Madeline Horn edited Once_we_had_taken_values__.tex  over 8 years ago

Commit id: 7ffac75dac6fb65a75d0a282f21fedf84600b94e

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Once we had taken values across the photo-diode from $0$ to $120$ mV in steps of $10$ mV, had our error in all the values, we were able to start calculating the current from both multi-meters. In order to find the current from the voltage across the photo-diode, we only had to use Ohm's Law (V = IR) because we knew the value of R, $10$K ohms. Finding the current for $V_{sq}$ was harder because we had to account for the change through the amplifiers and multiplier. The equation we used to find the filtered current was: $10 V_{sq} = \left( 2 e i_{dc} \Delta f \right) \left( G_1 G_2 R \right)^{2} $, where e is the charge of the electron that we are looking for, R is the resistance of $10$K ohms, and G2 varied as stated above. We solved for $2 \textrm { idc}$ and plotted $2\textrm{ idc}$ versus the current found from the photo-diode. We found the slope of the data points because we wanted to find e. Our value was $1.64 \cdot 10^{-19}$ 10^{-19} \pm 2.1 \cdot 10^{26}$  coulombs. The discrepancy between our value and the currently accepted standard value for the charge on the electron is less than $2.4 \%$.