Lecture 14 Capacitance and Conductance

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1 Lecture 14 Capacitance and Conductance ections: 6.3, 6.4, 6.5 Homework: ee 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 N P D Eds EdL, F Gauss law E Q P r E E V N Q LECTURE 14 slide

electrical free charge under certain voltage C Q V,

3 Capacitance of Parallel-Plate Capacitor a D s n a n 0 E s a z C P s n a n a D E ds Ez A A C N Ez d d EdL 0, F LECTURE 14 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 (of C 1 and of C ) are biased with voltage V = 1 kv, what would be their respective charges (Q 1 and Q )? Q1 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 14 slide 4

If the two capacitors (of C 1 and of C ) are biased with voltage V = 1 kv, what would be their respective charges (Q 1 and Q )? Q1 Q 3.

5 4. In the previous example: (a) Find the ratios of the free-to-total charge at the insulator-electrode interface. (b) Find the ratios of bound-to-total charge. (c) Compare the total (free + bound) charge values. sf sb LECTURE 14 slide 5

(b) Find the ratios of bound-to-total charge.

6 Capacitance and tored Energy 1 general energy expression (Lecture 9) 1 We ( vf 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 VQ (1) (1) e 1 sf W V ds V Q () () e sf 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 14 slide 6

electrodes 1 1 W V ds VQ (1) (1) e 1 sf 1 1 1 1 W V ds V Q () () e sf the capacitor is assumed charge

7 Capacitance and tored Energy total energy of the two electrodes (1) () 1 We We We Q ( V1V ) We 1 QV We ( CV ) V CV Q 1 Q 1 Q We Q C C C V W e V 1 C Q W e LECTURE 14 slide 7

) We 1 QV 1 1 1 We ( CV ) V CV 1 1 1 Q 1 Q 1

8 Capacitance Example 1: Double-Layer Plate Capacitor voltage between plates V Ed 1 1 Ed why 1 at the dielectric interface Dn1 Dn E E1 V V E1 s D1 1E1 d d ( / ) ( d / ) ( d / ) Q s C V V C 1 1 d1 d 1 1 C C 1 1 capacitors in series s s LECTURE 14 slide 8

Dn E E1 V V E1 s D1 1E1 d d ( / ) ( d / ) ( d / ) 1 1 1 1 Q s C

9 Equivalence of Metallic tructures 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 14 V 6 V V 0 V 9

uniqueness theorem: potential values at the boundary surfaces remain the same V 10 V V 6 V E

10 V Capacitance Example : pherical Capacitor d Q 4r Q D s E b E 1 Q 4 r a r Q 1 1 Q ( b a) E d L 4 a b 4 ab a Q ab C 4, F V ( b a ) E Q Q a r b ab single sphere capacitance: Ca lim C lim 4 4 a, F b b ( b a) LECTURE 14 slide 10

11 Capacitance per Unit Length: Parallel-Plate Line C A h wl h, F C w C, F/m l h h w l LECTURE 14 slide 11

12 Capacitance per Unit Length: Coaxial Cable apply Gauss law to find E field cross-section 1 b l Q E, V/m a l find voltage from E b Q b V0 E d ln, V a l a find capacitance from voltage Q l C, F V ln( b / a ) 0 find PUL capacitance C C, F/m l ln( b/ a) V 0 V 0 l LECTURE 14 slide 1 E

d ln, V a l a find capacitance from voltage Q l C, F V ln( b / a ) 0

13 Capacitance per Unit Length: Twin-Lead Cable 1 (optional) tep 1: Find equation of equipotential lines of two line charges at x= s and x =s. l s l s at observation point P: V ln, V ln 1 l 1 V( P) V V ln y h h P r l s V s equipotentials r x l LECTURE 14 slide 13

14 Capacitance per Unit Length: Twin-Lead Cable (optional) at an equipotential line V = V c we have 1 Vc 1 Vc ln K exp l l K at P(x,y) K 1 ( s x) y ( sx) y squaring and re-arranging we obtain the equation of a circle 1 Ks K xs y K 1 K 1 x-coordinate of center h r radius LECTURE 14 slide 14

1 ( s x) y ( sx) y squaring and re-arranging we obtain the equation of

15 Capacitance per Unit Length: Twin-Lead Cable 3 (optional) the equation of the equipotential line V = V c 1 Ks K xs 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 tep : 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 14 slide 15

s s h r K 1 tep : Construct an equivalent problem of wires of finite radius r by placing the wires so that their

16 Capacitance per Unit Length: Twin-Lead Cable 4 potential at the positive wire l V1 ln K potential at the negative wire l V ln K potential difference between wires l C V 1 h h l h h V1 V1 V ln 1 r r, F/m ln 1 r r K C, h h r ln r LECTURE 14 slide 16

V ln K potential difference between wires l C V 1 h h l h h

17 Analogy between Capacitance and Conductance Q C V N E d P E ds L, F I G V N E d P E ds L, for a given geometry the expressions for capacitance and conductance are identical except for the material constant Examples: 1) coaxial capacitor/resistor l l ln( b/ a) C, F G, ; R, ln( b/ a) ln( b/ a) l ) homework: conductance G of a parallel-plate resistor LECTURE 14 slide 17

for the material constant Examples: 1) coaxial capacitor/resistor l l ln( b/ a) C, F G, ; R,

18 Conductance per Unit Length 1 3) coaxial cable with lossy (dissipative) insulator (σ d 0) d C, F/m G, /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 10: R, /m m a c b no skin effect taken into account! I metal NOTE: G 1/ R I metal I diel LECTURE 14 slide 18

insulator (current flows through the insulator, described by Gꞌ) the loss in the metallic leads of a coaxial cable Rꞌ was

19 Conductance per Unit Length 4) twin-lead cable with lossy insulator (σ d 0) C d, F/m G, /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 10 1 R, /m A m G 1/ R 5) homework: derive the conductance per unit length Gꞌ of a parallel-plate line LECTURE 14 slide 19

twin-lead cable was obtained in Lecture 10 1 R, /m A m G 1/ R 5) homework:

20 Parameters per Unit Length in Circuit Models of TLs in in 1 Rl Ll v N Gl Cl vn 1 l z LECTURE 14 slide 0

21 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 14 slide 1

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