电势差为F=Q2 d+ Q Qd260=(28+6 s3 38 s 38 Q E0S38 2E+E0 C 38 0(2E+60) 介质板内电场能量 d od W=E=E()2S= 3 68S10 介质板内电场能量 W E V 2 2 1 =  2 3 1 2 d ) S S Q (  =  S Q d 6 2 = (2 ) 3 0 0     + = = d S V Q C (2 ) 3 0 0    + = C C d S Q d S Q V 3 1 3 2 0   电势差为 = + ) 3 3 2 ( 0 0    = + S Qd ) 3 2 ( 0 0     + = S Qd