(1)由动能定理得
mgLcosa-qE(L+Lsina)=0 由A-B电场力做负功,电场力水平向右,小球带负电
(2)qE=mgcosa/(1+sina)
由动能定理得
mgL-qEL=0.5mv^2
F-mg=mv^2/L
F=3mg-2qE=mg[3-2cosa/(1+sina)]
(1)由动能定理得
mgLcosa-qE(L+Lsina)=0 由A-B电场力做负功,电场力水平向右,小球带负电
(2)qE=mgcosa/(1+sina)
由动能定理得
mgL-qEL=0.5mv^2
F-mg=mv^2/L
F=3mg-2qE=mg[3-2cosa/(1+sina)]