①②⑤⑥→σ1=(QA+QB)/2S 2=(QA-QB)/2S 3=-(QA-QB)/2S =(QA+QB)/2S [讨论]①若QB=0 则σ1=04=Q4/2S o2=-o3=Q4/2S 两板内、外侧均有电荷！ ①②⑤⑥→ 1= (QA+QB )/2S 2= (QA-QB )/2S 3= -(QA-QB )/2S 4= (QA+QB )/2S [讨论]①若QB=0 则1=4=QA/2S 2= -3=QA/2S ——两板内、外侧均有电荷!