2.45伏,上15伏。 B C 解题思路:作积分路径A→B: B 0.2 E·d= 45 0.1 4兀E0 4兀Eo(0.10.2 作积分路径C→B: =-r,E.d=-E.c=E·dr B 0.2 E dl dr 15 034兀Enr 4zEo(0.3022. 45伏， −15伏。. q A B C 解题思路：作积分路径A → B: 45 0 2 1 0 1 1 4 4 0 0 2 0 1 2 0  =      =  = = −   . . q dr r q U E dl . . B A A     作积分路径C → B: (dl = −dr E dl = −E dr = E dr)       ， 15 0 2 1 0 3 1 4 4 0 0 2 0 3 2 0  = −      =  = = −   . . q dr r q U E dl . . B C C    