An electron is projected with an initial speed `v_(0)=1.60xx10^(6)ms^(-1)` into the uniform field between the parallel plates as shown in fig. Assume that the field between the plates is uniform and directed vertically downward, and that the field outside the plates is zero. the electrons enters the field at a point midway between the plates. Mass of electron is `9.1xx10^(-31)kg`.

The vertical displacement traveled by the proton as it exits the region between the plated is (mass pf proton is `1.67xx10^(-27)kg`).

Options

A. `124NC^(-1)`
B. `364NC^(-1)`
C. `224NC^(-1)`
D. `520NC^(-1)`

Answer

B

Explanation

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.