Irodov 3.204: A Fifth of the Charge – Amit & Deepali Sinha Foundation

Problem Irodov 3.204

A capacitor of capacitance $C = 2.00$ $\mu$ F has charges $+q_0$ and $-q_0$ on its plates, with $q_0 = 1.00$ mC. Its plates are then joined through a resistance $R = 5.0$ M $\Omega$ . Find:

the charge that passes through the resistance during a time interval $\tau = 2.00$ s
the heat released in the resistance during the same time

Before the connection: the capacitor C with charges q_0 (drawn for the upper plate positive) and the resistor R; the terminals are then joined to close one loop. — Before the connection: the capacitor $C$ with charges $\pm q_0$ (drawn for the upper plate positive) and the resistor $R$ ; the terminals are then joined to close one loop.

Given

$C = 2.00~\mu$ F
$q_0 = 1.00$ mC
$R = 5.0$ M $\Omega$
$\tau = 2.00$ s
interval counted from the connection (assumed)

Find

(a) the charge

\Delta q

through

R

in the time

\tau

, (b) the heat

Q

released in it

Solution

Step 1The capacitor discharges exponentially

After the connection: one loop, the current I leaving the positive plate through R. — After the connection: one loop, the current $I$ leaving the positive plate through $R$ .

Let $q$ be the charge on the capacitor at time $t$ . The voltage $q/C$ across the plates drives the current through $R$ , and that current is the rate at which the plates lose charge:

\begin{aligned} I &= \frac{q}{RC}\\ I &= -\frac{dq}{dt} \end{aligned}

Eliminating $I$ and integrating from $q_0$ at $t = 0$ :

\begin{aligned} \frac{dq}{q} &= -\frac{dt}{RC}\\ \ln\frac{q}{q_0} &= -\frac{t}{RC} \end{aligned}

q(t) = q_0\,e^{-t/RC}

(1)

The time constant is five times the interval $\tau$ :

RC = (2.00\times10^{-6}~\text{F})(5.0\times10^{6}~\Omega) = 10~\text{s}

Step 2(a) The charge through the resistor

The capacitor's charge decays with the time constant RC; whatever it has lost by a time t has passed through R. — The capacitor’s charge decays with the time constant $RC$ ; whatever it has lost by a time $t$ has passed through $R$ .

Every charge that leaves the plates passes through $R$ , so the charge through it in the time $\tau$ , counted from the connection (assumed), is the loss from (1):

\Delta q = q_0 - q(\tau) = q_0 - q_0\,e^{-\tau/RC}

Numerically, with $\tau/RC = 0.200$ and $e^{-0.200} = 0.819$ :

\Delta q = 1.00~\text{mC} \times (1 - 0.819)

\Delta q = q_0\left(1 - e^{-\tau/RC}\right) \approx 0.18~\text{mC}

(2)

Step 3(b) The heat, from the Joule power

The current follows from differentiating (1):

I = -\frac{dq}{dt} = \frac{q_0}{RC}\,e^{-t/RC} \tag{3}

The heat is the Joule power $I^2R$ integrated over the interval, with (3):

\begin{aligned} Q = \int_0^\tau I^2R\,dt &= \frac{q_0^2}{RC^2}\int_0^\tau e^{-2t/RC}\,dt\\ &= \frac{q_0^2}{RC^2}\cdot\frac{RC}{2}\left(1 - e^{-2\tau/RC}\right) \end{aligned}

Numerically, with $q_0^2/2C = 0.250$ J and $e^{-0.400} = 0.670$ :

Q = 0.250~\text{J} \times (1 - 0.670)

Q = \frac{q_0^2}{2C}\left(1 - e^{-2\tau/RC}\right) \approx 82~\text{mJ}

(4)

Check

Energy balance. The capacitor’s field energy falls from $q_0^2/2C$ to $q(\tau)^2/2C$ , and with (1) the drop is $(q_0^2/2C)(1 - e^{-2\tau/RC})$ , exactly (4): all the lost field energy appears as heat in $R$ . As $\tau \to \infty$ the two answers tend to $q_0$ and $q_0^2/2C$ , the whole charge and the whole stored energy.

Remark

In the first fifth of a time constant the capacitor loses $18$ per cent of its charge but $33$ per cent of its energy. The energy goes as $q^2$ , so it decays with half the time constant, $RC/2$ , and the early part of a discharge, when the voltage is highest, is where most of the heat is made.