Answer:
We can use the formula for the discharge of a capacitor through a resistor:
Q(t) = Q0 * e^(-t/(RC))
where Q(t) is the charge on the capacitor at time t, Q0 is the initial charge on the capacitor, R is the resistance, C is the capacitance, and e is the mathematical constant e.
Setting Q(t) to 30.0 μC, Q0 to 30.0 μC, R to 1.70 kΩ, and C to 30.0 μF, we get:
30.0 μC = 30.0 μC * e^(-t/(1.70 kΩ * 30.0 μF))
Simplifying, we get:
1 = e^(-t/(51.0 s))
Taking the natural logarithm of both sides, we get:
ln(1) = ln(e^(-t/(51.0 s)))
0 = -t/(51.0 s)
Solving for t, we get:
t = 0 s
This means that the capacitor is already discharged to 30.0 μC, so it took no time for this to happen.