Question

An electron is accelerated through a uniform electric field
of magnitude 2.5 3 102 N/C with an initial speed of 1.2 3
106 m/s parallel to the electric field, as shown in Figure 10.
(a) Calculate the work done on the electron by the field
when the electron has travelled 2.5 cm in the field.
(b) Calculate the speed of the electron after it has travelled 2.5 cm in the field.
(c) If the direction of the electric field is reversed, how far
will the electron move into the field before coming to
rest?

Answer 1

F = charge in Coulombs * E field strength
a = F/m= force from above / mass of electron
W = work done = F * 2.5*10^-2 meters
(1/2) m Vf^2 = (1/2) m Vi^2 + W

for ( c ), final speed Vf = 0
Wnew = - W = -Fd
and so
F * stopping distance = (1/2) m Vi^2