Displacement Current in a Parallel-Plate Capacitor: 04-BS-9, May 2013, Q5

Question 5: Displacement Current in a Parallel-Plate Capacitor (20 marks)

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

Given.

Quantity	Value
Plate radius $R$	5 cm $=0.05$ m
Air gap	0.1 mm (does not enter either answer below)
Drive current	$i(t)=I_0\cos(\omega t)$, $I_0=1$ A (peak)
Frequency $f$	$10^6$ Hz

Find. (i) RMS charge on the capacitor. (ii) RMS magnetic field intensity $H$ at the rim ($r=R$) of the gap.

Parallel circular plates; the conduction current charges the plates while an equal displacement current $J_d$ crosses the gap; $H$ circulates around an Ampèrian loop at the rim, $r=R$.

Approach. The plate current is the time-derivative of the stored charge, so charge is found by integrating the given current; $H$ at the rim follows from Ampère's law with the displacement current uniformly spread over the plate area (Maxwell's correction), evaluated at $r=R$ where the full current is enclosed.

Angular frequency. $$\omega=2\pi f = 2\pi(10^6)=6.283\times10^6\ \text{rad/s}.$$
Charge from the drive current. Since $i(t)=dq/dt=I_0\cos\omega t$, $$q(t)=\int i\,dt=\frac{I_0}{\omega}\sin\omega t,$$ a sinusoid of peak $q_0=I_0/\omega$. Its RMS value is $$q_{\text{rms}}=\frac{I_0}{\omega\sqrt2}=\frac{1}{(6.283\times10^6)(1.4142)}$$ $$q_{\text{rms}}=\boxed{112.5\ \text{nC}}$$
Displacement current density and $H$ at the rim. The plate current spreads uniformly over the plate area $\pi R^2$, so $J_d(t)=i(t)/\pi R^2$. Applying Ampère's law to a circular loop of radius $r$ concentric with the plates, the enclosed displacement current for $r\le R$ is $J_d\cdot\pi r^2$, so $$H(r,t)\,2\pi r = J_d(t)\,\pi r^2 \;\Rightarrow\; H(r,t)=\frac{i(t)\,r}{2\pi R^2}.$$ At the rim ($r=R$) this simplifies to $H(R,t)=i(t)/2\pi R$, peak value $$H_0=\frac{I_0}{2\pi R}=\frac{1}{2\pi(0.05)}=3.183\ \text{A/m}.$$
RMS field at the rim. $$H_{\text{rms}}=\frac{H_0}{\sqrt2}=\boxed{2.25\ \text{A/m}}$$

Quantity	Result
$\omega$	$6.283\times10^6$ rad/s
RMS charge $q_{\text{rms}}$	112.5 nC
Peak rim field $H_0$	3.18 A/m
RMS rim field $H_{\text{rms}}$	2.25 A/m

Concept & Theory

Topic. Maxwell's displacement current — magnetic field inside a charging capacitor.

Key relations. $i=dq/dt$; Ampère–Maxwell law $\oint\vec H\cdot d\vec l = I_{\text{enc}}+\dfrac{d}{dt}\displaystyle\int\vec D\cdot d\vec A$; RMS of a sinusoid $=$ peak$/\sqrt2$.

Why this works. No conduction current actually crosses the gap between the plates, yet $\vec H$ circulates there exactly as if it did — Maxwell's displacement-current term $\partial\vec D/\partial t$ supplies the missing source term so Ampère's law stays consistent (and continuous) whether the loop is drawn through the wire or through the gap. Because the plates are circular and the current is spread uniformly, symmetry makes $H$ depend only on $r$, growing linearly out to the rim then falling as $1/r$ beyond it (like the field of a straight wire).

Common pitfall. Treating "1A intensity" as already an RMS value and skipping the $/\sqrt2$ step — the question explicitly asks for the RMS of a stored quantity, signalling the given current is the sinusoid's amplitude, not its RMS. Also, forgetting that at $r=R$ the whole displacement current is enclosed, so $H(R)$ has the same simple form as the field just outside a long straight wire carrying the plate current.

Source. Sadiku, Elements of Electromagnetics, Ch. 9 (Maxwell's Equations, Displacement Current).

Question 5 of 8: Displacement Current in a Parallel-Plate Capacitor

Question 5: Displacement Current in a Parallel-Plate Capacitor (20 marks)