Interpenetrating charged balls

The idea here is that once the spheres interpenetrate, chunks of positive charge in one sphere overlap with chunks of negative charge in the other sphere, diminishing the Coulomb attraction that pulls on the un-penetrated portions of the opposing sphere toward the center.

To begin with, outside either sphere, there is a potential $\displaystyle \phi(r) = \frac{Q}{r}$ , ignoring factors of $k$ .

The total charge inside radius $r$ is $Q(r)$ or $\displaystyle \frac43 \pi r^3\rho = r^3/R^3$ .

Inside one of the spheres at a radius $r$ , we have a field of magnitude $E(r) = \frac{Q_{tot}}{S_2(r)} = \frac{Q_{tot}r}{R^3}$ where $S_2(r)$ is the surface area of a sphere of radius $r$ .

We can find the potential at a distance $r$ from the center by considering the work required for a test charge to be pulled in to $r$ from a point infinitely far away from the sphere, i.e.

$\begin{aligned} \phi_1(r) &= \int_{\infty}{r} E(r) dr \\ &= -\int\limits_{\infty}^{R} \frac{Q}{r^2} dr - \int\limits_R^r \frac{Qr}{R^3} dr \\ &=\frac{Q}{2R}\left(3-\frac{r^2}{R^2}\right) \end{aligned}$

Using this potential, we can find the total energy of the spheres' interaction. We just integrate the potential due to one sphere over the charge distribution of the other sphere. I.e.

$U = \int\limits_{\Sigma_2} \phi_1(r) \rho d^3r$

The integral is a bit tricky. As we've written the potential in terms of the distance from the center of each sphere, we can build the interaction out of spherical shells.

Consider the diagram below, the surface area of a small patch of curved surface is given by $\displaystyle SA = a^2 \sin \theta d\phi d\theta$ . For the spherical caps we're adding, we can sum them over a rotation in $\phi$ and out to the intersection of the spherical patch with the second sphere. I.e. $\displaystyle SA = a^2 2\pi \left(1-\cos\theta\right)$

We use the law of cosines to find $\displaystyle\cos\theta = \frac{R^2 - r^2 - a^2}{2ar}$

The volume element is then given by:

$\begin{aligned} dV &= SA(a) da \\ &= 2\pi a^2 \left(1 - \cos\theta\right) da\\ &= 2\pi a^2 \frac{R^2 +2ar - r^2 - a^2}{2ar} da \\ &= \pi \frac{a}{r} \left[R^2 -\left(r-a\right)^2\right] da \end{aligned}$

With all this in hand, we can find the interaction energy

$\begin{aligned} U &= \rho\int\limits_{\Sigma_2} \phi_1(a)\, dV \\ &= \rho\int\limits\limits_{a=r-R}^{a=R+r} \phi_1(a)\, dV \\ &= \rho\left(\int\limits\limits_{a=r-R}^{a=R} \phi_1(a)\, dV + \int\limits\limits_{a=R}^{a=R+r} \phi_1(a)\, dV\right) \\ &= \rho k\left(\int\limits\limits_{r-R}^{R} \frac{Q}{2R}\left(3-\frac{a^2}{R^2}\right)\pi \frac{a}{r} \left[R^2 -\left(r-a\right)^2\right]da + \int\limits\limits_{R}^{R+r} \frac{Q}{a}\pi \frac{a}{r} \left[R^2 -\left(r-a\right)^2\right]da\right) \\ &= \frac{kQ^2}{160R^6}\left(192R^5-80R^3r^2+30R^2r^3-r^5\right) \end{aligned}$

As energy is conserved, we have $\frac{d}{ds}\left[\frac12 M\dot{s}^2 + \frac12 M\dot{s}^2 + U(r)\right] = 0$ which implies $\displaystyle\ddot{s} = -\frac{1}{M}\frac{d}{dr}U(r)$ (with $r =2s$ ).

As we're not given $Q$ we must solve for it. It is actually more convenient to solve for $\displaystyle k\frac{Q^2}{MR}$ . When the spheres are very far apart, the only energy in the system is the self-interaction energy of each sphere. This can be found by pretending that each sphere is built in spherical shells , the result is $\displaystyle U_{self} = \frac35 k\frac{Q^2}{R}$ .

When the spheres are perfectly overlapped, there is no interaction energy (the effective charge density is 0 at every point) and all the energy is kinetic, i.e. $\frac12 M_{tot} v_{max}^2$ ). Thus we have $\displaystyle v_{max} = \sqrt{\frac{6}{5}\frac{kQ^2}{MR}}$ .

We're given $v_{max} = 10 \mbox{ m/s}$ , so $\displaystyle \lvert\frac{kQ^2}{MR}\rvert = \frac{250}{3}$ .

We have $\displaystyle -\frac{1}{M}\frac{d}{dr}U(r) = -\frac{1}{160R^6} k\frac{Q^2}{M} (-160 rR^3 + 90 R^2r^2 - 5 r^4)$

We can plot $\displaystyle -\frac{1}{M}\frac{d}{dr}U(r) = \ddot{s}$ as a function of $r$ , the separation distance between the two spheres.

The acceleration has a clear maximum when the spheres are overlapped halfway, i.e. $r = R$ .

At this point, we have

$\begin{aligned} \ddot{s} &= -\frac{1}{M}\frac{d}{dr}U(r) \\ &= -\frac{1}{160R^5} k\frac{Q^2}{MR} (-160 R^4 + 90 R^4 - 5 R^4) \\ &= k\frac{Q^2}{MR} \frac{15}{32} \\ &= \frac{625}{16}\si[per-mode=symbol]{\meter\per\second\squared}\\ & \approx 39.1\mbox{ m/s}^2 \end{aligned}$