Electromagnetism

Q1 — Charged disk

(a) Ring of radius $s$, width $ds$: charge $dq = \sigma\,2\pi s\,ds$. On-axis field of a ring (axial component only, by symmetry): $dE_z = \frac{1}{4\pi\varepsilon_0}\frac{dq\, z}{(s^2+z^2)^{3/2}}.$ (2 for ring set-up) Integrate $s:0\to R$: $E_z = \frac{\sigma z}{4\varepsilon_0}\int_0^R \frac{2s\,ds}{(s^2+z^2)^{3/2}} = \frac{\sigma z}{4\varepsilon_0}\left[-\frac{2}{\sqrt{s^2+z^2}}\right]_0^R.$ (2 for integral) $\boxed{E_z = \frac{\sigma}{2\varepsilon_0}\left(1 - \frac{z}{\sqrt{z^2+R^2}}\right)}.$ (2 result)

(b) As $R\to\infty$, $\dfrac{z}{\sqrt{z^2+R^2}}\to0$, so $E_z\to \dfrac{\sigma}{2\varepsilon_0}$ — the infinite plane field. (2)

(c) Close to a large sheet ($z\ll R$) the second term is negligible, giving a constant $\sigma/2\varepsilon_0$ independent of $z$: contributions from the far edges nearly cancel in the transverse direction and the geometry looks effectively infinite, so the field is uniform. (2)

Q2 — Uniform sphere

(a) Charge density $\rho = \dfrac{Q}{\frac43\pi R^3}$. Inside ($r<R$): enclosed $Q_{enc}=Q\,r^3/R^3$; Gauss $E\,4\pi r^2 = Q_{enc}/\varepsilon_0$: $E_{in} = \frac{Q\,r}{4\pi\varepsilon_0 R^3}.$ (3) Outside ($r>R$): $E\,4\pi r^2 = Q/\varepsilon_0$: $E_{out}=\frac{Q}{4\pi\varepsilon_0 r^2}.$ (2)

(b) $V(r) = -\int_\infty^r E\,dr$. Outside integral from $\infty$ to $R$ gives $V(R)=\dfrac{Q}{4\pi\varepsilon_0 R}$. (2) For $r<R$: $V(r)=V(R)-\int_R^r E_{in}\,dr' = \frac{Q}{4\pi\varepsilon_0 R} - \frac{Q}{4\pi\varepsilon_0 R^3}\cdot\frac{r^2-R^2}{2}$: $V(r)=\frac{Q}{8\pi\varepsilon_0 R}\left(3-\frac{r^2}{R^2}\right).$ (2) Centre: $V(0)=\dfrac{3Q}{8\pi\varepsilon_0 R}$. (1)

(c) The charge is spread through a volume (no surface sheet), so there is no surface charge density to produce a discontinuity; $E_{in}(R)=E_{out}(R)=\frac{Q}{4\pi\varepsilon_0R^2}$. A conductor concentrates charge on the surface giving a jump $\sigma/\varepsilon_0$. (2)

Q3 — RC discharge

(a) KVL: $V_C + IR = 0$ with $I = -C\,dV/dt$ (discharge). So $RC\,\dfrac{dV}{dt} = -V$: $V(t) = V_0 e^{-t/RC}, \qquad I(t)=\frac{V_0}{R}e^{-t/RC}.$ (5) (2 DE setup, 1 sign, 2 solutions)

(b) $P_R = I^2R = \frac{V_0^2}{R}e^{-2t/RC}$. $W=\int_0^\infty \frac{V_0^2}{R}e^{-2t/RC}dt = \frac{V_0^2}{R}\cdot\frac{RC}{2}=\tfrac12 CV_0^2.$ (3)

(c) (2)

R,C,V0=1e3,1e-6,5.0
tau=R*C; dt=tau/1000; V=V0; t=0
while t<tau:
    V += -V/(R*C)*dt   # dV/dt = -V/RC
    t += dt
print(V)   # ~ V0*e^-1 ≈ 1.84

