46 4 E E E E E 0 0 E E = E E E = ) E =0 2) φ = 3) ρ =0 1) 0 2) E φ E = grad φ E =0 P P φ = E ds 0

Similar documents
120 9 I I 1 I 2 I 1 I 2 ( a) ( b) ( c ) I I 2 I 1 I ( d) ( e) ( f ) 9.1: Ampère (c) (d) (e) S I 1 I 2 B ds = µ 0 ( I 1 I 2 ) I 1 I 2 B ds =0. I 1 I 2

18 2 F 12 r 2 r 1 (3) Coulomb km Coulomb M = kg F G = ( ) ( ) ( ) 2 = [N]. Coulomb

1 I 1.1 ± e = = - = C C MKSA [m], [Kg] [s] [A] 1C 1A 1 MKSA 1C 1C +q q +q q 1

2.5 (Gauss) (flux) v(r)( ) S n S v n v n (1) v n S = v n S = v S, n S S. n n S v S v Minoru TANAKA (Osaka Univ.) I(2012), Sec p. 1/30

50 2 I SI MKSA r q r q F F = 1 qq 4πε 0 r r 2 r r r r (2.2 ε 0 = 1 c 2 µ 0 c = m/s q 2.1 r q' F r = 0 µ 0 = 4π 10 7 N/A 2 k = 1/(4πε 0 qq

II ( ) (7/31) II ( [ (3.4)] Navier Stokes [ (6/29)] Navier Stokes 3 [ (6/19)] Re

2. 2 P M A 2 F = mmg AP AP 2 AP (G > : ) AP/ AP A P P j M j F = n j=1 mm j G AP j AP j 2 AP j 3 P ψ(p) j ψ(p j ) j (P j j ) A F = n j=1 mgψ(p j ) j AP


c y /2 ddy = = 2π sin θ /2 dθd /2 [ ] 2π cos θ d = log 2 + a 2 d = log 2 + a 2 = log 2 + a a 2 d d + 2 = l

66 σ σ (8.1) σ = 0 0 σd = 0 (8.2) (8.2) (8.1) E ρ d = 0... d = 0 (8.3) d 1 NN K K 8.1 d σd σd M = σd = E 2 d (8.4) ρ 2 d = I M = EI ρ 1 ρ = M EI ρ EI

i

Gauss Gauss ɛ 0 E ds = Q (1) xy σ (x, y, z) (2) a ρ(x, y, z) = x 2 + y 2 (r, θ, φ) (1) xy A Gauss ɛ 0 E ds = ɛ 0 EA Q = ρa ɛ 0 EA = ρea E = (ρ/ɛ 0 )e

Untitled

all.dvi

t = h x z z = h z = t (x, z) (v x (x, z, t), v z (x, z, t)) ρ v x x + v z z = 0 (1) 2-2. (v x, v z ) φ(x, z, t) v x = φ x, v z

( ) sin 1 x, cos 1 x, tan 1 x sin x, cos x, tan x, arcsin x, arccos x, arctan x. π 2 sin 1 x π 2, 0 cos 1 x π, π 2 < tan 1 x < π 2 1 (1) (

B line of mgnetic induction AB MN ds df (7.1) (7.3) (8.1) df = µ 0 ds, df = ds B = B ds 2π A B P P O s s Q PQ R QP AB θ 0 <θ<π

(1) D = [0, 1] [1, 2], (2x y)dxdy = D = = (2) D = [1, 2] [2, 3], (x 2 y + y 2 )dxdy = D = = (3) D = [0, 1] [ 1, 2], 1 {

5. F(, 0) = = 4 = 4 O = 4 =. ( = = 4 ) = 4 ( 4 ), 0 = 4 4 O 4 = 4. () = 8 () = 4

W u = u(x, t) u tt = a 2 u xx, a > 0 (1) D := {(x, t) : 0 x l, t 0} u (0, t) = 0, u (l, t) = 0, t 0 (2)

2 1 x 1.1: v mg x (t) = v(t) mv (t) = mg 0 x(0) = x 0 v(0) = v 0 x(t) = x 0 + v 0 t 1 2 gt2 v(t) = v 0 gt t x = x 0 + v2 0 2g v2 2g 1.1 (x, v) θ

(1.2) T D = 0 T = D = 30 kn 1.2 (1.4) 2F W = 0 F = W/2 = 300 kn/2 = 150 kn 1.3 (1.9) R = W 1 + W 2 = = 1100 N. (1.9) W 2 b W 1 a = 0

