Question

The trajectory of an electron when it moves perpendicular to the electric field is _______.

Answer 1

Hint:The trajectory of an electron is described as the path of motion of the electron in an electromagnetic field. Here the electron is taken as a point mass with a negative charge. The equation of motion of the electron is needed here to find the displacement in terms of acceleration and time. The relation between the displacement and time will show the type of path of the electron.

Formula Used:
The displacement of the electron, $y = \dfrac{1}{2}a{t^2}$
$a$ is the acceleration, $a = \dfrac{{eE}}{m}$, $t$ is the time.
$e$ is the charge of the electron, $m$ is the mass of the electron, $E$ is the electric field.

 Complete step by step solution:
Let, an electron of charge $e$and mass $m$moves normally in an electric field. If the electric field intensity is $E$, the acceleration of the electron will be $a = \dfrac{F}{m}$
Where, the force acting on the electron $F = eE$
$\therefore a = \dfrac{{eE}}{m}$
Now, the equation of motion of the electron is,
$y = \dfrac{1}{2}a{t^2}$
$ \Rightarrow y = \dfrac{1}{2}\dfrac{{eE}}{m}{t^2}$
This equation is in terms of the equation of a parabola.
Hence, the trajectory of the electron will be parabolic.

Note: A parabola is a U-shaped plane curve. In a parabola, every point maintains equidistant from a certain point and a fixed straight line. This point is called the focus and the straight line is known as the directrix. The simple equation that describes a parabola is ${y^2} = 4ax$ in a $y - x$ curve.
We get, the equation of motion of an electron is $ \Rightarrow y = \dfrac{1}{2}\dfrac{{eE}}{m}{t^2}$
\[ \Rightarrow {t^2} = \dfrac{{2m}}{{eE}}y\]
This implies the equation of a parabola for a time-displacement graph.