Special theory of relativity > Motion of relativistic particles in electric and magnetic fields > Problem 14.4.2

Statement

$14.4.2$ An electron flying at a velocity $v$ into an extended homogeneous electric field that moves towards it at a velocity $u$ leaves the field after a time $\tau$. The electric field is directed along the electron velocity. Determine its value.

Solution

In the Earth frame, the electron takes a time $\tau$ to leave the field. Half of this time is the time it takes for the electron to reach zero velocity. In the moving frame (the frame of the field), the electron has a velocity

\begin{equation}
v_1 = \frac{v - (-u)}{1 - \frac{(-u)v}{c^2}} = \frac{v + u }{1 + \frac{uv}{c^2}}
\end{equation}

The time needed by the electron to reach zero velocity in the moving frame is:

\begin{equation}
t_{1/2} = \frac{\tau \sqrt{1-u^2/c^2}}{2}
\end{equation}

The initial momentum of the electron in the moving frame is:

\begin{equation}
p = \frac{m_e v_1}{\sqrt{1-v_1^2/c^2}}
\end{equation}

Applying Newton's second law, we have:

\begin{equation}
\frac{\Delta p}{\Delta t} = \frac{- p}{t_{1/2}} =- e E \rightarrow E = \frac{p}{e t_{1/2}}
\end{equation}

Finally:

\begin{equation}
E = \frac{2 m_e ( v + u) }{e \tau (1-u^2/c^2) \sqrt{1- v^2/c^2}}
\end{equation}

Answer

\begin{equation}
E = \frac{2 m_e ( v + u) }{e \tau (1-u^2/c^2) \sqrt{1- v^2/c^2}}
\end{equation}