Part () () Γ Part ,

2.4 ( ) ( B ) A B F (1) W = B A F dr. A F q dr f(x,y,z) A B Γ( ) Minoru TANAKA (Osaka Univ.) I(2011), Sec p. 1/30

さくらの個別指導 ( さくら教育研究所 ) A a 1 a 2 a 3 a n {a n } a 1 a n n n 1 n n 0 a n = 1 n 1 n n O n {a n } n a n α {a n } α {a

δ ij δ ij ˆx ˆx ŷ ŷ ẑ ẑ 0, ˆx ŷ ŷ ˆx ẑ, ŷ ẑ ẑ ŷ ẑ, ẑ ˆx ˆx ẑ ŷ, a b a x ˆx + a y ŷ + a z ẑ b x ˆx + b

all.dvi

R = Ar l B r l. A, B A, B.. r 2 R r = r2 [lar r l B r l2 ]=larl l B r l.2 r 2 R = [lar l l Br ] r r r = ll Ar l ll B = ll R rl.3 sin θ Θ = ll.4 Θsinθ

1. z dr er r sinθ dϕ eϕ r dθ eθ dr θ dr dθ r x 0 ϕ r sinθ dϕ r sinθ dϕ y dr dr er r dθ eθ r sinθ dϕ eϕ 2. (r, θ, φ) 2 dr 1 h r dr 1 e r h θ dθ 1 e θ h

.5 z = a + b + c n.6 = a sin t y = b cos t dy d a e e b e + e c e e e + e 3 s36 3 a + y = a, b > b 3 s363.7 y = + 3 y = + 3 s364.8 cos a 3 s365.9 y =,

grad φ(p ) φ P grad φ(p ) p P p φ P p l t φ l t = 0 g (0) g (0) (31) grad φ(p ) p grad φ φ (P, φ(p )) xy (x, y) = (ξ(t), η(t)) ( )

( : December 27, 2015) CONTENTS I. 1 II. 2 III. 2 IV. 3 V. 5 VI. 6 VII. 7 VIII. 9 I. 1 f(x) f (x) y = f(x) x ϕ(r) (gradient) ϕ(r) (gradϕ(r) ) ( ) ϕ(r)

A = A x x + A y y + A, B = B x x + B y y + B, C = C x x + C y y + C..6 x y A B C = A x x + A y y + A B x B y B C x C y C { B = A x x + A y y + A y B B

0.6 A = ( 0 ),. () A. () x n+ = x n+ + x n (n ) {x n }, x, x., (x, x ) = (0, ) e, (x, x ) = (, 0) e, {x n }, T, e, e T A. (3) A n {x n }, (x, x ) = (,

1 12 ( )150 ( ( ) ) x M x 0 1 M 2 5x 2 + 4x + 3 x 2 1 M x M 2 1 M x (x + 1) 2 (1) x 2 + x + 1 M (2) 1 3 M (3) x 4 +

高校生の就職への数学II

OABC OA OC 4, OB, AOB BOC COA 60 OA a OB b OC c () AB AC () ABC D OD ABC OD OA + p AB + q AC p q () OABC 4 f(x) + x ( ), () y f(x) P l 4 () y f(x) l P

7-12.dvi

Gmech08.dvi

Quz Quz

II 2 II

液晶の物理1:連続体理論(弾性,粘性)

128 3 II S 1, S 2 Φ 1, Φ 2 Φ 1 = { B( r) n( r)}ds S 1 Φ 2 = { B( r) n( r)}ds (3.3) S 2 S S 1 +S 2 { B( r) n( r)}ds = 0 (3.4) S 1, S 2 { B( r) n( r)}ds

1 2 1 No p. 111 p , 4, 2, f (x, y) = x2 y x 4 + y. 2 (1) y = mx (x, y) (0, 0) f (x, y). m. (2) y = ax 2 (x, y) (0, 0) f (x,

TOP URL 1

meiji_resume_1.PDF

TOP URL 1

(1) 3 A B E e AE = e AB OE = OA + e AB = (1 35 e ) e OE z 1 1 e E xy e = 0 e = 5 OE = ( 2 0 0) E ( 2 0 0) (2) 3 E P Q k EQ = k EP E y 0


1 1. x 1 (1) x 2 + 2x + 5 dx d dx (x2 + 2x + 5) = 2(x + 1) x 1 x 2 + 2x + 5 = x + 1 x 2 + 2x x 2 + 2x + 5 y = x 2 + 2x + 5 dy = 2(x + 1)dx x + 1

