homework: derive the conductance per unit length Gꞌ of a parallel-plate line LECTURE 18 slide 3

24 Parameters per Unit Length in Circuit Models of TLs i i + 1 N N N N+1 R l L l v N G l C l v N + 1 l z LECTURE 18 slide 4

25 You have learned: what capacitance is and what capacitance per unit length is how to calculate capacitance from the field distribution how capacitance relates to the stored electric energy how to calculate the capacitance per unit length and the conductance per unit length of a parallel-plate line, coaxial cable and twin-lead cable LECTURE 18 slide 5