5. (10 points) A negatively charged particle with kinetic energy of 2 x 10^-16 j

Question

5. (10 points) A negatively charged particle with kinetic energy of 2 x 10^-16 j enters a region of uniform electric field E. If the charge q = -1.602 x 10^-19 C, mass m = 9.11 x 10^-31 kg, the electric field strength E = 2 x10^4 N/C, and width, w = 4 cm. (a) How far is the particle from the axis (X) when it emerges from the other side? (b) What is the velocity (v) of the particle when it emerges from the other side? (C) At what angle is the particle moving with respect to the axis when it emerges from the other side? Hint: Ignore the gravity. v is a vector so you have to find the vx and vy components when the particle emerges from the other side.

Explanation / Answer

speed of the particle is v = sqrt(2*K/m) = sqrt(2*2*10^-16/9.11*10^-31) = 20.95 *10^6 m/s

this speed is initilly in the horizontal direction

hence there is no initial vertical velocity

vox = 20.9*10^6 m/s

voy = 0 m/s

A) distance from the axis is y = -0.5*a*t^2

t is the time taken = w/v = 0.04/(20.9*10^6) = 1.91*10^-9 S = 1.91 ns

accelaration a = F/m = qE/m = (1.602*10^-19*2*10^4)/(9.11*10^-31) = 3.51*10^15 m/s^2
then y = 0.5*3.51*10^15*1.91*1.91*10^-18 = 6.4*10^-3 m

B) vx = 20.9*10^6 m/s

vy = a*t = 3.51*10^15*1.91*10^-9 = 6.7041*10^6

v = sqrt(20.9^2+6.7041^2)*10^6 = 21.94*10^6 m/s

C) tan (alpha) = vy/vx = 6.7041/20.9

alpha = 17.8 degrees below the X-axis

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