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Madeline Horn edited Once_we_had_taken_values__.tex
over 8 years ago
Commit id: a2fe271bfee3693e8b53f43e61962083ab16ac3b
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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, $10 k \Omega$. 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}, where $e$ is the charge of the electron that we are looking for,
R $R$ is the resistance of $10 k \Omega$, 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. $e$.
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 is less than $2.4 \%$.