Q4 — Solenoid

(a) Amperian rectangle: one side length $\ell$ inside (field $B$, parallel), opposite side outside ($B\approx0$), short sides perpendicular (no contribution). $\oint\vec B\cdot d\vec l = B\ell$. Enclosed current $= nI\ell$. Ampère: $B\ell=\mu_0 nI\ell$: $B=\mu_0 n I.$ (4)

(b) Flux per turn $\Phi=BA=\mu_0 nIA$. Linkage $N\Phi = \mu_0 n N I A$ with $n=N/\ell$: $L=\frac{N\Phi}{I}=\frac{\mu_0 N^2 A}{\ell}.$ (3)

(c) $\mathcal{E} = -L\frac{dI}{dt} = -L I_0\omega\cos\omega t = -\frac{\mu_0 N^2 A}{\ell}I_0\omega\cos\omega t.$ Sign: while current rises, induced EMF opposes the increase (opposes change in flux) per Lenz's law; the minus sign encodes this opposition. (3)

Q5 — Displacement current

(a) $\oint \vec B\cdot d\vec l = \mu_0\left(I_{enc}+\varepsilon_0\frac{d\Phi_E}{dt}\right).$ For a charging capacitor, no conduction current flows between the plates; a surface bounded by the same loop passing between plates carries $I=0$. Without the displacement term Ampère's law would be inconsistent (ambiguous depending on chosen surface). The changing $\vec E$ flux supplies the missing term. (4)

(b) Between plates $E=\sigma/\varepsilon_0=Q/(\varepsilon_0 A)$, so $\Phi_E=EA=Q/\varepsilon_0$. $I_d=\varepsilon_0\frac{d\Phi_E}{dt}=\varepsilon_0\cdot\frac{1}{\varepsilon_0}\frac{dQ}{dt}=\frac{dQ}{dt}=I.$ (3)

(c) $c=1/\sqrt{\varepsilon_0\mu_0}=1/\sqrt{8.85\times10^{-12}\cdot4\pi\times10^{-7}}\approx3.0\times10^8$ m/s. Matching the measured speed of light showed light is an EM wave. (3)

Q6 — Poynting vector

(a) $E = V/L$ (along axis, driving current). At surface, Ampère: $B=\mu_0 I/(2\pi a)$ (azimuthal). (4)

(b) $S = EB/\mu_0 = \frac{V}{L}\cdot\frac{\mu_0 I}{2\pi a\mu_0}=\frac{VI}{2\pi a L}$, directed radially inward ($\vec E\times\vec B$ points into the resistor). Surface area $=2\pi a L$: $\oint \vec S\cdot d\vec A = \frac{VI}{2\pi aL}\cdot 2\pi aL = IV.$ Energy flows in and equals dissipation $IV$. (4)

[
{"claim":"Disk field reduces to sigma/2eps0 as R->inf","code":"z,R,s,eps=symbols('z R s eps',positive=True); E=integrate(sigma:=1; s/(s**2+z**2)**Rational(3,2),(s,0,R)); Edisk=z*E; result=simplify(limit(Edisk,R,oo)-1)==0"},
{"claim":"Potential at centre of uniform sphere is 3Q/(8 pi eps0 R)","code":"Q,R,r,eps0=symbols('Q R r eps0',positive=True); Vc=Q/(8*pi*eps0*R)*(3-r**2/R**2); result=simplify(Vc.subs(r,0)-3*Q/(8*pi*eps0*R))==0"},
{"claim":"Energy dissipated in RC discharge equals half C V0^2","code":"t,Rr,C,V0=symbols('t R C V0',positive=True); W=integrate(V0**2/Rr*exp(-2*t/(Rr*C)),(t,0,oo)); result=simplify(W-Rational(1,2)*C*V0**2)==0"},
{"claim":"Poynting inflow equals IV","code":"V,L,I,a,mu0=symbols('V L I a mu0',positive=True); S=(V/L)*(mu0*I/(2*pi*a))/mu0; total=S*2*pi*a*L; result=simplify(total-I*V)==0"}
]