Capacitance of a Simple Circuit

The solution for a simple circuit with a resistance, R, and a capacitance, C, in series with a battery voltage, V, is
Q = C*V*(1 - exp(-T/(R*C))),
where Q is the charge on the capacitor and T is the time needed to get the charge. Solving for C as the unknown X, it pays to magnify the vertical scale by 10^6:

The following is based on a cue card that can be processed directly in any word processor, using MATHSERV. Simply copy the lines in courier font to any word processor, select, and click the spinning globe.

FX = "10^6*(X*V*(1 - exp(-T/(R*X))) - Q)"
**** required charge in coulombs, Q = 10^-5 := 0.00001
**** battery volts, V = 10
**** resistance in ohms, R = 2000
**** required time in seconds, T = 4*10^-3 := 0.004
C = bisect(FX, 0.000001, .000002, .00000001) := 1.2578E-6
* test Q = C*V*(1 - exp(-T/(R*C))) := 0.00001001

You can easily modify this to use the online solver. For example, if your values are as above, you would set up the equation:
1000000*(x*10*(1-exp(-.004/(2000*x)))-.00001)
in the f(x) box and enter 0.0000005 for xmin, 0.000003 for xmax, and 0.000000001 for tolerance.

What you enter for ymin and ymax depends on whether you used the magnification factor 10^6, in which case -6 and +6 will do nicely.

The answer should be 0.000001255.