A particle of mass `9xx10^(-31)` kg having a negative charge of `1.6xx10^(-19)C` is projected horizontally with a velocity of `10^(6)ms^(-1)` into a region between two infinite horizontal parallel plates of metal. The distance between the plates is `d=0.3cm` and the particle enters 0.1cm below the top plate, THe top and bottom plates are connected, respectively to the positive and negative terminal of a 30V battery. Find the components of the velocity of the particle just before it hits one of the plates.

Explanation

We known that between two parallel plates electric field can be given as
`E=(V)/(d)`
Here V = 30 volt and d = 0.3 cm `=3xx10^(-3)m`
Thus we have `E=(30)/(3xx10^(-3))=10^(4)N//C`
Force on the particle of negative charge moving between the plates
` F = e × E = 1.6 × 10^( 19) × 10^(4) = 1.6 × 10^(15)` newton The direction of force will be towards the positive plate i.e., upward.
Now acceleration of the particle is
or `a=(eE)/(m)`
or `a=(1.6xx10^(-15))//(9xx10^(-31))`
or `a = 1.77xx10^(15)m//sec^(2)`
As the electric intensity E is acting in the vertical direction the horizontal velocity v of the particle remains same. if y is the displacement of the particle, in upward direction, we have
`y = 1/2 at ^2`
Here, `y = 0.1 cm = 10^-3m,a =1.77xx10^15m//sec^2`
Thus, `10^-3=1/2xx(1.77xx10^(-15))(t^2)`
Solving we get `= 1.063 × 10^-10` second component of velocity in the direction of field is given by
`V_(y)=at`
`=(1.77xx10^(15))(1.063xx10^(-10))`
`=1..881xx10^(4)m//s` .