The electric field in the region between horizontal, oppositely charged, flat me

Question

The electric field in the region between horizontal, oppositely charged, flat metal plates is approximately uniform. Suppose an electron of charge -e is projected horizontally into this electric field with velocity 3 x 106i m/s. The field strength E = 200 j N/C. Here i andj represent appropriate unit vectors. Take the origin (0,0) to be the point where the electron enters the region of electric field. Find the velocity of the electron after the it travels a horizontal distance l = 0.15 m in the field.

Select one:

a. 3.48 x 106j m/s

b. -3.48 x 106j m/s

c. 3 x 106i m/s - 1.8 x 106j m/s

d. 1.2 x 106i m/s

e. 3 x 106i m/s + 1.2 x 106j m/s

Explanation / Answer

here ,

electric field , E = 200 j N/C

v = 3 *10^6 i m/s

let the acceleration of electron is a

- e * E = m * a

- 1.602 * 10^-19 * 200 J = 9.109 *10^-31 * a

solving for a

a = -3.517 *10^13 j m/s^2

Now ,time taken to travel 0.15 m

t = 0.15/3 *10^6

t = 5 *10^-8 s

Now, using frst equation of motion

vy = 5*10^-8 * -3.517 *10^13 j

Vy = -1.8 *10^6 m/s

hence , the final velocity is c. 3 x 106 i m/s - 1.8 x 106 j m/s

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