Noesis
Level 5 — MasteryElectromagnetism

Electromagnetism

⏱ 90 minutes60 marksprintable — key stays hidden on paper

Time limit: 90 minutes Total marks: 60 Instructions: Attempt all three questions. Full derivations, physical reasoning, and (where indicated) algorithmic/computational descriptions are required. Use ε0=8.854×10−12 F/m\varepsilon_0 = 8.854\times10^{-12}\,\mathrm{F/m}, μ0=4π×10−7 H/m\mu_0 = 4\pi\times10^{-7}\,\mathrm{H/m}, c=3.00×108 m/sc = 3.00\times10^8\,\mathrm{m/s} where numerical values are needed.

Question 1 — From Gauss to Energy in a Dielectric-Loaded Capacitor (20 marks)

A spherical capacitor consists of an inner conducting sphere of radius aa carrying charge +Q+Q and a concentric conducting shell of inner radius bb carrying −Q-Q. The gap is filled with a linear dielectric whose permittivity varies radially as ε(r)=ε0 κ0 ar,a≤r≤b,\varepsilon(r) = \varepsilon_0\,\kappa_0\,\frac{a}{r}, \qquad a \le r \le b, where κ0\kappa_0 is a dimensionless constant.

(a) Using Gauss's law in a dielectric (with the free charge QQ), derive the electric field E(r)E(r) in the gap. (4 marks)

(b) Derive the potential difference VV between the spheres and hence show that the capacitance is C = \frac{4\pi\varepsilon_0\kappa_0\,a}{\ln(b/a)}. \tag{4 marks}

(c) Derive the total electrostatic energy stored, first via U=12QVU=\tfrac12 QV and independently via the field-energy integral U=∫12ε(r)E2 dVU=\int \tfrac12 \varepsilon(r) E^2\, dV, and confirm the two agree. (6 marks)

(d) Taking a=1.0 cma = 1.0\,\mathrm{cm}, b=3.0 cmb = 3.0\,\mathrm{cm}, κ0=5\kappa_0 = 5, Q=2.0 nCQ = 2.0\,\mathrm{nC}, compute CC (in pF) and UU (in nJ). (3 marks)

(e) Explain physically why the rr-dependent permittivity removes the usual 1/r21/r^2-type non-uniformity from the potential integrand, and interpret the appearance of a logarithm (normally a cylindrical signature) in a spherical geometry. (3 marks)

Question 2 — Maxwell, Displacement Current, and a Poynting Audit (22 marks)

A parallel-plate capacitor (circular plates of radius RR, separation d≪Rd\ll R, vacuum gap) is being charged by a constant current II delivered by straight leads on the axis.

(a) Starting from the Ampère–Maxwell law in integral form, ∮B⋅dℓ=μ0(Ienc+ε0dΦEdt),\oint \mathbf{B}\cdot d\boldsymbol{\ell} = \mu_0\left(I_{\text{enc}} + \varepsilon_0\frac{d\Phi_E}{dt}\right), show that the magnetic field at radius r<Rr<R between the plates (no conduction current there) is B(r) = \frac{\mu_0 I r}{2\pi R^2}. \tag{5 marks}

(b) Compute the electric field E(t)E(t) between the plates in terms of II, tt, RR, ε0\varepsilon_0 (plates initially uncharged), and hence the Poynting vector S\mathbf{S} magnitude at the rim r=Rr=R. State its direction. (5 marks)

(c) By integrating S\mathbf{S} over the cylindrical rim surface of the gap, show that the total power flowing into the field-filled volume equals ddt(12ε0E2⋅πR2d)\dfrac{d}{dt}\left(\tfrac12 \varepsilon_0 E^2 \cdot \pi R^2 d\right), i.e. the rate of increase of stored energy. (6 marks)

(d) Write a short algorithm (pseudocode, ~10 lines) that numerically verifies part (c): it should choose sample parameters, compute the rim Poynting influx and the time-derivative of stored energy at time tt, and return whether they agree to within a tolerance. State clearly what quantities are compared. (4 marks)

(e) Explain why displacement current is not merely a bookkeeping trick but is required for consistency of ∇⋅J\nabla\cdot\mathbf{J} with charge conservation. (2 marks)

Question 3 — LC/RLC Oscillation as SHM, with a Faraday Twist (18 marks)

An ideal LCLC loop has inductance LL and capacitance CC; at t=0t=0 the capacitor holds charge Q0Q_0 and the current is zero.

