问题 问答题

如图所示,带等量异种电荷的两平行金属板竖直放置(M板带正电,N板带负电),板间距为d=80cm,板长为L,板间电压为U=100V.两极板上边缘连线的中点处有一用水平轻质绝缘细线拴接的完全相同的小球A和B组成的装置Q,Q处于静止状态,该装置中两球之间有一处于压缩状态的绝缘轻质小弹簧(球与弹簧不拴接),左边A球带正电,电荷量为q=4×10-5C,右边B球不带电,两球质量均为m=1.0×10-3kg.某时刻,装置Q中细线突然断裂,A、B两球立即同时获得大小相等、方向相反的速度 (弹簧恢复原长).若A、B之间弹簧被压缩时所具有的弹性势能为1.0×10-3J,小球A、B均可视为质点,Q装置中弹簧的长度不计,小球带电不影响板间匀强电场,不计空气阻力,取g=l0m/s2.求:

(1)为使小球不与金属板相碰,金属板长度L应满足什么条件?

(2)当小球B飞离电场恰好不与金属板相碰时,小球A飞离电场时的动能是多大?

(3)从两小球弹开进入电场开始,到两小球间水平距离为30cm时,小球A的电势能增加了多少?

答案

小球分离的过程中机械能守恒:E=2×

1
2
m
v20

小球分离时获得的初速度:v0=1m/s

进入电场后A球在水平方向做匀减速运动,B做匀速运动,所以是B先碰到极板.B向右做平抛运动,时间:

t=

d
2v0
=0.4s

竖直方向的位移:y=

1
2
gt2=0.8m

即为使小球不与金属板相碰,金属板的长度:L<0.8m

(2)水平方向A球向左做匀减速运动,其加速度a=

qE
m
=
qU
md
=5m/s2,方向向右;

A球飞离电场时的位移:s1=v0t-

1
2
at2=0

由动能定律,A球离开电场时的动能:

EK=

1
2
m
v20
+
qU
D
s1+mgL=8.5×10-3J

(3)两小球进入电场后,竖直方向都做自由落体运动,因此两小球在运动的过程中始终位于同一水平线上,

当两球间的距离是30cm时,v0

t′ 
+(v0t′-
1
2
at2)=s

解得:

t′1
=0.2s,
t′2
=0.6s
(舍去)

此时,A球水平位移为;sA=v0

t′1
-
1
2
Ud
dm
t′21
=0.1m

球A的电势能增加:△E=qEsA=5×10-4J

答:(1)为使小球不与金属板相碰,金属板长度L应满足L<0.8m;

(2)当小球B飞离电场恰好不与金属板相碰时,小球A飞离电场时的动能是8.5×10-3J;

(3)从两小球弹开进入电场开始,到两小球间水平距离为30cm时,小球A的电势能增加了5×10-4J.

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