Verified answer

voltage on a cap, discharging

v = v₀e^(–t/τ)

v₀ is the initial voltage on the cap

v is the voltage after time t

R is resistance in ohms,

C is capacitance in farads

t is time in seconds

RC = τ = time constant

Energy in a Capacitor in Joules

E = ½CV²

Assume you start with a voltage of V

for energy to drop by 1/2, the voltage has to be 1/√2 of it's initial value. or 0.707V

τ = 4m x 4k = 16 sec

0.707V = Ve^(–t/16)

e^(–t/16) = 0.707

–t/16 = –0.347

t = 5.55 sec