USAPhO 1999, Полуфинал — Задача 2. A2

• $Q_2 = -\frac{c}{b}q \text{ at } x = d = \frac{c^2}{b}$

Assume a charge of $Q_2$ at $x = d$ along the x-axis. The total potential a distance c from the origin is

where k is Coulomb's constant and $R_1$ and $R_2$ can be found using the law of cosines.

and

Since V = 0,

Therefore

This must be true for all points a distance c from the origin. In particular at θ = 0 or x = +c we have

At θ = 180° = π or x = -c we have

Solving equations (2) and (3) simultaneously for $Q_2$ and $d$, we have

Substituting $d$ into $R_2$, we have

for any angle $\theta$, i.e. any point a distance c from the origin. Substituting $Q_2$ and $R_2$ into (1),