Problem 21.48

21.48. a)
The charge on the capacitor as a function of time is given by: q = C V (1 - e^-t/RC)
Dividing by the capacitance, we find that
$\frac{q}{C} ~=~ V(t) ~=~ V_{0} (1 - e^{-t/RC})$
So, plugging in the numbers, we find that
$4 V ~=~ (10 V) (1 - e^{-3/(R)(10 \times 10^{-6} F)})$
Solving for Solving for R, we find
$.6 = e^{\frac{3 \times 10^{5}}{R}}$
$\frac{3 \times 10^{5}}{R} = .511$
$R = 587 k\Omega$