Question 6 of 7: Parallel-Plate Capacitor with a Lossless Dielectric

Question 6: Parallel-Plate Capacitor with a Lossless Dielectric (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. Square plates $4\ \text{mm}\times4\ \text{mm}$ ($A=16\ \text{mm}^2$), separation $d=0.25$ mm, lossless dielectric $\varepsilon_r=10.6$; DC potential $V=5$ V (parts b, c); AC source $V_{rms}=5$ V, frequency unstated (part d).

Find. $C$; $E$ and $D$ between the plates at $V=5$ V; the surface charge density; the RMS displacement and conduction currents under a 5 V$_{rms}$ AC excitation.

Approach. Use the parallel-plate formula for $C$; $E=V/d$ then $D=\varepsilon E$ for the field quantities; the boundary condition $\rho_s=D_n$ at a perfect conductor for the surface charge; and the definition of displacement current ($J_d=\partial D/\partial t$) versus conduction current ($J_c=\sigma E$) for part (d).

1. Part (a) — Capacitance. $A=(4\times10^{-3})^2=1.6\times10^{-5}\ \text{m}^2$, $d=0.25\times10^{-3}\ \text{m}$: $$C=\frac{\varepsilon_0\varepsilon_r A}{d}=\frac{(8.854\times10^{-12})(10.6)(1.6\times10^{-5})}{0.25\times10^{-3}}=\boxed{6.01\ \text{pF}}.$$
2. Part (b) — Electric field and flux density at $V=5$ V. $$E=\frac{V}{d}=\frac{5}{0.25\times10^{-3}}=\boxed{2.00\times10^4\ \text{V/m} = 20\ \text{kV/m}},$$ $$D=\varepsilon_0\varepsilon_r E=(8.854\times10^{-12})(10.6)(2\times10^4)=\boxed{1.877\ \mu\text{C/m}^2}.$$
3. Part (c) — Surface charge density. At a perfectly conducting plate the boundary condition gives $\rho_s=D_n$ directly (no free surface current, all of $D$ terminates on the plate charge): $$\boxed{\rho_s=\pm1.877\ \mu\text{C/m}^2}$$ ($+$ on the higher-potential plate, $-$ on the other).
4. Part (d) — Displacement vs. conduction current under AC excitation. The problem states the dielectric is LOSSLESS, i.e. its conductivity is $\sigma=0$ identically — not approximately small, but exactly zero. The conduction current density is $J_c=\sigma E$, so $$\boxed{I_{c,rms}=0\ \text{(exactly, at any frequency)}.}$$ With no stated drive frequency, the displacement current is left symbolic in $\omega=2\pi f$: for a sinusoidal source the capacitor's total current is purely displacement current, $I_{d,rms}=\omega C\,V_{rms}$, so $$\boxed{I_{d,rms}=\omega\,(6.01\ \text{pF})(5\ \text{V}) = (30.0\ \text{pF}\cdot\text{V})\,\omega = 30.0\,\omega\ \text{pA (}\omega\text{ in rad/s)}.}$$ (For illustration only, at a representative $f=1$ MHz this would be $I_{d,rms}=2\pi(10^6)(6.01\times10^{-12})(5)=189\ \mu\text{A}$ — check against the actual test frequency if one is specified elsewhere on the exam.)