A particle having a charge of `1.6 x 10^(-19) C` enters midway between the two plates of a parallel plate capacitor. The initial velocity of particle is parallel to the plates. A potential difference of300 Vis applied between the two plates. lfthe length of the plates is 10 cm and they are separated by 2 cm, calculate the greatest initial velocity for which the particle will not be able to come out ofthe plates. The mass of the particle is `12 xx 10^(-24) kg`
`[10^(4) m//s]`

Explanation

The situation is shown in figure.
Here we know the electric field can be given as
`E=(V)/(d)=(300)/(2//100)=15000v//m`
As the particle does not come out, its maximum

deflection in vertical direction can be
` y = 1 cm = 10^2 m`
we known that `y=1/2at^2=1/2.(qE)/(m)(l/u)^2`
`["As a "= (qE)/(m)and t=l/u]`
or `u^2=1/2.(qE)/(my).x^(2)`
`=1/2((1.6xx10^9-19)(15000))/((12xx10^(-24))(10^(-2)))(1/10)^(2)=10^8`
`u=10^4m//s`