# Is it possible to determine the specific charge of the electron using crossed electric fields?

Experimentally, we tried to find the specific charge, $\frac{e}{m}$ for an electron by using thermionic emission to emit electrons from a cathode, accelerating them using an anode, and then deflecting them using an electric field. Assuming that our apparatus was in a vacuum, and taking the kinetic energy of the electron to be totally from the acceleration through the anode, we use the following equations; $eV=\frac{1}{2}mv^2$

The vertical force on the electrons is $Ee$, where $E=\frac{V_p}{d}$, where $V_p$ is the potential difference between the two plates generating the electric field and $d$ is their separation.

Since $F=ma$, vertical acceleration $a$, $=\frac{V_pe}{dm}$.

Using $t=\frac{l}{v}$, where $l$ is the distance it took for the electron beam to be deflected through a vertical distance $d/2$, and assuming initial vertical velocity is $0$, we get:

$\frac{d}{2}=\frac{1}{2}at^2$

Substituting everything in gives $\frac{e}{m}=\frac{d^2v^2}{V_pl^2}$

However, if you take $V_a$ as the anode voltage in the electron gun, the electron has kinetic energy where $eV_a=\frac{1}{2}mv^2$. If this is substituted back in, it eliminates $\frac{e}{m}$ from the equation.

Is there a way in which we can use the measurements $d$, $V_a$, $V_p$ and $l$ to calculate the specific charge? Or is it necessary to use a magnetic field and electric field together to equate the two forces and then find the specific charge?

• I think you have shown quite conclusively that you need to add a magnetic field deflection to measure the specific charge of the electron. – Farcher Mar 11 '17 at 13:22