Discharging a Capacitor – Magnetic‑Field Aspects (Cambridge A‑Level Physics 9702)
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.
1. Syllabus Context
19.3 Discharging a Capacitor – recall the time constant \(\tau =RC\), exponential decay of voltage, charge and current, analyse related graphs and calculate \(\tau\).
20 Electromagnetic Induction – apply the full Maxwell–Ampère law, understand displacement current, and relate the magnetic field produced by a discharging capacitor to induced emf in surrounding loops.
2. The RC‑Discharge Circuit
Components: capacitor \(C\) (initial voltage \(V_{0}\)), resistor \(R\), ideal switch.
Procedure: Switch is closed at \(t=0\); current flows and the charge on the plates decays exponentially.
Simple RC discharge circuit. A Hall‑probe or a small circular coil can be placed around the capacitor to record the magnetic field.
Solution (exponential decay):
\[
Q(t)=Q_{0}e^{-t/RC},\qquad
I(t)=\frac{dQ}{dt}=-\frac{Q_{0}}{RC}e^{-t/RC},
\]
where \(Q_{0}=CV_{0}\) and \(\tau =RC\).
Voltage across the capacitor:
\[
V(t)=\frac{Q(t)}{C}=V_{0}e^{-t/\tau}.
\]
Electric field between the plates (plate separation \(d\)):
\[
E(t)=\frac{V(t)}{d}= \frac{Q(t)}{C\,d}.
\]
Displacement‑current density (Maxwell’s addition):
\[
J_{D}= \varepsilon_{0}\frac{\partial E}{\partial t}.
\]
4. Maxwell–Ampère Law – Core Idea
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.
Why the displacement term? Inside the capacitor there is no conduction current, yet the line integral of \(\mathbf B\) must be continuous. The term \(\varepsilon_{0}d\Phi_{E}/dt\) supplies the missing “current”.
During the discharge the conduction current in the external circuit and the displacement current between the plates are equal at every instant:
\[
I_{D}= \varepsilon_{0}\frac{d\Phi_{E}}{dt}=I(t).
\]
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.
5. Magnetic Field Outside the Capacitor (around the connecting wire)
Apply Ampère’s law to a circular path of radius \(r\) centred on the straight wire carrying the discharge current.
Find the magnetic field at a point \(r=1.0\;\text{cm}\) inside the capacitor after \(t=2\tau\).
Solution:
Time constant: \(\tau =RC = (5\times10^{3})(10\times10^{-6}) = 5.0\times10^{-2}\;\text{s}\).
Current at \(t=2\tau\):
\[
I(2\tau)=\frac{V_{0}}{R}e^{-2}
=\frac{12}{5\times10^{3}}e^{-2}
\approx 2.4\times10^{-3}\times0.135
=3.2\times10^{-4}\;\text{A}.
\]
Magnetic field inside the capacitor:
\[
B(r,t)=\frac{\mu_{0}I(t)}{2\pi a^{2}}\,r
=\frac{4\pi\times10^{-7}\times3.2\times10^{-4}}{2\pi(0.02)^{2}}\times0.01
\approx 1.6\times10^{-9}\;\text{T}.
\]
Direction: using the right‑hand rule, the field points into the page for the chosen current direction.
11. Common Misconceptions
“No magnetic field because the current is decreasing.” – The magnetic field depends on the instantaneous current, not on its rate of change.
“Displacement current is only a mathematical trick.” – It is a real physical term required for the continuity of \(\mathbf B\) in regions without conduction current.
“Magnetic field inside the capacitor must be zero because there are no wires.” – The changing electric field creates a displacement current, which generates a magnetic field exactly as a wire does.
12. Practice Questions (AO2)
Derive the expression \(B_{\text{inside}}(r,t)=\dfrac{\mu_{0}I(t)}{2\pi a^{2}}r\) assuming a uniform electric field between the plates.
A \(22\;\mu\text{F}\) capacitor discharges through a \(10\;\text{k}\Omega\) resistor. Calculate the magnetic field at \(r=5\;\text{mm}\) from the connecting wire after one time constant.
Explain qualitatively how the direction of the magnetic field changes if the polarity of the initially charged capacitor is reversed before discharge.
Design a simple experiment (using a Hall probe or a small coil) to verify the linear dependence of \(B\) on \(r\) inside the capacitor.
13. Further Reading (AO1)
Full set of Maxwell’s equations – especially the Maxwell–Ampère law and its role in electromagnetic waves.
Induced emf in a loop surrounding a discharging capacitor – relationship to Faraday’s law.
Energy stored in magnetic fields: \(U_{B}= \frac{1}{2\mu_{0}}\int B^{2}\,dV\) and its conversion from electric‑field energy.
Experimental techniques for measuring weak magnetic fields (Hall probes, search‑coil method).
14. Actionable Review – Alignment with Cambridge Syllabus 9702
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. \(t\).
Your generous donation helps us continue providing free Cambridge IGCSE & A-Level resources,
past papers, syllabus notes, revision questions, and high-quality online tutoring to students across Kenya.