Question

An electron with a speed of 4.28

Answer 1

a. The electric field retards the electron's motion. So if the electron travels far enough in the field to stop then the work done by the electric field on the electron has to equal the electron's intial kinetic energy.

K = 1/2mv^2

W=work = qEd wher d= distance, q = charge on electron and E = field strength

So W =K ---> qEd = 1/2 mv^2 and solve for d ---> d = 1/2 (mv^2/(Eq))

we have given that m = mass of electron = 9.1 x 10^-31 kg and charge q of electron = 1.6022 x 10^-19

E = 1.57 x 10³ N/C and v = 4.28 x 10⁸ cm/sec = 4.28 x 10⁶ m/sec

d = 0.06627 m = 6.627 cm.................................Ans.

b). We know the initial speed and the final speed (0 m/s). We cna find the acceleration from:

F = qE = ma ---> a = qE/m

So v = -at + v0 = -qEt/m +v0

solve for t setting v = 0 ---> t = mv0/(qE) where v0 = 4.28 x10^6 m/s

t = 1.5504 x10^-8 sec ...........................Ans.

c.). Find the work done by the electic field over the 4.59 mm.

W =qEd ---> set d = 4.59 mm = 0.00459 m ---> W = 1.15x10^-18 J

This is the kinetic energy lost by the electron. The fraction is:

f = (qEd)/[1/2mv^2] ---> v= 4.28x10^6 m/s ---->, f = 0.1379 = 13.79%.............................Ans.