(a) From Kirchhoff's voltage law, derive the differential equation for q(t)q(t) and show it is SHM with ω0=1/LC\omega_0 = 1/\sqrt{LC}. Solve for q(t)q(t) and i(t)i(t). (4 marks)

(b) Prove that total energy U=q22C+12Li2U = \frac{q^2}{2C} + \frac12 L i^2 is conserved, and identify the electrical analogs of mass, spring constant, position, and velocity. (4 marks)

(c) Now a small resistance RR is added in series. Derive the condition on RR, LL, CC for underdamped oscillation, and give the decay envelope of the charge amplitude. (5 marks)

(d) A single-turn circular loop of the same circuit (radius ρ\rho) sits in a uniform external field B(t)=B0sin⁡ΩtB(t) = B_0\sin\Omega t perpendicular to its plane. Using Faraday's and Lenz's laws, write the additional induced EMF term and state (with sign reasoning) whether at t=0+t=0^+ the induced current opposes or aids the capacitor-driven current if Ωt\Omega t is just past zero. (5 marks)

End of paper.

Answer keyMark scheme & solutions

Question 1

(a) Gauss's law in a dielectric uses ∮D⋅dA=Qfree\oint \mathbf{D}\cdot d\mathbf{A} = Q_{\text{free}}. Spherical symmetry: D(r) 4πr2=Q⇒D=Q4πr2D(r)\,4\pi r^2 = Q \Rightarrow D = \dfrac{Q}{4\pi r^2}. Then E=D/ε(r)=Q4πε0κ0a r2⋅r=Q4πε0κ0a rE = D/\varepsilon(r) = \dfrac{Q}{4\pi \varepsilon_0\kappa_0 a\, r^2}\cdot r = \dfrac{Q}{4\pi\varepsilon_0\kappa_0 a\, r}. E(r)=Q4πε0κ0a r\boxed{E(r) = \frac{Q}{4\pi\varepsilon_0\kappa_0 a\, r}} (D from Gauss: 2; divide by ε(r)\varepsilon(r) correctly, note the rr cancels one power: 2.)

(b) V=∫abE dr=Q4πε0κ0a∫abdrr=Q4πε0κ0aln⁡ba.V = \int_a^b E\,dr = \frac{Q}{4\pi\varepsilon_0\kappa_0 a}\int_a^b \frac{dr}{r} = \frac{Q}{4\pi\varepsilon_0\kappa_0 a}\ln\frac{b}{a}. C=QV=4πε0κ0aln⁡(b/a).✓C = \frac{Q}{V} = \frac{4\pi\varepsilon_0\kappa_0 a}{\ln(b/a)}. \checkmark (Integral setup 2; log result and C=Q/VC=Q/V: 2.)

(c) Via U=12QVU=\tfrac12 QV: U=Q28πε0κ0aln⁡ba.U = \frac{Q^2}{8\pi\varepsilon_0\kappa_0 a}\ln\frac{b}{a}. Via field integral, dV=4πr2drdV = 4\pi r^2 dr:

