Past exam of the physics 2 paper of the University of Oxford / 2016 / 9

Spherical symmetry and the vacuum Gauss's law imply $\mathbf{E}=A\hat{\mathbf{r}}/r^2$ in the gap, where $A$ is constant. Integrating the electric field between the two equipotentials givesConsequentlyThe inner charge is $Q=4\pi {ϵ}_{0}A$, which reproduces the spherical capacitor capacitance $4\pi {ϵ}_{0}ab/(b-a)$. The radial direction is outward for positive $V_1$ and inward for negative $V_1$; the formula represents the signed field, not just its magnitude.

The electric field magnitude falls as $1/r^2$ throughout the vacuum gap, so its maximum occurs just outside the inner sphere:ThusInside the metal the electric field is zero. The one-sided gap value is the relevant value for dielectric breakdown; the ideal zero-thickness conducting boundary itself should not be assigned a single field value across its surface-charge discontinuity.

Inserting an independently biased conducting sphere makes two vacuum gaps, each with its own constant radial electric flux. Let $A_1$ and $A_2$ be the coefficients of $1/r^2$. The two potential drops areThereforeThe middle conductor can carry net electric charge, so $A_1$ and $A_2$ need not be equal. Its charge is $4\pi {ϵ}_0(A_2-A_1)$, consistent with the jump in the normal electric field. For $0<V_2<V_1$ both fields point outwards; outside that interval one of the gap fields reverses direction.

The largest field in each gap lies at that gap's smaller radius. Thus the quantity to minimize isFor positive $V_1$, projecting any $V_2$ outside $[0,V_1]$ back to that interval reduces both absolute potential drops, so the optimum lies inside it. Within the interval, one candidate maximum decreases linearly with $V_2$ and the other increases linearly. Their maximum is minimized at their intersection, not by making the potential drops equal.

With $a=0.1\mathrm{m}$, $c=0.2\mathrm{m}$, $b=0.4\mathrm{m}$ and $V_1=1000\mathrm{V}$, the two maxima are $20(1000-V_2)$ and $10V_2$ in $\mathrm{V}{\mathrm{m}}^{-1}$. Equating them givesThe maximum is halved relative to the two-sphere system and is attained at both $a^{+}$ and $c^{+}$. At $c^{-}$ and $b^{-}$ the optimized field is only about $1667\mathrm{V}{\mathrm{m}}^{-1}$; at $c$ the charged middle shell produces a jump.

For plotting, the two-sphere electric potential is $V^{(2)}(r)=V_1(1/r-1/b)/(1/a-1/b)$. In the three-sphere system it isThe electric potential remains continuous at $c$, while its radial derivative changes there. The graphs show both the continuous potentials and the one-sided gap fields; the marked endpoints identify all maxima.