Question

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 × 10–31 kg.)

Answer 1

If the electron is released just near the negatively charged plate , then we have to use formula,

$\bf{y=\frac{qEL^2}{2mv_x^2}}$ [ from Exercise 1.33 ]

here, q is charge on electron. i.e., q = 1.6 × 10^-19 C

E is electric field between plates, i.e., E = 9.1 × 10² N/C

y is seperation between plates. i.e., y = 0.5 cm

m is mass of electron i.e., m = 9.1 × 10^-31 Kg

$v_x$ is velocity of electron in horizontal direction. i.e., $v_x$ = 2 × 10^6 m/s

and L is the horizontal displacement.

so, $L^2=\frac{2mv_x^2}{qE}$

= {2 × 9.1 × 10^-31 × (2 × 10^6)²}/{1.6 × 10^-19 × 9.1 × 10² }

= 2.5 × 10^-4

so, L = 1.6 × 10^-2 m or 1.6 cm

hence, answer is 1.6 cm

Answer 2

Answer:

• 1.6 cm

Explanation:

• Velocity of the particle, $v_{x} = 2.0 \times 10^6 m/s$
• Separation of the two plates, d = 0.5 cm = 0.005 m
• Charge on an electron, q = 1.6×10$^{-19}$C
• Mass of an electron, $m_{e}$ = 9.1×10$^{-31}$kg

$s = \frac{QEL^2}{1mv^{2}_{x}}$

$L = \sqrt{\frac{2dmv^{2}_{x}}{qE}}$

$=\sqrt{\dfrac{2 \times0.005 \times 9.1 \times 10^{-31}\times (2.0 \times10^{6})^{2}}{1.6 \times 10^{-10} \times 9.1 \times10^2}}$

$= \sqrt{0.025\times10^{-10}}= \sqrt{2.5\times10^{-4}}$

= 1.6 × 10$^{-2}$ m

= 1.6 cm

Hence

• Electron will strike the upper plate after travelling 1.6 cm.