Verified answer

The time constant τ is C*R, thats 1s in this case.

After one τ the charge of a capacitor will have dropped to e^-1 (0.367879) times the original voltage.

So to calculate the voltage after n time constants you can do V = Vorig * (e^-1)^n = Vorig * e^(-n).

That means n = -ln(V/Vorig)

And then t = n*τ

V/Vorig = 1/2

So -ln(0.5) = 0.6931

0.6931 * 1s = 0.6931s

The time constant is 1 s.

The capacitor wil discharge to 1/2 Q(o) in 0.693 s