Answer to Question #291918 in Electricity and Magnetism for Sabiha

Question

An electron of charge e is moving at a constant velocity u, along the x-axis. It enters a region of constant electric field E, which is pointing perpendicular to the x-axis. The electron moves in a parabola. Which of the following represents the focal length of the parabola? Neglect any effects due to gravity. (for a parabola of type x2=4ay, a is the focal length)

Answer 1

The force acting on the electron:

$$F=eE.$$

As the electron enters the region of the electric field, the force causes acceleration:

$$a=F/m=eE/m.$$

Height above the sheet can be found as

$$y=\dfrac12at^2.$$

Horizontal range can be found as

$$x=ut→t=x/u.$$

Hence:

$$y=\dfrac12ax^2/u^2=\dfrac{ax^2}{2u^2}=\dfrac{eEx^2}{2mu^2},\\\space\\ x^2=\dfrac{2mu^2}{eE}y=\dfrac{2mu^2}{eE}y=4·\dfrac{mu^2}{2eE}·y,$$

the focal length is

$$f=\dfrac{mu^2}{2eE}.$$