Answer:
Q=107μC
Explanation:
Given an RC series circuit
Input voltage V=8V
Resistor =90Ω
And the capacitor=20 μF
Charge Q? After t=0.002
In an RC circuit,
The capacitor voltage is given as
Vc=V(1-e^-t/τ )
Where τ is time constant
τ =RC
τ =90×20×10^-6
τ=0.0018s
Therefore,
Vc=V(1-e^-t/τ)
Given that V=8V and t=0.002
Then,
Vc=8(1-e^-0.002/0.0018)
Vc=8(1-e^-1.111)
Vc=8(1-0.329)
Vc=8×0.671
Vc=5.37Volts
Then the voltage across the capacitor is 5.37V
From the relation
Q=CV, we can calculate the charge
Q=20×10^-6×5.37
Q=1.07×10^-4 C
Q=107μC
