Charge in Uniform Electric Field

Find the trajectory of a particle of mass m, charge e, in a uniform electric field E, assuming zero velocity parallel to E at t = 0. Sketch the trajectory in the plane of motion.

SOLUTION

The plane of motion of a particle will be defined by its initial velocity v and the direction of the electric field E. Let the initial velocity coincide with the x axis and E with the y axis. We may write the equations of motion for a charge in an electric field

(1)

where p is the momentum of the particle. Obviously, since there is no force in the direction perpendicular to the x - y plane, the particle will move in this plane at all later times. We can write (1) in the form

(2)

(3)

Integrating (2) and (3) yields

(4)

(5)

The energy ε of the particle (without the potential energy due to the field) is given by

(6)

where is the initial energy of the particle. The work done by the electric field changes the energy of the particle

(7)

or

(8)

Equations (6) and (8) result in

(9)

which yields

(10)

and

(11)

On the other hand

(12)

Substituting p_x = p_y and p_y = eEt into (12) and using t from (11), we find

(13)

Integrating (13), we obtain

For the initial conditions x₀ = y₀ = 0

(14)

So the particle in a constant electric field moves along a catenary (see Figure 1.1, where we took e > 0). If the velocity of the particle v << c, then p₀ = mv₀, ε₀ = mc² and expanding cosh (eEx/p₀c), we obtain

which gives the classical result for a charged particle in an electric field. Also note that (10) coincides with the result for uniformly accelerated  motion in the proper reference frame, where the acceleration ω₀ = eE/m and p₀ = 0. Under Lorentz transformations for frames moving with velocities parallel to the electric field E, the field is unchanged.

Figure 1.1