[解法二]按照电势能定义式求 打g打4+% 42=0 4π8R 4π6R2 [解法二]按照电势能定义式求 ] ∫ = dquW 2 1 e ∫ = 11d 2 1 qu ∫ + 22d 2 1 qu q R q R q ⎟ ⎟ ⎠ ⎞ ⎜ ⎜ ⎝ ⎛ − = + 10 π 20 42 π 4 1 εε )(0 2 1 −××+ q ⎟ ⎟ ⎠ ⎞ ⎜ ⎜ ⎝ ⎛ = − 210 2 11 8 π RR q ε 10 20 1 4 π 4 π R q R q u εε − += u 2 = 0 R 2 R 1 q −q