電網(wǎng)專(zhuān)業(yè)課考點(diǎn)總結(jié)電磁場(chǎng)考綱新增第二章

SteadyElectricgg=1?rS/

fi

注 CurrentinConductiveMedia1.1. =d d 體電荷v

J=AI=sJA A AII=l( en)dl線電荷vdI=t = I= =

JEOhm’sLawRR= d = =Edd說(shuō) ①J與E成正比，且方向一致PP=UI dPdP=(JdS)(E= EdV

JEJp= E=γE γ

EEfeeqSourceEMFand0therField1.1. ~ E=Ec+EeJ=g(Ec+Ee

e=lE

Edl=(Ec+Ee)dl= dl+

Ee =0+e=BasicEquations?Boundary①Ee+cl= e+cl= dl+lEe)((·E)dS=l·E)dS=l ·E=

dS=dS=-sI=?q?t==sJdS J=

JdV=JdJdSs 微分形JJ= dl=·E=J=gEJJ=gE·E=

E=-

g=J=0fi(E)=-gf+E 注

JdS lldl=dl+ldl==U1+U2+L=U

lEdl=lsDdS=

=E2 J1n=J

Edl=lsJdSl

=J2

=g2E2 -D2 =

e -e = 2

-e2

E1n

=g

-g

2理想介質(zhì)交界面i0，只有速縛電荷而ga討 ga

= 222afi2

E1t=E2tfi 2g?1討

2=0，J2=0J1n=J2n=0空氣中E2 =J2n

0?0

E1n=D2n-D1n=2E2n=

E1t=E2t=J1t/1?討

fi

J1=1E1=E1=0 E=

E2=E2==0

fSfS=f= 例例解解g 1區(qū)域g2f112f11r2 =j22f212r2j2=j=0=f1j=π=U24 1=212

通解f1=Aj+B 電位f1

p(12

(121+f2

41U0 p(1+2E1=-

ejE2=

41U0 p(1+2

p(1+2=010E2

40U0p1

(1

·E=D=D=eE=-2f= q=sD

·E=J=J=E=-2f= I=s

E1t=E2t =D2f=

E1t=E2tJ1n=Jf=

=e1 解

U= eEDEeED

r

f= lnR2=

ln

ln

fi

J= e2

E=J= gEJg2prgEJU=

ln2πg(shù)

ln 結(jié) (I=1-g2I,1+2

22 I1+2 實(shí)驗(yàn)方法：

（如導(dǎo)電紙模擬（如電解槽模擬>>ConductanceandGround

G= =JdS=γE

E E①

G=I②

③ 2f=

C= U=sDdS=esEdS=e

s=g

E 例例解設(shè)IfiJ解

fiE

U=Edl=r1 dr=I lnr2 r2rg

電導(dǎo) U

r

2pe

,

Ce

R=1 ln 例解f= j=q時(shí),f=U0解r2

ff=(U0

E=-f=-?fej=- J=E=-U rqI=JdS

b0

(-e)hdr(-e

U0h arq

電導(dǎo)G=

=hln

(SU q2.接地電阻(Ground

~ R=f= (接地器的電位 例例IfiJ

fiE=

￥4r￥

dr

4a

R=f= C= C=4pea

G=1=aR例例解

當(dāng)h較大f

1(1+14pag

例例解設(shè)解E ￥

J=r2