=\int_a^b \tfrac12\Big(\varepsilon_0\kappa_0\frac{a}{r}\Big)\Big(\frac{Q}{4\pi\varepsilon_0\kappa_0 a r}\Big)^2 4\pi r^2\,dr.$$ Simplify integrand: $\tfrac12\cdot\dfrac{Q^2}{4\pi\varepsilon_0\kappa_0 a}\cdot\dfrac{1}{r}$, so $$U=\frac{Q^2}{8\pi\varepsilon_0\kappa_0 a}\int_a^b\frac{dr}{r}=\frac{Q^2}{8\pi\varepsilon_0\kappa_0 a}\ln\frac{b}{a}.$$ Identical. $\checkmark$ *(Each method 3 marks; must show the $r$-powers collapse to $1/r$.)* **(d)** $\ln(3)=1.0986$. $$C=\frac{4\pi(8.854\times10^{-12})(5)(0.01)}{1.0986}=\frac{5.563\times10^{-12}}{1.0986}=5.06\times10^{-12}\,\mathrm{F}\approx 5.06\,\mathrm{pF}.$$ $$U=\frac{Q^2}{2C}=\frac{(2.0\times10^{-9})^2}{2(5.06\times10^{-12})}=\frac{4.0\times10^{-18}}{1.013\times10^{-11}}=3.95\times10^{-7}\,\mathrm{J}\approx 395\,\mathrm{nJ}.$$ *(C: 1.5; U: 1.5.)* **(e)** The permittivity $\propto 1/r$ multiplies $D\propto 1/r^2$ to give $E\propto 1/r$; the single inverse power integrates to a logarithm. Physically the dielectric "softens" the field growth toward small $r$, converting the spherical $1/r^2$ falloff into an effective $1/r$ (cylindrical-like) profile, hence the log. *(Concept of cancellation 2; interpretation 1.)* --- ## Question 2 **(a)** Choose an Amperian circle radius $r<R$ coaxial with the plates. No conduction current pierces it. The E-field between plates is $E=\sigma/\varepsilon_0 = Q_{\text{plate}}/(\varepsilon_0\pi R^2)$; flux through the circle $\Phi_E = E\cdot\pi r^2$. Displacement current enclosed $=\varepsilon_0\dfrac{d\Phi_E}{dt}=\varepsilon_0\pi r^2\dfrac{dE}{dt}$. Since $\varepsilon_0\pi R^2\dfrac{dE}{dt}=I$ (total), the fraction $=I\,r^2/R^2$. Then $$B\,2\pi r = \mu_0 I \frac{r^2}{R^2}\Rightarrow B=\frac{\mu_0 I r}{2\pi R^2}.\checkmark$$ *(Displacement current 2; enclosed fraction 2; solve 1.)* **(b)** $Q(t)=It$ so $E(t)=\dfrac{It}{\varepsilon_0\pi R^2}$. $B(R)=\dfrac{\mu_0 I}{2\pi R}$. Poynting magnitude $S=\dfrac{EB}{\mu_0}=\dfrac{1}{\mu_0}\cdot\dfrac{It}{\varepsilon_0\pi R^2}\cdot\dfrac{\mu_0 I}{2\pi R}=\dfrac{I^2 t}{2\pi^2\varepsilon_0 R^3}$. Direction: $\mathbf{S}=\mathbf{E}\times\mathbf{B}/\mu_0$ points radially **inward** (into the gap). *(E(t): 1; B(R):1; S:2; direction:1.)* **(c)** Rim area $=2\pi R\,d$. Influx power $$P=S\cdot 2\pi R d=\frac{I^2 t}{2\pi^2\varepsilon_0 R^3}\cdot 2\pi R d=\frac{I^2 t\,d}{\pi\varepsilon_0 R^2}.$$ Stored energy $U=\tfrac12\varepsilon_0 E^2\cdot(\pi R^2 d)=\tfrac12\varepsilon_0\dfrac{I^2t^2}{\varepsilon_0^2\pi^2R^4}\pi R^2 d=\dfrac{I^2 t^2 d}{2\pi\varepsilon_0 R^2}$. $$\frac{dU}{dt}=\frac{I^2 t\,d}{\pi\varepsilon_0 R^2}=P.