Problem 30.33 (Y-F 15th Edition)

Clear["Global`*"]

L = 85.0 * (10^(-3 )); (* Henries *)
c = 3.20 * (10^(-6)) ; (* Farads *)
imax = 0.850 * (10^(-3)) (* Amperes)*);

UL[i_] := (1/2)L (i^2)

Uc[q_] := ((q^2)/(2 c))

(a)

Umax = UL[imax]

qmax = q/. Solve[Uc[q] == Umax && q > 0, q][[1]]

(b)

i = 0.500 * (10^(-3)) (* Amperes)*);

Clear[q]
q = q /. Solve[Umax == UL[i] + Uc[q] && q > 0, q][[1]]

Alternatively, look at how (q) and (i) vary with time:

ω = (1/(Sqrt[L c]))

T = ((2 π)/ω)

qosc[t_] := qmax Cos[ω t]
iosc[t_] := imax Sin[ω t]

Plot[qosc[t],  {t, 0, T}, PlotLegends -> {"Charge"}, PlotStyle -> Purple,
LabelStyle -> Larger, AxesLabel -> {"t (s)", "q"}, ImageSize->Scaled[0.7]]

Plot[iosc[t],  {t, 0, T}, PlotLegends -> {"Current"},
LabelStyle -> Larger, AxesLabel -> {"t (s)", "i"}, ImageSize->Scaled[0.7]]

When does the current = 0.500 mA?

Clear[t]
t = t/. FindRoot[iosc[t] == 0.500 * (10^(-3)), {t, 0}][[1]]

What is the charge at that time?

qosc[t]