Past exam of the physics 2 paper of the University of Oxford / 2013 resit / 7

Put $q_1$ at $x=-L$, $q_2$ at $x=+L$, and $q_3$ at $x=fL$, with $-1<f<1$. For a nonzero test charge, the force vanishes when the electric field vanishes:For equal-sign nonzero $q_1,q_2$, take the positive square root $s=\sqrt{q_2/q_1}$. The distances are positive, so $1-f=s(1+f)$ andFor $s>1$ the zero moves toward the weaker charge $q_1$; equal charges give $f=0$.

With $q_2=-q_1$, both field contributions between the charges point the same way. Their signed force on $q_3$ isSince $f$ is dimensionless, the small-displacement expansion is $|f|\ll 1$, givingThe force at the midpoint is nonzero and the first variation is quadratic, not linear. In particular, the midpoint is not an equilibrium supporting a linear restoring-force approximation. The printed comparison of $f$ with $L$ is dimensionally inconsistent; the expansion above keeps their distinct roles.

Here $q_1=3.8q_3$ and $q_2=-1.9q_3$. The part of the electrostatic potential energy that changes isAt the two endpoints, $U(-0.9)=37k_e{q}_3^2/L$ and $U(0.9)=-17k_e{q}_3^2/L$. The fixed-source interaction energy is unchanged, and conservation of energy givesThe force points toward increasing $f$ throughout this interval: the left charge repels $q_3$ and the right charge attracts it, so release from rest reaches the stated endpoint without a turning point.