Discharging a Capacitor

Electricity and Magnetism Level 2

A capacitor of capacitance $C$ is charged until its plates contain charge $\pm Q_0$ . It is then hooked up in a circuit to a resistor of resistance $R$ and allowed to discharge starting at time $t=0$ . Find the charge on the capacitor as a function of time, $Q(t)$ .

Q_0 (1-e^{-\frac{t}{RC}})

Q_0 e^{\frac{t}{RC}}

Q_0 e^{-\frac{t}{RC}}

Q_0 (e^{-\frac{t}{RC}}-1)

1 solution

Matt DeCross
Feb 22, 2016

The equation for the amount of charge on the capacitor comes from charge conservation / Kirchoff's voltage law (closed loops) -- the current through the circuit is the time rate of change of charge on the capacitor. So:

$\dot{Q} R + \frac{1}{C} Q = 0.$

Using the initial conditions, this is solved by:

$Q(t) = Q_0 e^{-\frac{t}{RC}}$

as claimed.

Discharging a Capacitor

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