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.
