Question 6 of 7: Parallel-plate capacitor and displacement current

Question 6: Parallel-plate capacitor and displacement current (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. A square parallel-plate capacitor with a lossless high-permittivity filling and a plate spacing sixteen times smaller than the plate edge, so fringing is negligible.

Find. The capacitance; the electric field and flux density under 5 V DC; the plate surface charge density; and the rms displacement and conduction currents under a 5 V rms sinusoid.

Approach. Use the ideal parallel-plate model for $C$, get E from the uniform-field assumption, apply the constitutive relation for D, use the conductor boundary condition for $\rho_s$, and finally separate the two current mechanisms in Ampère's law.

1. Capacitance of the parallel-plate geometry. With $w/d = 4/0.25 = 16$, fringing contributes a fraction of a percent and the ideal formula applies: $$C = \frac{\varepsilon_0\varepsilon_r A}{d} = \frac{(8.854\times 10^{-12})(10.6)(16\times 10^{-6})}{2.5\times 10^{-4}} = \boxed{6.01\ \text{pF}}.$$
2. Electric field under the DC potential. The field between closely spaced plates is uniform and normal to them, so the potential is simply $V = Ed$: $$E = \frac{V}{d} = \frac{5}{2.5\times 10^{-4}} = \boxed{2.00\times 10^{4}\ \text{V/m} = 20.0\ \text{kV/m}}$$ directed from the positive plate to the negative plate.
3. Electric flux density. The constitutive relation for a linear isotropic dielectric gives $$D = \varepsilon_0\varepsilon_r E = (8.854\times 10^{-12})(10.6)(2.00\times 10^{4}) = \boxed{1.877\ \mu\text{C/m}^{2}}$$ in the same direction as E. Note that D is set by the free charge alone, which is why it carries the permittivity while E does not.
4. Surface charge density on the plates. At the face of a perfect conductor the boundary condition is $\rho_s = \mathbf{D}\cdot\hat{\mathbf{n}} = D_n$, so $$\boxed{\rho_s = \pm 1.877\ \mu\text{C/m}^{2}}$$ positive on the plate held at the higher potential and negative on the other. Cross-checking through the lumped model, the stored charge is $Q = CV = (6.01\ \text{pF})(5\ \text{V}) = 30.0\ \text{pC}$, and $Q/A = 30.0\times 10^{-12}/16\times 10^{-6} = 1.877\ \mu\text{C/m}^{2}$ — identical, as it must be.
5. (d)(i) RMS displacement current. The displacement current is the surface integral of $\partial\mathbf{D}/\partial t$ across the gap. For a sinusoid, differentiation in the phasor domain multiplies by $j\omega$, so $$I_d = \left|\int_S \frac{\partial \mathbf{D}}{\partial t}\cdot d\mathbf{S}\right| = \omega\,\frac{\varepsilon_0\varepsilon_r A}{d}\,V_{\text{rms}} = \omega C V_{\text{rms}} = 2\pi f\,(6.01\ \text{pF})(5\ \text{V}).$$ The question does not state the source frequency, so the answer is necessarily a function of it: $$\boxed{I_d = 2\pi f C V_{\text{rms}} = \left(1.887\times 10^{-10}\ \text{A per Hz}\right)\times f}$$ which evaluates to 188.7 $\mu$A at 1 MHz and 188.7 mA at 1 GHz. Equivalently the displacement current density is $J_d = \omega\varepsilon_0\varepsilon_r V_{\text{rms}}/d$, and $J_dA$ reproduces the same number.
6. (d)(ii) RMS conduction current. Conduction current is $\mathbf{J}_c = \sigma\mathbf{E}$, and the dielectric is stated to be lossless: $\sigma = 0$. Therefore $$\boxed{I_c = \sigma E A = 0\ \text{A}}$$ no matter what the frequency or the applied voltage is. This is the whole point of Maxwell's addition to Ampère's law: a conduction current $I_d$ flows in the external leads and stops dead at the plate, yet Ampère's circuital law must give the same answer for a surface passing through the wire and for one passing between the plates. The displacement current supplies exactly the missing term, so the lead current equals $I_d$ even though no charge crosses the gap. A real capacitor with a small loss tangent $\tan\delta$ would carry $I_c = \tan\delta\cdot I_d$; here $\tan\delta = 0$ by hypothesis.