Lecture Notes

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Notes on Experiment 16: Electric Field and Electric Potential

Sujeet Akula

Error Analysis

Concentric Electrodes

You should have parametrized the apparatus with the following (one-time) measurements: radius of the inner electrode $a$, radius of the outer electrode $b$, potential difference between the two electrodes $V_0 = V_a-V_b$. With our analysis, we should be able to reproduce the geometrical and electrical configuration of the apparatus $(V_0,a,b)$. This is because, theoretically, we have that \begin{align} V(r) &= \frac{V_0}{\ln\left(\dfrac{b}{a}\right)}\left(\ln b - \ln r\right) \\ \end{align}and,\begin{align} E(r) &= \left(\frac{V_0}{\ln\left(\dfrac{b}{a}\right)}\right) \frac{1}{r} \mathrm{.} \end{align} Now, your raw data for each point should be a voltage $V$, and a set of radii $\{r_i\}$. First, we must compute the average and standard deviation $r$, and $\sigma_r$. This gives \begin{align} \delta r &= \sqrt{\sigma_r^2 + \Delta_r^2} \mathrm{,} \end{align}where $\Delta_r$, the system error, should estimate how far the probe can move, and give the same voltage reading (typically 0.5\,cm). Then,\begin{align} \delta \left(\ln r\right) &= \frac{\delta r}{r} \mathrm{.} \end{align} So, you will be able to plot $V$ on the $y$-axis against $\ln r$ on the $x$-axis with $\delta\left(\ln r\right)$ as your error bars in $x$. Compute the slope and uncertainty using http://freeboson.org/slope/ and compare to the expected slope from eqn. (1), $V_0 / \ln\left(b/a\right)$. Next, the implications of our data on the electric field is a more complicated analysis. From our full set of data: $\left\{V^i,r^i,\delta r^i\right\}_{i=1}^{i=n}$ (we will use superscript indices here, since the $\left\{r^i\right\}$ are averages over the raw $\left\{r_i\right\}$ data -- be sure you understand this before moving on), we will compute data relevant to the electric field analysis $\left\{E_j, 1/\rho_j, \delta \left(1/\rho_j\right)\right\}_{j=1}^{j=n-1}$. It is important to note that from the $n$ voltage data points, we only have $n-1$ electric field data points, this is because each electric field datum lies somewhere "in-between'' two voltage data points. (Think of some fence between two fence-posts.) \begin{align} \end{align}Since\begin{align} \vect{E} &= -\grad V \mathrm{,} \\ \end{align}discretely, the radial component of the electric field is\begin{align} E &= -\frac{\Delta V}{\Delta r} \mathrm{.} \\ \end{align}So, for our data, we let\begin{align} E_j &= -\frac{V^{j+1}-V^j}{r^{j+1}-r^j} \\ \end{align}which is the electric field defined at a radial length somewhere in the interval $\left[r^j,r^{j+1}\right]$. We will approximate the point to be the midpoint of the interval, meaning\begin{align} \rho_j &= \frac{r^j+r^{j+1}}{2} \mathrm{.} \end{align}It logically follows that\begin{align} \delta \rho_j &= \frac{\delta r^j + \delta r^{j+1}}{2} \mathrm{.} \end{align}Finally, this gives\begin{align} \delta \left( \frac{1}{\rho_j} \right) &= \frac{\delta \rho_j}{\rho_j^2} \mathrm{.} \end{align} This will allow you to plot the electric field data $\left\{E_j, 1/\rho_j, \delta \left(\dfrac{1}{\rho_j}\right)\right\}_{j=1}^{j=n-1}$, with $E$ on the $y$-axis, $1/\rho$ on the $x$-axis, with $\delta \left(1/\rho\right)$ for $x$ error bars. Again, compute the slope and slope error, and compare to the expected slope from eqn. (2), $V_0 / \ln\left(b/a\right)$.

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