deletions | additions       diff --git a/subsection_Analysis_Once_we_had__.tex b/subsection_Analysis_Once_we_had__.tex  index b2ece13..48ef616 100644  --- a/subsection_Analysis_Once_we_had__.tex  +++ b/subsection_Analysis_Once_we_had__.tex 
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\subsection{Analysis}  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 $R$, $10 k \Omega$. Finding %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: \begin{equation}  10 V_{sq} = \left( 2 e i_{dc} \Delta f \right) \left( G_1 G_2 R \right)^{2}   \end{equation}, \end{equation}  where $e$ is the charge of the electron that we are looking for, $R$ is the resistance of $10 k \Omega$, $G1$ was $X100$, $\times 100$,  and $G$2 $G2$  varied as stated above (and shown in Table \ref{table:shot_noise}). We solved for $2 \textrm { idc}$ i_{dc}}$  and plotted $2\textrm{ idc}$ i_{dc}}$  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 $1.649 \pm 0.007 \cdot 10^{-19} \textrm{ Coulombs}$, 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 ($1.649 \pm 0.007 \cdot 10^{-19} \textrm{ Coulombs}$ - accepted value from: \href{http://physics.nist.gov/cgi-bin/cuu/Value?e}{NIST value with uncertainty}) is less than $2.4 \%$.