En fy E 12p E B 4丌r E 1 E EAE 4丌Er 结论 E∝pE∝3 0 2 4 1 r p EA     = 结论 3 1 r E  3 4 0 1 r p EB     = − l r y x • • B A l  r E+  E+  E−  E−  EB  EA  E  p