This topic connects the RC‑discharge of a capacitor (Syllabus 19.3) with the Maxwell–Ampère law (Syllabus 20). It shows how a changing electric field produces a magnetic field (displacement current) and why this is essential for electromagnetic induction.
\[
\frac{dQ}{dt}+\frac{Q}{RC}=0 .
\]
\[
Q(t)=Q_{0}e^{-t/RC},\qquad
I(t)=\frac{dQ}{dt}=-\frac{Q_{0}}{RC}e^{-t/RC},
\]
where \(Q{0}=CV{0}\) and \(\tau =RC\).
\[
V(t)=\frac{Q(t)}{C}=V_{0}e^{-t/\tau}.
\]
\[
E(t)=\frac{V(t)}{d}= \frac{Q(t)}{C\,d}.
\]
\[
J{D}= \varepsilon{0}\frac{\partial E}{\partial t}.
\]
For any closed Amperian loop,
\[
\oint\mathbf B\!\cdot\!d\mathbf l
=\mu{0}\!\left(I{\text{cond}}+\varepsilon{0}\frac{d\Phi{E}}{dt}\right)
\tag{1}
\]
where \(\Phi_{E}\) is the electric flux through the surface bounded by the loop.
\[
I{D}= \varepsilon{0}\frac{d\Phi_{E}}{dt}=I(t).
\]
Link to Topic 20 – Electromagnetic Induction
The magnetic field produced by the changing electric flux inside the capacitor is identical to that produced by a real current of magnitude \(I(t)\). Consequently a loop placed around the capacitor experiences an induced emf \(\mathcal{E}= -d\Phi_{B}/dt\) exactly as in a conventional inductive circuit.
Apply Ampère’s law to a circular path of radius \(r\) centred on the straight wire carrying the discharge current.
\[
\oint\mathbf B\!\cdot\!d\mathbf l = B(2\pi r)=\mu_{0}I(t)
\]
\[
\boxed{B{\text{outside}}(r,t)=\frac{\mu{0}I(t)}{2\pi r}
=\frac{\mu{0}Q{0}}{2\pi rRC}\,e^{-t/RC}}
\]
Consider a circular Amperian loop of radius \(r\) (\(r
\[
\Phi_{E}(r)=E(t)\,\pi r^{2}.
\]
\[
I{D}(r)=\varepsilon{0}\frac{d\Phi_{E}}{dt}
=I(t)\frac{r^{2}}{a^{2}} .
\]
\[
B(2\pi r)=\mu{0}I{D}(r)
\;\Longrightarrow\;
\boxed{B{\text{inside}}(r,t)=\frac{\mu{0}I(t)}{2\pi a^{2}}\,r\qquad(r\le a)} .
\]
The magnetic field varies linearly with radius inside the plates – a direct consequence of the displacement‑current concept.
\[
U{E}= \tfrac12 CV{0}^{2}.
\]
\[
U{B}= \frac{1}{2\mu{0}}\int B^{2}\,dV .
\]
(For a parallel‑plate geometry the integral gives a value proportional to \(I^{2}t^{2}\), showing the continual conversion between electric and magnetic forms.)
\[
B{\text{inside}}(r,t)=\frac{\mu{0}I(t)}{2\pi a^{2}}\,r .
\]
| Quantity | Expression | Notes (Cambridge Syllabus) |
|---|---|---|
| Time constant | \(\tau =RC\) | Controls the rate of decay (19.3). |
| Charge on capacitor | \(Q(t)=Q_{0}e^{-t/\tau}\) | \(Q{0}=CV{0}\). |
| Voltage across capacitor | \(V(t)=V_{0}e^{-t/\tau}\) | Directly measurable; used to plot exponential decay. |
| Current in circuit | \(I(t)= -\dfrac{Q{0}}{\tau}e^{-t/\tau}= \dfrac{V{0}}{R}e^{-t/\tau}\) | Instantaneous magnitude determines magnetic field. |
| Displacement current | \(I{D}(t)=\varepsilon{0}\dfrac{d\Phi_{E}}{dt}=I(t)\) | Ensures continuity of \(\oint\mathbf B\!\cdot\!d\mathbf l\) (20). |
| Magnetic field outside the wire | \(B{\text{outside}}(r,t)=\dfrac{\mu{0}I(t)}{2\pi r}\) | Same form as for a steady current. |
| Magnetic field inside the capacitor (r ≤ a) | \(B{\text{inside}}(r,t)=\dfrac{\mu{0}I(t)}{2\pi a^{2}}\,r\) | Linear with radius – result of displacement current. |
| Electric‑field energy | \(U_{E}= \tfrac12 CV^{2}\) | Decreases as the capacitor discharges. |
| Magnetic‑field energy | \(U{B}= \dfrac{1}{2\mu{0}}\int B^{2}\,dV\) | Transiently stores part of \(U_{E}\) during discharge. |
\[
I(2\tau)=\frac{V_{0}}{R}e^{-2}
=\frac{12}{5\times10^{3}}e^{-2}
\approx 2.4\times10^{-3}\times0.135
=3.2\times10^{-4}\;\text{A}.
\]
\[
B(r,t)=\frac{\mu_{0}I(t)}{2\pi a^{2}}\,r
=\frac{4\pi\times10^{-7}\times3.2\times10^{-4}}{2\pi(0.02)^{2}}\times0.01
\approx 1.6\times10^{-9}\;\text{T}.
\]
| Syllabus Requirement | How the Notes Meet It | Suggested Improvement |
|---|---|---|
| 19.3 Discharging a Capacitor – recall \(\tau=RC\), exponential decay of \(V,Q,I\); analyse graphs; calculate \(\tau\). | All governing equations, explicit \(\tau\) definition, example calculation, and a reminder to plot \(V(t)\) and \(I(t)\). | Add a small “Graph Sketch” box showing the expected shape of \(V\) vs. \(t\) and \(I\) vs. \(t\). |
| 20 Electromagnetic Induction – apply Maxwell–Ampère law, understand displacement current, relate to induced emf. | Full derivation of the magnetic field inside and outside the capacitor, clear link to induced emf, and practical AO3 activity. | Include a concise statement: “The induced emf in a surrounding loop equals \(\mu_{0}I(t)\) for a loop that encloses the capacitor.” |
| Use of mathematical techniques – differentiate, integrate, manipulate exponential functions. | Derivations of \(I(t)\), \(I_{D}\), and \(B\) use differentiation; energy expressions involve integration. | Insert a short “Mathematical tip” after the energy section reminding students of the integral \(\int B^{2} dV\) for a cylindrical volume. |
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