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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?

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  • $\begingroup$ I think you have shown quite conclusively that you need to add a magnetic field deflection to measure the specific charge of the electron. $\endgroup$ – Farcher Mar 11 '17 at 13:22
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The simple answer is no, and if you think about this right, the reason is simple.

Take your experiment and assume any value you like for e and m. Now repeat the experiment for a new m (M), where M = 4m.

Particle velocity will be v/2, rather than v. This means that the particle will spend twice as long in deflecting region, but it will deflect at a rate 1/4 as great. So when you calculate d/2, the factors cancel, and the deflection remains the same.

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