Illuminated Junction With A Capacitor

 Daniel and I have been messing around with the differential equation (HW6 P.5) Zack gave us in class. Instead of attempting to solve it analytically we plugged it into Mathematica and got a solution and some cool graphs.

This is the equation Zack turned into a DE

\(I = I_{ill} + I_o(1-e^\frac{-eVa}{kT})\)

where \(I = \frac{dQ}{dt}\) and \(Va = \frac{Q}{C}\)  our \(Va\) is negative and \(C = 10 Farads\)

So this becomes

\(\frac{dQ}{dt} = I_{ill} + I_o(1-e^\frac{eQ(t)}{CkT})\) and the exponential becomes \(e^{4Q(t)}\)

This is the code for Wolfram Mathematica:

DSolve[{Q'[t] == 12.8*10^-3 + 3*10^-11*(1 - Exp[4 Q[t]]), Q[0] == 0}, Q, {t}]

Then it gives you this solution:

\(Q(t) = -0.25 Ln[2.34375*10^{-9} + 7.8125*10^{-10} * e^{20.9701 - 0.0512 t}]\)

 Here is the solution graphed

(Time is in seconds and Charge is in coulombs)

It takes around 400 seconds to charge

 Here we used a value of \(I_o = 3x10^{-7}A\)

Notice the cap doesn't fully charge and steady state is achieved faster.

Here is also a link to some work Jimmy Layne did on this. It looks like he got an analytic solution for the time dependence.

https://drive.google.com/file/d/0B_GIlXrjJVn4aHJFaDhvd1o5X3M/view?usp=sharing

Extra Special Bonus Pictures (not related to things above) 

 This is a three dimensional graph of E vs K where the energy surface is displayed

\(E(k_x,k_y)\)

\(E(k_x,k_y,k_z)\) Contour plot (Surfaces of constant energy)

  This picture is interesting anyone have some cool thoughts about it?