A 20.0 \mu F capacitor initially charged to 20.0 \mu C is discharged through a 1.90 k\Omega resistor. How long does it take to reduce the capacitor's charge to 5.00 \mu C?

The charge of the discharging capacitor, as a function of t, is:

Q(t) = Qo* e^(-t/RC)

Qo is the initial charge, RC is the time constant, which in this case is 1900*20*10^-6 = 0.038 seconds

So, solve

5*10^-6 = 20*10^-6*e^(-t/0.038)

ln(1/4) = -t/0.038

t = ?

To solve this problem, we can use the formula for the charge on a capacitor during discharge:

Q(t) = Q(0)e^(-t/RC)

Where:
Q(t) is the charge on the capacitor at time t
Q(0) is the initial charge on the capacitor
e is Euler's number (approximately 2.71828)
t is the time in seconds
R is the resistance in ohms
C is the capacitance in farads

In this case, we are given:
Q(0) = 20.0 μC
Q(t) = 5.00 μC
R = 1.90 kΩ = 1.90 * 10^3 Ω
C = 20.0 μF = 20.0 * 10^(-6) F

We can rearrange the formula to solve for time, t:

t = -RC * ln(Q(t)/Q(0))

Substituting the given values:

t = (-1.90 * 10^3 Ω * 20.0 * 10^(-6) F) * ln(5.00 * 10^(-6) C / 20.0 * 10^(-6) C)

Now, you can solve for t using a calculator or a programming language that supports logarithms.

Note: The ln() function represents the natural logarithm.