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 position of the electron after it travels a horizontal distance l = 0.15 m in the field.

a. 0.44 i m + 0.15 j m

b. -0.44 i m + 0.15 j m

c. 0.15 i m + 0.44 j m

d. 0.15 i m - 0.044 j m

e. 0.3 i m - 0.088 j m

Explanation / Answer

The question can be solved without working out. x distance does not changes so x = 0.15 i. From the given answers we have option c and d only with 0.15 i. As the y direction is negative so correct option is d

Here is the solution

The time taken to horizontal travel of the electron is

t = L / Vx = 0.15 / 3x10^6

t = 5x10^-8 s

now the acceleration is

a= e E / m = (1.6x10^-19)(200)/(9.11x10^-31)

a = 3.513x10^13 m/s^2

The y positions is

y = -(1/2) a t^2

   = (1/2)( 3.513x10^13 m/s^2 )(5x10^-8 s )

   = -0.044 m j

So the answer is 0.015m i - 0.044 j , correct answer is (d)

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