# In quest of the specific charge of an electron

In the 1890th the knowledge about electrons was limited. Even the word electron was not very popular. But in this time some physicists (Herz, Perrin, Thomson, Kaufmann) worked hard on the point cathode rays. So the electron beam, produced by an experimental setup like the electron gun, was called in this time. They explored the cathode rays with different experiments. So Perrin showed that cathode rays are transporting electric charge.
J.J. Thomson tried to develop quantitative statements. He examined an experiment equal to the experiment, shown in this learning environment. He realized that the Lorentz force is equal to the centripetal force:$$$$F_{\rm{Lorentz}}= F_{\rm{Zentripetal}}$$$$ $$$$e\cdot v_0 \cdot B = m\frac{{v_0}^2}{r}$$$$ $$$$e \cdot B = m\frac{{v_0}}{r}$$$$ In this approach, $v_{0}$ must now be replaced by a measurable quantity in the experiment. For this purpose, it is used that, due to the conservation of energy, the work performed in the electron gun's E-field corresponds to the kinetic energy of the electron on leaving it:$$$${e\cdot V_{\rm{a}} = \frac {1}{2}\cdot m\cdot v_0^2}\quad \Leftrightarrow \quad v_0=\sqrt{\frac{2\cdot e \cdot V_{\rm{a}}}{m}}$$$$ Equations $(3)$ and $(4)$ lead in squared form to $$$$e^2 \cdot B^2 = m^2\frac{{\frac{2\cdot e\cdot V_{\rm{a}}}{m}}}{r^2}$$$$ Separation of measurable and unknown quantities leads to$$$$\frac{e}{m}=\frac{2\cdot V_{\rm{a}}}{\left(B \cdot r\right)^2}$$$$ So Thomson was able to determine the mass-to-charge ratio of an electron just using measurable quantities.
In general, this ratio $\frac {q}{m}$ is called the specific charge of a particle.