electricity and magnetism?

Question

An electron is released from rest in a uniform electric field of magnitude 8 * 10^4 v/m directed along the positive x-axis. The electron is displaced 0.5m in the direction opposite E.

a. Find the change in the electric potential between the point of start and finish.

b. Find the change in the potential energy of the electron.

c.Find the speed of the electron then convert the energy values into eV.

Answer 1

a)Potential difference PD is work that can be done by moving a charge by distance d in electric field E.

PD= E d

PD= 8 E+4 x 0.5= 4 E+4 Volts

b) Potential energy Pe equal to the potential difference times the chargge q moved (or to be moved)
Pe= PD q
q= 1.602 E-19 C
Pe= 4E+4 x 1.602 E-19 =6.4 E-15 Joules

c) Potential energy equal to kinetic energy
Pe=Ke and
Ke= 0.5 mV^2
m of an electron = 9.109 E–31 kg
so

V= sqrt( 2 Pe/m)
V= sqrt(2 x 6.4 E-15/9.109 E–31)
V=1.2 E+8 m/s