Problem — Two equal charges and a test charge

Problem

Two charges $+q$ are fixed at a distance $d$ apart. A test charge $+q'$ is placed at the midpoint. Discuss the equilibrium position and its stability, distinguishing displacements along the line joining the charges from displacements perpendicular to it.

Solution

The midpoint is an equilibrium point. At the midpoint the two charges $+q$ are at the same distance $d/2$ from $q'$ and repel it with forces of equal magnitude and opposite direction along the line joining them. The resultant is zero:

$\vv{F}_{\text{net}} = \vv{F}_1 + \vv{F}_2 = \vv 0$

Displacement along the joining line (axis $x$). If $q'$ moves towards one of the two charges, it gets closer to that one and farther from the other. The nearer charge repels it with greater force ($1/r^2$), the farther one with less: a net force arises that pushes it back towards the centre. The equilibrium is stable in this direction.

Displacement perpendicular to the joining line (axis $y$). If $q'$ moves perpendicularly, both charges $+q$ are now on the same side relative to the new position and repel it, pushing it further away from the axis. There is no restoring force: the equilibrium is unstable in the transverse direction.

Conclusion: saddle point. The equilibrium is stable along one direction and unstable along the other. It is a typical saddle point, consistent with Earnshaw’s theorem: with electrostatic forces alone there is no point of equilibrium that is stable in every direction.

$\ev{\text{saddle point: stable axially, unstable transversally}}$

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Topics: Electrostatics Concepts: Coulomb’s law Skills: Superposition principle