Past exam of the physics 2 paper of the University of Oxford / 2017 resit / 7

The integral and differential forms of Gauss's law in electrostatics areHere $\partial \mathcal{V}$ is a closed surface bounding a volume $\mathcal{V}$, $d\mathbf{A}=\hat{\mathbf{n}}dA$ is its outward area vector, $\mathbf{E}$ is the electric field, $\rho$ is total volume charge density, and $Q_{\mathrm{enc}}={\int }_{\mathcal{V}}\rho d^{3}r$ is enclosed electric charge. The divergence theorem converts one form to the other. For ideal sheets or point charges, interpret the differential equation with their distributional charge densities.

Write the uniform surface charge density on the annulus as $\sigma =Q/[\pi (b^2-a^2)]$, and let $x$ be signed distance along its axis. A ring of radius $r$ and width $dr$ contains electric charge $dq=2\pi \sigma rdr$. Its transverse electric field components cancel in pairs, and its electric potential is $dV=dq/[4\pi {ϵ}_0\sqrt{x^2+r^2}]$. Setting $V(\infty )=0$ and integrating gives the axial field of a uniformly charged annulus:At $x=0$, each element's electric field cancels that of the diametrically opposite element, so $\mathbf{E}=0$. There is no surface charge at the centre because $a>0$; the axis electric field is continuous there.

For positive $Q$ and $x>0$, differentiate the signed axial electric field:Its stationary condition is $a^{4/3}(x^2+b^2)=b^{4/3}(x^2+a^2)$. Factoring with $A=a^{2/3}$ and $B=b^{2/3}$ givesThe derivative starts positive, has exactly one positive zero and is negative at large $x$, so this is the positive-side maximum. For either sign of $Q$, the electric field magnitude has equal maxima at $x=\pm x_{\max }$. As $b\to a$, the thin-ring limit is $x_{\max }\to a/\sqrt{2}$.

Let $m$ be the sliding point charge's mass. Expanding the axial field of a uniformly charged annulus for $|x|\ll a,b$ givesIts electric charge is $-Q$, so its electric force points toward the centre. Newton's second law becomesTo leading order this is simple harmonic motion, withThe mass is an additional parameter, since the source supplies no numerical value. The constraint is essential: stability along the axis does not imply unconstrained three-dimensional trapping by a static electric field.

For a steady filamentary electric current $I$, the Biot-Savart law isHere $\mathbf{r}$ is the observation position, ${\mathbf{r}}^{\prime }$ the source position, $d{\mathbf{\ell }}^{\prime }$ points in the direction of positive electric current, and ${\mu }_0$ is the vacuum permeability. A circular loop has centre magnetic field ${\mu }_{0}I/(2r)$: the cross product has magnitude $rd{\ell }^{\prime }$ and a constant axial direction, and $\int d{\ell }^{\prime }=2\pi r$.

Introduce the constant signed angular velocity $\Omega$ about the positive axis. A ring's electric charge $dq=2\pi \sigma rdr$ crosses any radial cut once per rotation, so its electric current is $dI=\Omega dq/(2\pi )=\sigma \Omega rdr$. Adding loop contributions yields the magnetic field of a rotating charged annulus:The direction follows the right-hand rule for positive electric charge; reversing either $Q$ or $\Omega$ reverses the magnetic field. The source supplies no rotation rate, so the answer necessarily retains $\Omega$.