deletions | additions       diff --git a/Once_we_had_taken_values__.tex b/Once_we_had_taken_values__.tex  index 3fca4a0..bbc840d 100644  --- a/Once_we_had_taken_values__.tex  +++ b/Once_we_had_taken_values__.tex 
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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 \begin{equation}  10  V_{sq} = \left( 2 e i_{dc} \Delta f \right) \left( G_1 G_2 R \right)^{2} $, \end{equation},  where e $e$  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 $1.64 \cdot 10^{-19} \pm 7.0 \cdot 10^{ -22}$ coulombs, where the error comes from the mean squared error. The discrepancy between our value and the currently accepted standard value for the charge on the electron is less than $2.4 \%$.