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

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.