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.

Options

A. `V_(x)=10^(5)m//s`
B. `V_(y)=1.89xx10^(6)m//s`
C. `V_(x)=0`
D. `V_(y)=10^(5)m//s`

Explanation

The particle will hit the upper plate because of attraction
Electric field = `(30)/(0.3xx10^(-2))=10000V//m`
Upward acceleration = `("force")/("mass")=(eE)/(m)=(1.6xx10^(-19)xx10000)/(9xx10^(-31))=1.78xx10^(15)m//s^(2)`
Take the initial direction of the particles as the x-axis and the field direction as the y-axis.
From, `v^(2)-v_(0)^(2)=2as`,
`v_(y)^(2)=2xx1.78xx10^(15)xx(0.1xx10^(-2))`
`v_(y)=1.89xx10^(6)m//s`
The initial velocity along the x-axis remains unchanged as there is no acceleration along the direction.
`therefore v_(y)=1.89xx10^(6)m//sandv_(x)=10^(5)m//s`