The charge on the capacitor after 2 s is 0.43 C.
Explanation:
The formula for finding the voltage while charging a capacitor is
Here , V₀ is the initial potential before charging and t is the time at which we have to determine the voltage, R is the resistance and C is the capacitance for the given circuit.
The given problem have given us the values for V₀ = 50 V and maximum current I is given as 500 mA.
Then, resistance can be determined using Ohm's law:
The capacitance is defined as the ratio of charge to the unit voltage.
Here T is the time constant which is given as 1 s and R is found to be 100 ohm, then capacitance will be
So, the values for parameters like V₀ = 50 V, R = 100 Ω, C = 10 mF and t = 2 s.
Then,
V= 43 V.
Then,
Thus, the charge on the capacitor after 2 s is 0.43 C.