Past exam of the physics 2 paper of the University of Oxford / 2016 / 1

Take the electrostatic potential energy to vanish for infinitely separated point charges, and write $k_e=1/(4\pi {ϵ}_0)$. The first charge costs no work to place. Bringing the second to distance $a$ costs $k_e{q}^2/a$; the third is distance $a$ from both existing charges and costs $2k_e{q}^2/a$. Thus the external quasistatic assembly work isEquivalently, sum over the three distinct pairs in the equilateral triangle. Counting both ordered pairs without a compensating factor of one half would double the answer. The positive result applies to either sign of $q$, since all three electric charges have the same sign.

By symmetry, the resultant Coulomb force on each charge points along the line from the centroid to its vertex. Their accelerations and subsequent speeds are equal, so the centroid remains fixed and the equilateral triangle expands without rotating. Each charge follows a straight radial trajectory; the separation grows and the repulsive force decreases.

For example, if the instantaneous side is $s$ and the centroid-to-charge distance is $r=s/\sqrt{3}$, the two pair forces have a $60^{\circ }$ angle between them. Their resultant is $\sqrt{3}{k}_e{q}^2/s^2=k_e{q}^2/(\sqrt{3}{r}^2)$, along the outward radius. This confirms that the symmetric shape persists.

Assuming release from rest, nonrelativistic motion and negligible radiation losses, energy conservation givesThe three velocity vectors sum to zero, consistent with momentum conservation. Their magnitudes approach the displayed value as the remaining electrostatic potential energy tends to zero.