An electron of charge `1.6xx10^(-19)C` and mass `9.1xx10^(-31)` kg, travelling along the X-axis with a uniform velocity of `10^(6)m.s^(-1)`, enters in a uniform electric field of `10^(3)V.m^(-1)` acting perpendicular to the X-axis. If the electric field extends over a length of 2 cm along the X-axis, what will be the deflection of the electron along the direction of the field when it emerges from it?

Explanation

Here, electric field `E=10^(3)` V/m is directed along Y-axis and effective upto a length L = 2 cm along X -axis [Fig. 2.31].
An electron is projected from O with an initial speed of `v_(x)=10^(6)` m/s along + X-axis. Charge of the electron, `e=1.6xx10^(-19)`C and mass of the electron, `m=9.1xx10^(-31)kg`.

If we consider the motion of the electron along Y-axis, the force on the electron due to the electric field, `f_(y)=eE`
`therefore` Acceleration of the electron along Y-axis,`a_(y)=(eE)/m`
Let the deflection of the electron when it is emitted from the electric field be y and the time taken to cover that vertical distance be t.
`thereforey=1/2cdota_(y)t^(2)`
or, `y=1/2cdot(eE)/mt^(2)` ...(1)
Along X-axis, force on the electron, `F_(x)=0`
`therefore` Acceleration, `a_(x)=0`
`therefore` Velocity, `v_(x)=10^(6)m//s` (constant)
Now distance covered along X-axis in time t is L =`v_(x)t`
`thereforet=l/(v_(x))` ..(2)
From equation (1) and (2),
`y=1/2cdot(eE)/mcdot(L^(2))/(v_(x)^(2))`
`=1/2xx(1.6xx10^(-19)xx10^(3)xx(.02)^(2))/(9.1xx10^(-31)xx(10^(6))^(2))=0.0351` m
`=3.51` cm
Thus, required deflection is 3.51 cm.