Suppose that the particle is an electron projected with velocity v_{x }= 2.0 × 10^{6} m s^{−1}. If E between the plates separated by 0.5 cm is 9.1 × 10^{2} N/C, where will the electron strike the upper plate? (|e| = 1.6 × 10^{−19} C, m_{e }= 9.1 × 10^{−31 }kg)

#### Solution

Velocity of the particle, v_{x} = 2.0 × 10^{6} m/s

Separation of the two plates, d = 0.5 cm = 0.005 m

Electric field between the two plates, E = 9.1 × 10^{2} N/C

Charge on an electron, q = 1.6 × 10^{−19} C

Mass of an electron, m_{e }= 9.1 × 10^{−31} kg

Let the electron strike the upper plate at the end of plate L, when deflection is s.

Therefore,

`"s" = ("qEL"^2)/(2"mv"_"x"^2)`

`"L" = sqrt((2"dmv"_"x"^2)/("qE"))`

= `sqrt((2 xx 0.005 xx 9.1 xx 10^-31 xx (2.0 xx 10^6)^2)/(1.6 xx 10^-10 xx 9.1 xx 10^2))`

= `sqrt(0.025 xx 10^-2)`

= `sqrt(2.5 xx 10^-4)`

= `1.6xx10^-2` m

= 1.6 cm

Therefore, the electron will strike the upper plate after travelling 1.6 cm.