Question

57. (II) An electron has an initial velocity o (8.0 x 10m/s)i /s)j It enters a region where E (2.0i +8.0j) x 10*N/C (a) Determine the vector acceleration of the electron as function of time. (b) At what angle θ is it moving (relative to its initial direction) at t = 1.0 ns?

A1. For problem 21.57, write a formula for the position vector of the electron as a function of time, r⃗(t)r→(t), using unit vector ("i-hat" and "j-hat") notation. Assume the electron starts at the origin of your coordinate system at t = 0, so r0→=0iˆ+0jˆ. (Hint: Use equation 3-13b from the textbook.)

--- Hint

Answer 1

answer) force on the electron is given by

F=eE..........1)

from newtons second law we have

F=me*a.......2

so we have

eE=me*a

a=eE/me

now we have x and y components of the acceleration is

ax=eEx/me=-1.6*10^-19*2*10⁴/9.1*10^-31kg=-3.5*10¹⁵m/s²

ay=eEy/me=-1.6*10^-19*8*10⁴/9.1*10^-31=-1.4*10¹⁶m/s²

the vector acceleration of the electron=-3.5*10¹⁵i-1.4*10¹⁶j (answer)

b) now for than angle we have

Vx=vxo+axt=8*10⁴m/s+(-3.5*10¹⁵*1*10^-9)=-3.4*10⁶m/s

vy=vyo+ayt=0-1.4*10¹⁶*10^-9=-1.4*10⁷m/s

so tan $\Theta$=vy/vx

$\Theta$=tan^-1(-1.4*10⁷/3.4*10⁶)=76.3

but since the initial velocity is going upwards

the angle =180+76.3=256.3^o

so the answer is 256^o or 256.3^o