Electromagnetics in Medical Physics

Transcription

1 Electromagnetics in Medical Physics Part I. Impulses in Nerve Tong In Oh Department of Biomedical Engineering Impedance Imaging Research Center (IIRC) Kyung Hee University Korea

2 Electricity 2

3 Atom and Charge 질량수 = 양성자수 + 중성자수 12 C 6 원소기호 원자번호 = 양성자수 = 중성원자의전자수 3

4 Periodic Table of Elements 4 그림출처 : Daum 통합검색

5 Property: Conductivity Insulator vs. Conductor Proton Electron Conductivity Conductivity 5

6 Charged Particle and Charge Density Free electron and hole are mobile Unbounded ion and molecule are mobile Bounded atom and molecule are immobile but may vibrate Polar molecule has no net charge but dipole moment and may rotate Mass Charge Size Position m, Q, r, d 6

7 Electric Field Space with nothing Space with a single charged particle Space with two charged particles z r Q r y Q 0 x 7

8 Coulomb s Law (a) The two glass rods were each rubbed with a silk cloth and one was suspended by thread. When they are close to each other, they repel each other. (b) The plastic rod was rubbed with fur. When brought close to the glass rod, the rods attract each other. Ch21, HRW, Fundamentals of Physics, Wiley 8

9 Coulomb s Law (a) Two charged rods of the same sign repel each other. (b) Two charged rods of opposite signs attract each other. Plus signs indicate a positive net charge, and minus signs indicate a negative net charge. Ch21, HRW, Fundamentals of Physics, Wiley 9

10 Induced Charge 10

11 쿨롱의법칙, 중첩 쿨롱의법칙 : F = 1 q 1 q 2 4πε 0 r Nm 2 C 2 4πε 0 where ε 0 = C 2 /N.m 2 is the permittivity constant. The ratio 1/4πε 0 is often replaced with the electrostatic constant (or Coulomb constant) k= N.m 2 /C 2. Thus k = 1/4πε 0. Multiple Forces: If multiple electrostatic forces act on a particle, the net force is the vector sum (not scalar sum) of the individual forces. 11

12 Coulomb s Law F QQ 1(r r1) 4 r r QQ2(r r 4 r r ) 3 QQ 4 0 N (r r r r N N ) 3 Q 4 0 N k1 Qk (r r r r k k 3 ) (4.8) F r 1 r Q 1 r F r 2 Q2 Source Points Q Field Point F r N r F r k r Q k Q N 12

13 Electric Field Electric Field The electric field E at any point is defined in terms of the electrostatic force F that would be exerted on a positive test charge q 0 placed there: Ch22, HRW, Fundamentals of Physics, Wiley 13

14 Electric Field E Q 1(r r1 ) Q 3 40 r r1 4 N 1 Qk (r rk ) 3 4 r r 0 k1 k 2 0 (r r2 ) r r 2 3 Q 4 N 0 (r rn ) r r N 3 (4.12) E r 1 r Q 1 r E r 2 Q2 Source Points Q=1C E r k r Q k Field Point E r N r Q N 14

15 Electric Flux ( 전기장의선속 ) Electric field vectors and field lines pierce an imaginary, spherical Gau ssian surface that encloses a partic le with charge +Q. Now the enclosed particle has charge +2Q. Can you tell what the enclos ed charge is now? Answer: -0.5Q Ch23, HRW, Fundamentals of Physics, Wiley 15

16 Electric Flux Electric flux dϕ through a patch element with area vector da Now we can find the total flux by integrating the dot produc t over the full surface. The total flux through a surface is given by The net flux through a closed surface (which is used in Gaus s law) is given by where the integration is carried out over the entire surface. Ch23, HRW, Fundamentals of Physics, Wiley 16

17 Flux through a Surface AS 17

18 Flux through a Surface S A cosds S : Surface S에서적분 A a n ds S A ds (3.13) S A ds : Surface S에서감싸서적분 (3.14) Closed Surface 18

19 Gauss Law Gauss law relates the net flux ϕ of an electric field through a closed surface (a Gaussian surface) to the net charge q enc that is enclosed by that surface. It tells us that We can also write Gauss law as Two charges, equal in magnitude but opposite in sign, and the fiel d lines that represent their net electric field. Four Gaussian surface s are shown in cross section. S1: Positive, S2: Negative, S3: Zero(q enc = 0), S4:Zero 전기장의선속 : E n ds = q ε 0 = 4πq 4πε 0 Ch23, HRW, Fundamentals of Physics, Wiley 19

20 Applying Gauss Law: Spherical Symmetry In the figure, applying Gauss law to surface S 2, for which r R, we would find that And, applying Gauss law to surface S 1, for which r < R, A thin, uniformly charged, spherical shell with total charge q, in cross section. Two Gaussian surfaces S1 and S2 are also shown in cross section. Surface S2 encloses the shell, and S1 encloses only the empty interior of the shell. 점전하 E n ds = E ds = 4πr 2 E = q ε 0, E = 점전하의경우에대한쿨롱의법칙유도 q 4πε 0 r 2 20

21 Applying Gauss Law: Cylindrical Symmetry Figure shows a section of an infinitely long cylindrical plastic rod with a uniform charge density λ. The charge distribution and the field have cylindrical symmetry. To find the field at radius r, we enclose a section of the rod with a concent ric Gaussian cylinder of radius r and height h. The net flux through the cylinder from Gauss Law reduces to yielding 세포막 ( 직선전하 ) dq = λdx, λ = 선전하밀도 (Cm 1 ) E න න ds = 2πrLE = λl ε 0, E = λ 2πε 0 r, E = 1 2λ 4πε 0 r A Gaussian surface in the form of a closed cylinder surrounds a sec tion of a very long, uniformly cha rged, cylindrical plastic rod. 21

22 Applying Gauss Law: Planar Symmetry Non-conducting Sheet Figure (a-b) shows a portion of a thin, infinite, nonconducting sheet with a uniform (positive) surface charge density σ. A sheet of thin plastic wrap, uniformly charged on one side, can serve as a simple model. Here, Is simply EdA and thus Gauss Law, becomes where σa is the charge enclosed by the Gaussian surface. This gives 22

23 Applying Gauss Law: Planar Symmetry Two conducting Plates Figure (a) shows a cross section of a thin, infinite conducting plate with excess positive charge. Figure (b) shows an identical plate with excess negative charge having the same magnitude of surface charge density σ1. Suppose we arrange for the plates of Figs. a and b to be close to each other and parallel (c). Since the plates are conductors, when we bring them into this arrangement, the excess charge on one plate attracts the excess charge on the other plate, and all the excess charge moves onto the inner faces of the plates as in Fig.c. With twice as much charge now on each inner face, the electric field at any point between the plates has the magnitude [ 교재그림 6.11 참조 ] 23

24 EOD