Past exam of the physics B course of the University of Cambridge / 2012 / Paper 2 / Section A

For the one-mole convention, $b$ is the excluded molar volume: finite molecular size reduces the volume available for translational motion from $V$ to $V-b$. It is an effective excluded volume, not simply the sum of the bare molecular volumes. $a/V^2$ corrects for attractive intermolecular forces, which reduce the pressure exerted on the wall. The attraction per molecule scales with number density, and the number of affected molecules also scales with density, producing the density-squared correction. The Van der Waals equation consequently raises the measured pressure to the corresponding repulsive-gas value before multiplying by the available volume. Both corrections vanish relative to the ideal-gas terms at sufficiently low density.

For inner radius $a$ and outer conductor's inner radius $b$, Gauss's law gives $E(s)=\lambda /(2\pi {ϵ}_0{ϵ}_{r}s)$. Integrating across the dielectric gives voltage $V=\lambda \ln (b/a)/(2\pi {ϵ}_0{ϵ}_r)$. Thus the capacitance per unit length isThe stated outer radius is interpreted as the radius of the dielectric-facing surface. Magnetic permeability affects inductance, but does not enter this electrostatic capacitance.

Kepler's third law for negligible planet mass gives $a_P/a_E=(P_P/P_E{)}^{2/3}$. Therefore $a_P=1.5\times 10^8(248{)}^{2/3}\simeq 5.92\times 10^9\mathrm{km}$. The furthest point is apoapsis, soThe supplied orbital period and eccentricity are sufficient; no present-day ephemeris is needed.

Neglect ends and take the exterior as vacuum. With polar angle measured from the applied field, regularity inside and the far-field condition allowContinuity of potential gives $B=1-A$; continuity of normal electric displacement, with no free surface charge, gives ${ϵ}_{r}B=1+A$. Hence the transverse electrostatic field inside a dielectric cylinder is uniform, parallel to the applied field, with magnitudeThis cylindrical result differs from the spherical depolarization factor. It returns $E_0$ for ${ϵ}_r=1$ and tends to zero for large positive ${ϵ}_r$.

Let $N$ be circumferential tension. A small arc subtending $d\theta$ has inward resultant $Nd\theta$ and mass $\rho Ard\theta$. Its centripetal acceleration gives $N=\rho A{\omega }^2{r}^2$. Circumferential stress is $N/A$ and the strain is $\Delta (2\pi r)/(2\pi r)=\Delta r/r$. Thus the centrifugal strain of a thin elastic hoop isUse the unstrained radius and cross-section at this order. The approximation requires $\rho {\omega }^2{r}^2/E\ll 1$, a slender hoop and linear elastic response; the area cancels.