Question

Question 1.34: Suppose that the particle in Exercise in 1.33 is an electron projected with velocity vx= 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, me = 9.1 × 10 −31 kg.)

Class 12 - Physics - Electric Charges And Fields Electric Charges And Fields Page-50

Answer 1

if the electron is released just near the negatively charged plate, then deflection in vertical direction , y = 0.5 cm .

use formula, $\large{\bf{\frac{qEL^2}{2mv_x^2}}}$
where L is displacement of particle in horizontal direction ( the point where electron will strike the upper plate ).
$v_x$ intial velocity of Particle in x - direction. E is electric field and m is the mass of Particle.

given, m = $9\times10^{-31}Kg,q=1.6\times10^{-11}C\\v_x=2\times10^6m/s,E=9.1\times10^2N/C$

now, L² = 2m$v_x^2$y/qE
= 2 × 9.1 × 10^-31kg × (2 × 10^6m/s)² × 0.5 × 10^-2m/(1.6 × 10^-19C × 9.1 × 10² N/C)

= 2.5 × 10^-4 m²
taking square root both sides,
L = 1.6 × 10^-2 m = 1.6 cm