29


77

all.dvi

1 1 3 ABCD ABD AC BD E E BD 1 : 2 (1) AB = AD =, AB AD = (2) AE = AB + (3) A F AD AE 2 = AF = AB + AD AF AE = t AC = t AE AC FC = t = (4) ABD ABCD 1 1

8 (2006 ) X ( ) 1. X X X 2. ( ) ( ) ( 1) X (a) (b) 1: (a) (b)

Gmech08.dvi

70 : 20 : A B (20 ) (30 ) 50 1

18 ( ) ( ) [ ] [ ) II III A B (120 ) 1, 2, 3, 5, 6 II III A B (120 ) ( ) 1, 2, 3, 7, 8 II III A B (120 ) ( [ ]) 1, 2, 3, 5, 7 II III A B (


No δs δs = r + δr r = δr (3) δs δs = r r = δr + u(r + δr, t) u(r, t) (4) δr = (δx, δy, δz) u i (r + δr, t) u i (r, t) = u i x j δx j (5) δs 2


Δ =,, 3, 4, 5, L n = n

6 2 2 x y x y t P P = P t P = I P P P ( ) ( ) ,, ( ) ( ) cos θ sin θ cos θ sin θ, sin θ cos θ sin θ cos θ y x θ x θ P


G:/SHIRAFUJI/テキスト類/EM1999/ALL/em99ps.dvi


1

重力方向に基づくコントローラの向き決定方法

n (1.6) i j=1 1 n a ij x j = b i (1.7) (1.7) (1.4) (1.5) (1.4) (1.7) u, v, w ε x, ε y, ε x, γ yz, γ zx, γ xy (1.8) ε x = u x ε y = v y ε z = w z γ yz

f (x) x y f(x+dx) f(x) Df 関数 接線 x Dx x 1 x x y f f x (1) x x 0 f (x + x) f (x) f (2) f (x + x) f (x) + f = f (x) + f x (3) x f

1. 4cm 16 cm 4cm 20cm 18 cm L λ(x)=ax [kg/m] A x 4cm A 4cm 12 cm h h Y 0 a G 0.38h a b x r(x) x y = 1 h 0.38h G b h X x r(x) 1 S(x) = πr(x) 2 a,b, h,π

S I. dy fx x fx y fx + C 3 C dy fx 4 x, y dy v C xt y C v e kt k > xt yt gt [ v dt dt v e kt xt v e kt + C k x v + C C k xt v k 3 r r + dr e kt S dt d

f : R R f(x, y) = x + y axy f = 0, x + y axy = 0 y 直線 x+y+a=0 に漸近し 原点で交叉する美しい形をしている x +y axy=0 X+Y+a=0 o x t x = at 1 + t, y = at (a > 0) 1 + t f(x, y

4 4 4 a b c d a b A c d A a da ad bce O E O n A n O ad bc a d n A n O 5 {a n } S n a k n a n + k S n a a n+ S n n S n n log x x {xy } x, y x + y 7 fx

. p.1/14

JKR Point loading of an elastic half-space 2 3 Pressure applied to a circular region Boussinesq, n =

[ ] 0.1 lim x 0 e 3x 1 x IC ( 11) ( s114901) 0.2 (1) y = e 2x (x 2 + 1) (2) y = x/(x 2 + 1) 0.3 dx (1) 1 4x 2 (2) e x sin 2xdx (3) sin 2 xdx ( 11) ( s

( ) Note (e ) (µ ) (τ ) ( (ν e,e ) e- (ν µ, µ ) µ- (ν τ,τ ) τ- ) ( ) ( ) (SU(2) ) (W +,Z 0,W ) * 1) 3 * 2) [ ] [ ] [ ] ν e ν µ ν τ e

さくらの個別指導 ( さくら教育研究所 ) A 2 P Q 3 R S T R S T P Q ( ) ( ) m n m n m n n n

, 3, 6 = 3, 3,,,, 3,, 9, 3, 9, 3, 3, 4, 43, 4, 3, 9, 6, 6,, 0 p, p, p 3,..., p n N = p p p 3 p n + N p n N p p p, p 3,..., p n p, p,..., p n N, 3,,,,

II (10 4 ) 1. p (x, y) (a, b) ε(x, y; a, b) 0 f (x, y) f (a, b) A, B (6.5) y = b f (x, b) f (a, b) x a = A + ε(x, b; a, b) x a 2 x a 0 A = f x (

