Irodov 3.349: The Lead Current Crosses the Gap – Amit & Deepali Sinha Foundation

Problem Irodov 3.349

A parallel-plate capacitor with air between its plates, each plate of area $S = 100$ cm $^2$ , is connected in series into an ac circuit. The current in the leads is sinusoidal with amplitude $I_m = 1.0$ mA and angular frequency $\omega = 1.6\cdot 10^7$ s $^{-1}$ . Find the amplitude of the electric field strength inside the capacitor.

The air capacitor in series with the ac circuit, seen edge on: two plates of area S, each joined to the circuit by a lead. — The air capacitor in series with the ac circuit, seen edge on: two plates of area $S$ , each joined to the circuit by a lead.

Given

$S = 100\ \text{cm}^2$
$I_m = 1.0\ \text{mA}$
$\omega = 1.6\cdot 10^{7}\ \text{s}^{-1}$
uniform field between the plates (assumed)
no steady part in the field (assumed)

Find

the field amplitude

E_m

Solution

Step 1The lead current continues as displacement current

No charge crosses the air gap, yet the current in the leads is the same on both sides of the capacitor: in the gap it continues as the displacement current $S\,\partial D/\partial t$ . With the edge effects neglected (assumed), the field between the plates is uniform and set by the plate charge $q$ through Gauss’s law, $\varepsilon_0 E = q/S$ . Differentiating $q = \varepsilon_0 S E$ in time:

I = \frac{dq}{dt} = \varepsilon_0 S\,\frac{dE}{dt} \tag{1}

Step 2From the current’s amplitude to the field’s

Take the current as $I = I_m\cos\omega t$ ; the phase origin does not affect an amplitude. Integrating (1) in time, with no steady part of the field in an ac circuit (assumed):

E = \frac{1}{\varepsilon_0 S}\int I_m\cos\omega t\,dt = \frac{I_m}{\varepsilon_0\omega S}\,\sin\omega t

The field oscillates a quarter period behind the current, and its amplitude is the coefficient of $\sin\omega t$ . Numerically, with $S = 1.00\cdot 10^{-2}$ m $^2$ ,

E_m = \frac{1.0\cdot 10^{-3}}{8.85\cdot 10^{-12}\cdot 1.6\cdot 10^{7}\cdot 1.00\cdot 10^{-2}}\ \text{V/m}

E_m = \frac{I_m}{\varepsilon_0\omega S} \approx 0.71\ \text{kV/m}

(2)

Check

Units: $I_m/(\varepsilon_0\omega S)$ is $\text{A}/(\text{F/m}\cdot\text{s}^{-1}\cdot\text{m}^2) = \text{A}\cdot\text{s}/(\text{F}\cdot\text{m}) = \text{C}/(\text{F}\cdot\text{m}) = \text{V/m}$ , a field strength. Limit: at a fixed current amplitude a lower frequency means each half cycle carries more charge onto the plates, and (2) grows as $1/\omega$ , as it should.

Remark

The plate separation never entered: the field is fixed by the charge per unit area, and the leads deliver that charge whatever the gap. The gap enters only the voltage, $E_m d$ , and so the capacitor’s reactance $1/(\omega C)$ .