How to find capacitance in a circuit with a resistor and a charging capacitor?

[nautica]

A capacitor, in a series with a 720 ohm resistor, is being charged. At the end of 10 ms its charge is half the final value. The capacitance is about:

a)9.6 uF
b)14 uF
c)20 uF
d)7.2 uF
e)10F

Now how in the heck and I suppose to know this one. I know ohms law and I know how to find capacitance. But what does either of these has to do with time?

 

[gnome]

The rate at which a capacitor is charged in an RC circuit is:

$$q(t) = Q_{max}(1-e^\frac{-t}{RC})$$

You know t and R.
You don't know q(t) or Q_max but you do know the ratio of each to the other, and that's all you need to solve for C.

 

[M98Ranger]

To find the capacitance in this scenario, we can use the formula Q = CV, where Q is the charge on the capacitor, C is the capacitance, and V is the voltage across the capacitor. In this case, we know that the charge on the capacitor at the end of 10 ms is half of the final value, so we can write the equation as 0.5Q = CV. We also know that the resistor in series with the capacitor has a resistance of 720 ohms, so we can use Ohm's Law to find the voltage across the capacitor. The voltage can be calculated as V = IR, where I is the current flowing through the circuit. Since the capacitor is being charged, the current gradually decreases to zero, and we can assume that the current at the end of 10 ms is half of the initial current. Therefore, we can write the equation as I/2 = V/720. Now we can substitute this into our previous equation to get 0.5Q = C(I/2)(720). Simplifying, we get C = 720Q/I. We can also express the current in terms of time as I = Q/t. Substituting this into our equation, we get C = 720Qt/I. Finally, we know that the charge on the capacitor is proportional to the capacitance and the voltage, so we can write Q = CV. Substituting this into our equation, we get C = (720Vt)/(V/720). Simplifying, we get C = (720Vt)/(1/720), which simplifies to C = 720^2t. Now we can plug in the values given in the problem to find the capacitance. Since the voltage across the capacitor is not given, we can assume it to be 1V for simplicity. Plugging in the values, we get C = (720^2)(10 ms) = 5.184 uF. Therefore, the closest option would be a) 9.6 uF.