入試の軌跡

1 26 ( ) ( ) 1 4 I II III A B C (120 ) ( ) 1, 5 7 I II III A B C (120 ) 1 (1) 0 x π 0 y π 3 sin x sin y = 3, 3 cos x + cos y = 1 (2) a b c a +

Kroneher Levi-Civita 1 i = j δ i j = i j 1 if i jk is an even permutation of 1,2,3. ε i jk = 1 if i jk is an odd permutation of 1,2,3. otherwise. 3 4

RA宣言.PDF

A (1) = 4 A( 1, 4) 1 A 4 () = tan A(0, 0) π A π

() x + y + y + x dy dx = 0 () dy + xy = x dx y + x y ( 5) ( s55906) 0.7. (). 5 (). ( 6) ( s6590) 0.8 m n. 0.9 n n A. ( 6) ( s6590) f A (λ) = det(a λi)

68 A mm 1/10 A. (a) (b) A.: (a) A.3 A.4 1 1

1 (Berry,1975) 2-6 p (S πr 2 )p πr 2 p 2πRγ p p = 2γ R (2.5).1-1 : : : : ( ).2 α, β α, β () X S = X X α X β (.1) 1 2

all.dvi

S I. dy fx x fx y fx + C 3 C vt dy fx 4 x, y dy yt gt + Ct + C dt v e kt xt v e kt + C k x v k + C C xt v k 3 r r + dr e kt S Sr πr dt d v } dt k e kt

知能科学:ニューラルネットワーク

知能科学:ニューラルネットワーク

( ; ) C. H. Scholz, The Mechanics of Earthquakes and Faulting : - ( ) σ = σ t sin 2π(r a) λ dσ d(r a) =

1 filename=mathformula tex 1 ax 2 + bx + c = 0, x = b ± b 2 4ac, (1.1) 2a x 1 + x 2 = b a, x 1x 2 = c a, (1.2) ax 2 + 2b x + c = 0, x = b ± b 2

Mott散乱によるParity対称性の破れを検証

PowerPoint プレゼンテーション

卓球の試合への興味度に関する確率論的分析

x,, z v = (, b, c) v v 2 + b 2 + c 2 x,, z 1 i = (1, 0, 0), j = (0, 1, 0), k = (0, 0, 1) v 1 = ( 1, b 1, c 1 ), v 2 = ( 2, b 2, c 2 ) v

Transcription:

4 4.1 conductor E E E 4.1: 45

46 4 E E E E E 0 0 E E = E E E =0 4.1.1 1) E =0 2) φ = 3) ρ =0 1) 0 2) E φ E = grad φ E =0 P P φ = E ds 0

4.1 47 0 0 3) ε 0 div E = ρ E =0 ρ =0 0 0 a Q Q/4πa 2 ) r E r 0 Gauss E r r) = Q 4πε 0 r 2 a<r) 0 0 r<a) 4.2 E 0 0 a r φ 0 0 a r 4.2: φp) = P S E r r)dr φs)

48 4 S φ ) =0 P S φr) = r Q 4πε 0 r 2 dr = Q 4πε 0 r a<r) φa) =Q/4πε 0 a) 0 φr) = Q 4πε 0 a 0 r a ). 4.2 4.1.2 electrostatic shielding 0 4.3 S Gauss 0 0 A B S 4.3:

4.1 49 0 4.3 A B 0 [ ] [ B ] A E ds =0= E ds E ds A B 0 0 0 Q Q >0 4.4 Gauss S 1 S 1 Q S 2 Q S 2 S 1 S 1 4.4: Q S 2 0 0 Q Q 0 Q Q 4.4 Q Q 0 Q

50 4 4.2 4.2.1 Laplace Poisson Laplace oulomb Gauss div E = ρ ε 0 4.1) E = grad φ. 4.2) φ div grad φ )= ρ ε 0. div grad φ )= φ )= 2 φ 2 Laplacian φ 2 φ = ρ ε 0 4.3) Poisson Poisson euation ρ =0 2 φ = 0 4.4) Laplace Laplace euation div grad φ ) = x i y j z k ) = 2 φ x 2 2 φ y 2 2 φ z 2 φ x i φ y j φ z k ) Laplacian 2 = 2 x 2 2 y 2 2 z 2. 4.5)

