An electron is fired into a parallel plate moving at 1.5 x 10^7 m/s. Determine at what velocity it leaves the parallel plates. The uniform electric..
field intensity is 15000 N/C. The length of the plate is 12.0 cm and the 2 plates are separated by a distance of 5 cm.
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