. A short time later the capacitor reaches its maximum potential difference of 62.0 V . What is
the value of the capacitance?
Solution
Imax = Qmax
C = Qmax/Vmax
==> C = (Imax/)/Vmax
==> C = Imax/Vmax
==> C (1/(LC)) = Imax/Vmax
==> (C^2) (1/LC) = (Imax/Vmax)^2
==> C = L (Imax/Vmax)^2
==> C = (9e-3) * (0.550/62)*(0.550/62)
==> C = 0.708e-6 F