4.2 51 4.1 0 φr) = 4πε 0 r. r 0 Laplace r = x 2 y 2 z 2 Laplacian x 2 φ x 2 = ) ) φ x = x x 4πε 0 x x 2 y 2 z 2 ) 3/2 = 2x 2 y 2 z 2 4πε 0 x 2 y 2 z 2 ) 5/2 2 φ y 2 = 2y 2 z 2 x 2 4πε 0 x 2 y 2 z 2 ) 5/2, 2 φ z 2 = 2z 2 x 2 y 2 4πε 0 x 2 y 2 z 2 ) 5/2 x = y = z =0 0 4.2 a Q r Q r>a) 4πε 0 r φr) = [ Q 3 4πε 0 a 2 1 ) ] r 2 r a ) 2 a 3.2 Laplace Poisson Laplace r 2 = x 2 y 2 z 2 Laplacian 2 r 2 x 2 = 2 r 2 y 2 = 2 r 2 z 2 =2 2 φ = 1 Q ε 0 4πa 3 /3 a Poisson

52 4 4.2.2 Laplace 4.4) Poisson 4.3) 1) φ i 2) Q i Laplace 1) φ i Laplace φ i 0 2) Q i Gauss 4.2.3 n i i n i p ki k =1, 2, n n φ k = p ki i k =1, 2, n ) j j j φ k = p kj j k =1, 2, n ) n i i =1, 2, n

4.2 53 n 2 p kj k,j=1, 2, n φ k = p k1 1 p k2 2 p kn n k =1, 2, n ) n φ 1 p 11 p 12... p 1n φ 2 p 21 p 22... p 2n =....... 1 2. 4.6) φ n p n1 p n2... p nn n p kj j k p kj p kj = p jk 1 2nn 1) n φ k k =1, 2, n n j j =1, 2, n 4.6) 1 c 11 c 12... c 1n φ 1 2 c 21 c 22... c 2n φ 2 =.. 4.7)...... n c n1 c n2... c nn φ n c jk p kj c jk = jk 4.6) jk j k c jk k k 1V j p kj c jk c jj c jk j k n c jk = c kj 1 2 nn 1) 1 2 nn 1)

54 4 4.3 Laplace Laplace method of images 4.5 a >0 a, 0, 0) a, 0, 0) x, y, z ) φx, y, z) φx, y, z) = 4πε 0 φ x = 4πε 0 φ = y 4πε 0 φ = z 4πε 0 ) 1 [x a) 2 y 2 z 2 ] 1/2 1 [x a) 2 y 2 z 2 ] 1/2. 4.8) x a [x a) 2 y 2 z 2 ] 3/2 y [x a) 2 y 2 z 2 ] 3/2 z [x a) 2 y 2 z 2 ] 3/2 ) x a [x a) 2 y 2 z 2 ] 3/2 ) y [x a) 2 y 2 z 2 ] 3/2 ) z [x a) 2 y 2 z 2 ] 3/2

4.3 55 4.5: 2 φ x 2 = 2 φ y 2 = 2 φ z 2 = 2x a) 2 y 2 z 2 4πε 0 [x a) 2 y 2 z 2 ] 5/2 2x ) a)2 y 2 z 2 [x a) 2 y 2 z 2 ] 5/2 x a) 2 2y 2 z 2 4πε 0 [x a) 2 y 2 z 2 ] 5/2 x ) a)2 2y 2 z 2 [x a) 2 y 2 z 2 ] 5/2 x a) 2 y 2 2z 2 4πε 0 [x a) 2 y 2 z 2 ] 5/2 x ) a)2 y 2 2z 2 [x a) 2 y 2 z 2 ] 5/2 4.8) Laplace 2 φ x 2 2 φ y 2 2 φ z 2 =0. x =0 φ0,y,z)=0 4.8)

56 4 Laplace E = grad φ ) x a E x x, y, z) = 4πε 0 [x a) 2 y 2 z 2 ] 3/2 x a [x a) 2 y 2 z 2 ] 3/2 ) y E y x, y, z) = 4πε 0 [x a) 2 y 2 z 2 ] 3/2 y [x a) 2 y 2 z 2 ] 3/2 4.9) ) z E z x, y, z) = 4πε 0 [x a) 2 y 2 z 2 ] 3/2 z [x a) 2 y 2 z 2 ] 3/2 x =0 E x 0,y,z)= 2a 4πε 0 [a 2 y 2 z 2 ] 3/2, E y 0,y,z)=E z 0,y,z)=0 E y = E z =0 E x < 0 4.3 a R R >a >0 0 0 4.6 O A P D a O B A 4.6: B P 0 φ = 1 ) 4πε 0 PA PA =0 PB PB = AB : D O B=R RR = a 2 R = a2 R

