# Effective Charge Laboratory

#### The definition of the effective charge

In this paper we will determine the effective charge of an electron. The effective charge is by definition the ratio (e/m).

Background Information and History

Thomson was the first scientist that used a selection of different gases to determine the ratio of the electric charge of a particle to its mass (e/m), although he was not able measure the electric charge or mass of such a particle (egglescliffe.org.uk). In 1909, Robert Millikan then determined the charge of the electron by the Oil Drop experiment. That was achieved by measuring the effects of the electrical field on the movements of many droplets. A conclusion of that experiment was that the charge of oil drops was always an integral multiple of 1.602 x 10-19 coulombs, which Millikan assumed that number was the charge of an electron. The mass of the electron (me), however, was calculated by combining Millikan’s value of electron charge with Thomson’s electron charge is me = 9.1094 x 10-31 kg (outreach.phy.cam.ac.uk).

#### Theory of the Experiment

For this experiment, the electron beam is generated by heating a filament with current from a 6 V AC supply. That heating, the thermal energy, will cause the electrons to escape from the metal into the surrounding vacuum, and that is known as thermionic emission. The acceleration for the electrons is caused by the potential difference ΔVa between a positively charged cylindrical anode and the filament as shown in Figure 1. After the electrons pass through the slit in the anode, their kinetic energy (½mv2) and their potential energy (e ΔVa) will be equal.

$1/2 mv^2=e\Delta Va$   or $mv^2=2e*\Delta Va$   or $v^2=(e/m)*2\Delta Va$   or $(e/m)=v^2/(2\Delta Va)$

Thus, the last equation can be used to determine the electron charge to mass ratio (e/m).

#### Methodology

First, the circuit was correctly connected. After that, the lights were turned off so that the electron beam could be observed clearly. Then, after the EHT power supply was turned on and set to 1000 V, the electron beam was deflected upwards due to the electric field. The previous step was made to check that the circuit was connected correctly and the electron could be observed.  After that, the DC supply for the magnetic coils was turned on after the anode voltage was reset to 1000 V and the electron beam was downwards under the effect of the magnetic force. Finally, the anode potential difference was set several times with the values of 1500, 2000, 2500, 3000, 4000 and 5000 and the current was found for each value.

#### Results

$I^2$ (A)       Current in magnetic field (A)   Anode potential difference (V)
0.04                                 0.2                                                    1500
0.0625                            0.25                                                  2000
0.09                                 0.3                                                    2500
0.1225                             0.35                                                  3000
0.16                                  0.4                                                    4000
0.2025                            0.45                                                  5000

Therefore $slope = 5*10^{-5}$. We find $(e/m)$ using  $em = ΔVa1.81*10-7(I^2) C/Kg$.
Error percentage is the absolute value of the next equation:

(estimated value – actual value / actual value )* 100 %

Anode potential difference <V>     Current in magnetic field <A>  $I^2$      (e/m) <C/kg>              Error

1500                                           0.2                                        0.04          $2.07182*10^{11}$                  18%
2000                                          0.25                                      0.0625       $1.76796*10^{11}$                 0%
2500                                          0.3                                        0.09            $1.53468*10^{11}$                13%
3000                                          0.35                                      0.1225        $1.35303*10^{11}$                23%
4000                                          0.4                                        0.16            $1.38122*10^{11}$                 22%
5000                                          0.45                                      0.2025       $1.36416*10^{11}$                 22%

Finding the effective charge (e/m) using the slope:
$Slope = 11.81*10^{-7}(e/m)$

$5*10^{-5} = 11.81*10^{-7}(e/m)$

$(5*10^{-5})*(1.81*10^{-7}) (e/m) = 1$

$(9.05*10^{-12}) (e/m) = 1$

$e/m = 19.05*10^{-12}$    or   $e/m = 1.1*10^{11} C/Kg$

$Error =1.1*10^{11}- (1.76*10^{11})(1.76*10^{11}) *100 = 37%$

#### Discussion

Results show that there was a big difference between the values of the effective charge (e/m). There were some gaps between the curtains and the light of the next room found a way through to our room. This could be the main reason because the electron beam was really hard to be observed. This explains why there was a big error percentage.

#### Conclusion

This experiment showed that the charge to mass ratio was measured by balancing the electric and magnetic forces acting on an electron. The charge to mass ratio for electron was found as:  $e/m = 1.1*10^{11}$  which is different from the accepted value of $1.76*10^{11}$ and that was explained in the discussion part.

#### References

Applied physics laboratory (photoelectric effect): Measurement module, University of South Australia, School of Electrical & Information Engineering (Applied Physics)

Giancoli, D 2005, Physics Principles With Applications, Pearson Education, United States of America.