Hi there!
We can use the equation for the charge of a charging capacitor:
[tex]q(t) = C\epsilon( 1 - e^{-\frac{t}{RC}})[/tex]
Using Capacitor equations:
[tex]C = \frac{Q}{V}}\\\\Q = CV[/tex]
Therefore, Cε equals the steady-state charge of the capacitor (the function approaches this value as t ⇒ ∞.
We can plug in the givens and solve.
[tex]\epsilon = 5 V\\\\C = 100\mu F = 0.0001 F[/tex]
[tex]Q = C \epsilon = (0.0001)(5) = \boxed{0.0005 C}[/tex]
The charge on the capacitor when the circuit has reached a steady-state is 0.0005C.
The capacitance of a conductor is defined as the ratio of the quantity of electric charge stored on it to the difference in electric potential.
Given that the resistance of the resistor is 20 kω, while the capacitance of the capacitor is 100 μf, and the voltage of the battery is 5V.
Now, the charge on the capacitor when the circuit has reached a steady-state can be written as,
[tex]Q=CV\\\\Q = 100\mu F \times 5V\\\\Q = \dfrac{100}{1,000,000}F \times 5V\\\\Q = 0.0005C[/tex]
Hence, the charge on the capacitor when the circuit has reached a steady state is 0.0005C.
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