Correct Answer - D
Passing between the charged plates, the electron feels a force upward and just misses the top plate. The distance it travels in upward and just misses the to plate. The distance it travels in the y-direction is 0.005m. Time of flight is `t=(0.0200m)/(1.60xx10^(6)ms^(-1))`
`=1.25zz10^(-8)s`
the initial y-velocity is zero.
Now, `y=v_(0y)t+1/2at^(2)`
so, `0.005m=1/2a(1.25xx10^(-8)s)^(2)`
or, `a=6.40xx10^(13)ms^(-2)`
But also `a=(F)/(m)=(eE)/(m_(e))`
`E=((9.1xx10^(-31)kg)(6.40xx10^(13)ms^(-2)))/(1.60xx10^(-19)C)`
`=364NC^(-1)`
Since the proton is more massive, it will acceletate less and not hit the plates. To find the vertical displacement when it exists the plates we use the kinematic equation again.
`y=1/2at^(2)=1/2(eE)/(m_p)(1.25xx10^(-8)s)^(2)`
`=2.73xx10^(-6)m`
As mentioned in (b), the proton will not hit one of the plates because although the electric force felt by the proton is the same as the electrons felt, a smaller accelaration produced by the electric force is much greater than g, it is reasonable to ignore gravity.