Let the direction of the electron entry be x and the direction perpendicular to it be y. Then the electric field is in the direction of y. Therefore force is acting in the direction y only and electron's movement in x direction will be left unhindered.

To find the force...

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Let the direction of the electron entry be x and the direction perpendicular to it be y. Then the electric field is in the direction of y. Therefore force is acting in the direction y only and electron's movement in x direction will be left unhindered.

To find the force acting, we need additional data, the charge of an electron and mass of an electron.

Charge, e is = 1.602 x 10^(-19) C

Mass m is, = 9.109 x 10^(-31) kg

Let's first analyse the movement in x direction,

V_x = 1.5 x 10^7 ms-1

The time taken to pass the parallel plates is t,

t = 0.12 m / 1.5 x 10^7 ms-1 = 8 x 10^(-9) seconds.

Now let's analyse the movemnt in y direction,

First I am going to calculate the force exerted by the electric field on the electron.

F = eE where E is lectric field.

F = 1.602 x 10^(-19) C x 15000 N/C

F = 2.403 x 10^(-15) N.

The acceleration on the electron is,

F = ma

a = F/m

a = 2.403 x 10^(-15) N / 9.109 x 10^(-31) kg

a = 2.638 x 10^15 ms-2.

Let's find the distance the elctron travels in y direction in time t,

For this s = ut + at^2/2 can be used.

U is initial velocity which is 0 in y direction,

s = 0 + (1/2) x 2.638 x 10^15 ms-2 x (8 x 10^(-9) s)^2

s = 0.08 m = 8 cm

**Therefore the electron doesn't exit the plates, it strikes the plate. The final velocity is,**

V_y = u+at

V_y = 0 + 2.638 x 10^15 ms-2 x 8 x 10^(-9) s

**V_y = 2.1 x 10^7 ms-1**

V = (v_x ^2 + V_y ^ 2) ^(1/2)

**V = 2.58 x 10^7 ms-1**

The final velocity is **2.58 x 10^7 ms-1**