Question

Suppose that the particle in the above question is an electron projected with velocity `v_x = 2.0 xx 10^(6) m s^(–1)`. If E between the plates separated by `0.5 cm` is `9.1 xx 10^(2) N//C`, where will be the electron strike the upper plate? `(|e|=1.6 xx 10^(-19) C, m_e = 9.1 xx 10^(–31) kg)`.

Answer 1

Velocity of the particle, `V_(X)=2.0xx10^(6)m//s`
Separation of the two plates, d = 0.5 cm = 0.005 m
Electric field between the two plates, E = `9.1xx10^(2)N//C`
Charge on an electron, q = `1.6xx10^(-19)C`
Mass of an electron, `m_(e)=9.1xx10^(-31)` kg
Let the electron strike the upper plate at the end of plate L, when deflection is s.
Therefore,
`s=(qEL^(2))/(2mv_(x)^(2))`
`L=sqrt((2dmv_(x)^(2))/(qE))`
`=sqrt((2xx0.005xx9.1xx10^(-31)xx(2.0xx10^(6))^(2))/(1.6xx10^(-10)xx9.1xx10^(2)))`
`=1.6xx10^(-2)`
= 1.6 cm
Therefore, the electron will strike the upper plate after travelling 1.6 cm.