An uncharged 5.0-µf capacitor and a resistor are connected in series to a 12-v battery and an open switch to form a simple rc circuit. the switch is closed at t = 0 s. the time constant of the circuit is 4.0 s. what is the charge on either plate after one time constant has elapsed?

For a direct current resistor-capacitor circuit where the capacitor is initially uncharged, the charge stored on one of the capacitor's plates is given by:

Q(t) = Cℰ(1-e^{-t/(RC)})

Q(t) is the charge, t is time, ℰ is the battery's terminal voltage, R is the resistor's resistance, and C is the capacitor's capacitance.

The time constant of the circuit τ is the product of the resistance and capacitance:

τ = RC

Q(t) can be rewritten as:

Q(t) = Cℰ(1-e^{-t/τ})

We want to know how much charge is stored when one time constant has elapsed, i.e. what Q(t) is when t = τ. Let us plug in this time value:

Q(τ) = Cℰ(1-e^{-τ/τ})

Q(τ) = Cℰ(1-1/e)

Q(τ) = Cℰ(0.63)

Given values:

C = 5.0×10⁻⁶F

ℰ = 12V

Plug in these values and solve for Q(τ):

Q(τ) = (5.0×10⁻⁶)(12)(0.63)

Q(τ) = 3.8×10⁻⁵C

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explanation: