o12304 2 2S 3 2S 4 2S 按电场叠加原理可求得: AB!C E Q A 28S B eOs C260 (2)第二板接地 则04与大地构成一导体4=0 E Q o1+o2 同理可得: 2+σ3=0 σ1+σ2+σ3=0 联立求解:1=002=s Q 3 EA=EC=O E A C 6按电场叠加原理可求得: 2 S Q E o A =− (2)第二板接地 则 4 与大地构成一导体 4 = 0 同理可得: S Q 1 + 2 = 2 + 3 = 0 1 + 2 + 3 = 0 联立求解: 1 = 0 S Q 2 = S Q 3 = − S Q E E 0 E o A C B = = = 1 2 3 4 .P A B C E Q 2S Q 1 = 2 = 2S Q 3 = − 2S Q 4 = 2 S Q E o B = 2 S Q E o C = 6