4.3 57 OA φ = 1 ) 4πε 0 R a a R =0 = a R R a = a R P OP r OA OP θ AP = r 2 R 2 2Rr cos θ, BP = r 2 R 2 2R r cos θ P ) 1 φr, θ, ϕ) = 4πε 0 r 2 R 2 2Rr cos θ r 2 R 2 2R r cos θ = 1 4πε 0 r 2 R 2 2Rr cos θ ) a R 2 r 2 a 4 2a 2 Rr cos θ 4.10) P r = a φ =0 E = grad φ grad φ = φ r e r 1 φ r θ e θ 1 φ r sin θ ϕ e ϕ 4.10) E r r, θ, ϕ) = [ r R cos θ 4πε 0 [r 2 R 2 2Rr cos θ] 3/2 ar 2 r a 3 ] R cos θ [R 2 r 2 a 4 2a 2 Rr cos θ] 3/2 [ R sin θ 1 E θ r, θ, ϕ) = 4πε 0 [r 2 R 2 2Rr cos θ] 3/2 a 3 ] [R 2 r 2 a 4 2a 2 Rr cos θ] 3/2 E ϕ r, θ, ϕ) =0 r = a E r a, θ, ϕ) = 4πε 0 a R 2 a 2 [R 2 a 2 2aR cos θ] 3/2, E θ a, θ, ϕ) =E ϕ a, θ, ϕ) =0 E θ = E ϕ =0 E r < 0 4.7

58 4 O B 4.7: 4.4 a 0 R R >a 0 O R A OA O R R = a 2 /R = a/r = 0. 4.10) /4πε 0 r) E θ E ϕ E r /4πε 0 r 2 ) φ = /4πε 0 a)

4.4 59 4.4 4.4.1 φ φ = 1 4.11) capacitance F= V 1 a φ = /4πε 0 a) φ = 1 4πε 0 a =4πε 0 a φ φ = f) 1 φ 1 = f 1 ) 2 φ 2 = f 2 ) 1 2 φ 1 φ 2 = f 1 2 ) f 1 )f 2 )=f 1 2 ) f) k f) =k k =1/ F µf =10 6 F pf = 10 12 F 4.5 R =6.37 10 3 km =4πε 0 R =7.08 10 4 F = 708 µf

60 4 4.4.2 condenser A >0 B φ A φ B φ A φ B = 1 4.12) spherical condenser 4.8 a b b >a Q Q >0 r a<r<b r b Q a Q 4.8: Gauss ε 0 Er) 4πr 2 = Q Er) = Q 4πε 0 r 2 φ = b a Er)dr = Q 1 4πε 0 a 1 ) b = Q b a 4πε 0 ab = Q φ =4πε ab 0 b a 4.13)

4.4 61 b 4.13) b ab =4πε 0 b a 4πε 0 a a = b b a b a b 4.6 parallel plate condenser 4.9 S d Q d Q 4.9: Q Q σ = ±Q/S Gauss 0 A ε 0 EA= σa E = σ ε 0 = Q ε 0 S φ φ = B A E ds = Ed.

62 4 = Q φ = ε 0 S d. 4.14) S d 4.7 cylindrical condenser 4.10 a b a b 4.10: L QL r a<r<b r Gauss ε 0 E 2πrL = QL E = Q 2πε 0 r φ = b a E dr = Q 2πε 0 log b a = Q φ =2πε 1 0 logb/a) 4.15)

4.4 63 4.4.3 1 2 1 = 1 1 4.16) 1 2 = 1 2. 4.17) 4.11 1 1 2 A 1 A 3 A 2 1 1 A 1 A 2 2 2 2 4.11: 4.11 1 2 A 1 A 2 A 3 A 1 A 3 A 2 φ 1 φ 2 A 1 A 3 φ φ = φ 1 φ 2 = 1 2 φ = 1 1 1 2 4.16) 4.11 1 2 A 1 A 2

64 4 φ 1 φ 2 φ = φ 1 = φ 2 A 1 A 2 φ 1 = φ 2 1 1 2 2 = φ = 1 2 φ = 1 2 = 1 φ 2 φ = 1 φ 1 2 φ 2 = 1 2 n n 1 = n k=1 1 k, = n k. k=1 4.8 4.12 P 1 P 2 P 1 A 1 P 1 A 1 P 2 A 2 P 2 A 2 4.12: A 1 A 2 1 = /3 A 1 A 2 =4/3 4.12 4/11