1、在A点,mg-FA=ma=3mg/4,则在A点电场力FA=mg/4.即kQq/h^2=mg/4
FB/FA=kQq/(0.25h)^2:kQq/h^2=16,则FB=4mg.在B点,FB-mg=ma,则a=3g,方向向上
2、动能定理:WAB+mg*(0.75h)=0-0,则电场力做功WAB=-0.75mgh=0.75*(4kQq/h^2)*h=qUAB,则UAB=3kQ/h
1、在A点,mg-FA=ma=3mg/4,则在A点电场力FA=mg/4.即kQq/h^2=mg/4
FB/FA=kQq/(0.25h)^2:kQq/h^2=16,则FB=4mg.在B点,FB-mg=ma,则a=3g,方向向上
2、动能定理:WAB+mg*(0.75h)=0-0,则电场力做功WAB=-0.75mgh=0.75*(4kQq/h^2)*h=qUAB,则UAB=3kQ/h