Suppose that the particle in Exercise in 1.33 is an electron projected with velocity vₓ = 2.0 × 10⁶ m s⁻¹. If E between the plates separated by 0.5 cm is 9.1 × 10² N/C, where will the electron strike the upper plate? (|e|=1.6 × 10⁻¹⁹ C, mₑ = 9.1 × 10 ⁻³¹kg.)

Velocity of the particle, V_x = 2.0 x 10⁶ m/s

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

Electric field between the two plates, E = 9.1 x 10² N/C

Charge on an electron, q = 1.6 x 10^–19 C

Mass of an electron, m_e= 9.1 x 10^-31 kg

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

s= (qEL²)/(2mV²_x)

=>L = √ (2smV²_x)/qE

=  √ 2 x 0.005 x 9.1 x 10^-31)/(1.6 x 10^–19 x 9.1 x 10²

= √ 0.00025  = 0.016 m=1.6 cm

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