\checkmark$$ *(P integral 3; U and derivative 3.)* **(d)** Pseudocode: ``` choose I, R, d, t, eps0 E = I*t/(eps0*pi*R**2) B = mu0*I/(2*pi*R) S = E*B/mu0 P = S*(2*pi*R*d) # rim influx U = 0.5*eps0*E**2*(pi*R**2*d) # stored energy # numerical derivative of U(t): dU = (U(t+h) - U(t-h))/(2h) # recompute U at t±h return abs(P - dU) < tol ``` Compares rim Poynting influx $P$ against $dU/dt$ (central difference). *(Compute S/P 2; dU/dt & comparison 2.)* **(e)** Taking divergence of $\nabla\times\mathbf{B}=\mu_0\mathbf{J}$ gives $0=\mu_0\nabla\cdot\mathbf{J}$, false when charge accumulates ($\nabla\cdot\mathbf{J}=-\partial\rho/\partial t\neq0$). Adding $\mu_0\varepsilon_0\partial\mathbf{E}/\partial t$ and using $\nabla\cdot\mathbf{E}=\rho/\varepsilon_0$ restores $\nabla\cdot\mathbf{J}+\partial\rho/\partial t=0$. So displacement current is demanded by charge conservation. *(2.)* --- ## Question 3 **(a)** KVL: $\dfrac{q}{C}+L\dfrac{di}{dt}=0$, $i=\dot q$, so $L\ddot q + q/C=0\Rightarrow \ddot q+\omega_0^2 q=0$, $\omega_0=1/\sqrt{LC}$. With $q(0)=Q_0,\ i(0)=0$: $q(t)=Q_0\cos\omega_0 t$, $i(t)=-Q_0\omega_0\sin\omega_0 t$. *(ODE 2; solution 2.)* **(b)** $\dfrac{dU}{dt}=\dfrac{q\dot q}{C}+Li\dot i=\dot q\!\left(\dfrac{q}{C}+L\ddot q\right)=0$. Conserved. Analogs: $L\leftrightarrow m$ (inertia), $1/C\leftrightarrow k$ (stiffness), $q\leftrightarrow x$, $i=\dot q\leftrightarrow v$. *(Proof 2; analogs 2.)* **(c)** With $R$: $L\ddot q+R\dot q+q/C=0$. Underdamped when discriminant negative: $$R^2<\frac{4L}{C}\quad\Longleftrightarrow\quad R<2\sqrt{L/C}.$$ Decay envelope $\propto e^{-Rt/2L}$; damped frequency $\omega_d=\sqrt{\tfrac{1}{LC}-\tfrac{R^2}{4L^2}}$. *(Equation 2; condition 2; envelope 1.)* **(d)** External flux $\Phi=B(t)\pi\rho^2=B_0\pi\rho^2\sin\Omega t$; induced EMF $\mathcal{E}_{\text{ind}}=-d\Phi/dt=-B_0\pi\rho^2\Omega\cos\Omega t$, added to the loop KVL: $L\ddot q+R\dot q+q/C=\mathcal E_{\text{ind}}$. At $t=0^+$, $\cos\Omega t\approx1$ so $\mathcal E_{\text{ind}}<0$; the external flux is increasing (positive $\dot\Phi$), so by Lenz's law the induced current opposes the increase — it drives current opposite to the direction that would augment $\Phi$. Relative to the capacitor-driven current, it therefore **opposes** the buildup that increases the flux. *(EMF term 2; sign 1; Lenz reasoning 2.)* ```verify [ {"claim":"Q1(b) capacitance formula matches Q/V from integrated field","code":"Q,e0,k0,a,b,r=symbols('Q e0 k0 a b r',positive=True); E=Q/(4*pi*e0*k0*a*r); V=integrate(E,(r,a,b)); C=simplify(Q/V); result= simplify