Exercises — Displacement current — Maxwell's addition to Ampere's law

Only (b), a changing electric field. Maxwell's insight is that $I_d={\epsilon }_{0}d{\Phi }_E/dt$ — the rate of change of flux, not the field itself. A steady field has $d{\Phi }_E/dt=0$, so $I_d=0$; a stationary charge has no current at all.

False. No charge crosses the gap — electrons pile onto one plate and leave the other. What crosses conceptually is the effect of the growing field, captured by $I_d$. This is the whole reason Maxwell needed a new term instead of just "more current".

Only $I_d$. The conduction current $I_c$ is zero there (no charge flows through the gap), but the electric flux through that surface is growing, so $I_d={\epsilon }_{0}d{\Phi }_E/dt\ne 0$. On that surface the whole magnetic field comes from $I_d$.

Since $Q=CV$ and $I_d=I_c=\frac{dQ}{dt}=C\frac{dV}{dt}$: $I_d=(2\times 10^{-6})(10^6)=2\text{A}.$ Notice we never touched the field — because $I_d$ always equals the conduction current feeding the capacitor.

Straight from the definition: $I_d={\epsilon }_0\frac{d{\Phi }_E}{dt}=(8.85\times 10^{-12})(4\times 10^6)=3.5\times 10^{-5}\text{A}.$ No charges appear anywhere — this is the pure-vacuum form that keeps working inside a light wave.

Flux is ${\Phi }_E=EA$ (uniform field ⟂ to area), so $\frac{d{\Phi }_E}{dt}=A\frac{dE}{dt}$. Then $I_d={\epsilon }_{0}A\frac{dE}{dt}=(8.85\times 10^{-12})(0.02)(5\times 10^9)=8.85\times 10^{-4}\text{A}.$

Symmetry (WHAT it looks like): by the circular symmetry of the setup (see figure), $\vec{B}$ is tangential and constant on a circle of radius $r$, so $\oint \vec{B}\cdot d\vec{l}=B(2\pi r)$. Enclosed $I_d$: the field is uniform across the plate, so the flux through a loop of radius $r$ is the fraction $\frac{\pi r^2}{\pi R^2}$ of the total. Thus $I_{d,\text{enc}}=I\frac{r^2}{R^2}.$ Apply the law: $B(2\pi r)={\mu }_{0}I\frac{r^2}{R^2}$, so $B=\frac{{\mu }_{0}Ir}{2\pi R^2}=\frac{(4\pi \times 10^{-7})(2)(0.03)}{2\pi (0.05{)}^2}=4.8\times 10^{-6}\text{T}.$ Identical in form to $B$ inside a wire carrying uniform current — the promised symmetry.

Inside ($r<R$): only the fraction $r^2/R^2$ of flux is enclosed, giving $B=\frac{{\mu }_{0}Ir}{2\pi R^2}$ — grows linearly with $r$. Outside ($r>R$): the loop now encloses the entire flux, so $I_{d,\text{enc}}=I$ and $B(2\pi r)={\mu }_{0}I\Rightarrow B=\frac{{\mu }_{0}I}{2\pi r}$ — falls as $1/r$. Maximum is at the edge $r=R$. Check both formulas there: $B_{\text{in}}(R)=\frac{{\mu }_{0}IR}{2\pi R^2}=\frac{{\mu }_{0}I}{2\pi R},B_{\text{out}}(R)=\frac{{\mu }_{0}I}{2\pi R}.$ They match — the field is continuous. Numerically $B_{\max }=\frac{(4\pi \times 10^{-7})(2)}{2\pi (0.05)}=8.0\times 10^{-6}\text{T}$.

$I_d=I_c=\frac{dQ}{dt}=\frac{Q_0}{\tau }{e}^{-t/\tau }$.

• At $t=0$: $I_d=\frac{Q_0}{\tau }=\frac{6\times 10^{-6}}{2\times 10^{-3}}=3\times 10^{-3}\text{A}=3\text{mA}$.
• As $t\to \infty$: $e^{-t/\tau }\to 0$, so $I_d\to 0$. The field stops changing once the capacitor is full — no more displacement current, hence no more $B$.

(a) Field between plates $E=\frac{Q}{{\epsilon }_{0}A}$ with $A=\pi R^2$. Differentiate: $\frac{dE}{dt}=\frac{1}{{\epsilon }_{0}A}\frac{dQ}{dt}=\frac{I}{{\epsilon }_0\pi R^2}=\frac{1.5}{(8.85\times 10^{-12})\pi (0.04{)}^2}=3.37\times 10^{13}\text{V/(m⋅s)}.$ (b) $I_d={\epsilon }_{0}A\frac{dE}{dt}={\epsilon }_0\pi R^2\cdot \frac{I}{{\epsilon }_0\pi R^2}=I=1.5\text{A}.$ The displacement current in the gap exactly equals the conduction current in the wire — the paradox is dissolved, quantitatively.

(i) At the edge the whole current is enclosed: $B=\frac{{\mu }_{0}I}{2\pi R}=\frac{(4\pi \times 10^{-7})(1.5)}{2\pi (0.04)}=7.5\times 10^{-6}\text{T}.$ (ii) $B(2\pi R)={\mu }_0{\epsilon }_0\frac{d{\Phi }_E}{dt}={\mu }_0{\epsilon }_0(\pi R^2)\frac{dE}{dt}$. Using $dE/dt$ from L4.1: $B=\frac{{\mu }_0{\epsilon }_{0}R}{2}\frac{dE}{dt}=\frac{(4\pi \times 10^{-7})(8.85\times 10^{-12})(0.04)}{2}(3.37\times 10^{13})=7.5\times 10^{-6}\text{T}.$ Same number — because ${\epsilon }_{0}d{\Phi }_E/dt$ is the current $I$.

Units: ${\mu }_0$ is $\text{T⋅m/A}={\text{kg⋅m/A}}^2{\text{s}}^2$ and ${\epsilon }_0$ is ${\text{A}}^2{\text{s}}^4/({\text{kg⋅m}}^3)$. Their product ${\mu }_0{\epsilon }_0$ has units ${\text{s}}^2/{\text{m}}^2$, so $1/\sqrt{{\mu }_0{\epsilon }_0}$ has units $\text{m/s}$ — a speed. $\frac{1}{\sqrt{{\mu }_0{\epsilon }_0}}=\frac{1}{\sqrt{(4\pi \times 10^{-7})(8.85\times 10^{-12})}}=3.00\times 10^8\text{m/s}.$ This is $c$, the Speed of Light. The very term Maxwell added ($I_d$) is what puts ${\epsilon }_0$ into the wave equation — without displacement current there is no $c$ and no light. See Electromagnetic Waves.

In vacuum $I_c=0$ everywhere, so the only source of $\vec{B}$ is the displacement term ${\mu }_0{\epsilon }_0\partial \vec{E}/\partial t$. This is exactly the mirror of Faraday's Law of Induction (changing $\vec{B}$ makes $\vec{E}$); together they make a self-feeding wave. Numerically, treating $\vec{E}\perp$ to the patch: $I_d={\epsilon }_{0}A\frac{dE}{dt}=(8.85\times 10^{-12})(1)(3\times 10^{11})=2.66\text{A}.$ A pure field change — no electrons — carrying a multi-amp "current". That is displacement current standing entirely on its own.