Question

2. An Electron between two parallel plates has an initial velocity of 1.6*10°m/s. In between the plates is a uniform Electric Field which points down (see picture). The plates are separated by 1 em and they are each 2cm long. You can assume the electron begins halfway between the plates. a) Suppose that the electron just clears the top plate as it emerges from the plates. In this case, find the value of the electric field. b) Now that we know the Electric field, find the electric potential between the plates. c) What is the magnitude and direction of the final velocity as the electron emerges from the plates?

Answer 1

The velocity of electron in x direction initially u_X = 1.6 times 10^6 m /s The displacement n x - direction (x) = 2 cm = 2 time 10^-2/ 1.6 times 10^6 rightarrow 1.25 times 10^-8 second now in y - direction s_y or Y = U_r t + 1/2 a_r t^2 where u_r = 0 m /s where u_y is initial velocity in Y - direction a_r - accelerations in Y - direction Y = 0.5 times 10^-2 m [as plate separation is km] so, (0.5 times 10^-2) = 1/2 a_u (1.25 times 10^-8)^2 a_r = 5 times 10^-2 times 2/(1.25 times 10^-8)^2 = 0.64 times 10^14 m /s^2 Now, the electro static force = ma_y where m is the mass of electron q elementof = ma_y electric filed acting in Y = -direction down ward. \r\n q = 1.6 times 10^-19 c: m = 9.1 times 10^-27 epsilon = ma_r /q = (9.1 times 10^-27) (64 times 10^14)/1.6 times 10^-19 epsilon rightarrow 3.64 times 10^4 N /c b) since dv = -epsilon^rightarrow . dr^rightarrow rightarrow (3.64 times 10^4) (1 times 10^-2) [dr rightarrow separation between plats)