Problem — Equilibrium of three charges on an axis

Problem

Two point charges $q_1 = +Q$ at $x = 0$ and $q_2 = +4Q$ at $x = L$ are fixed on the $x$ axis. Add a third charge $q_3$ so that all three are simultaneously in equilibrium (zero total force on each). Determine the position, sign and value of $q_3$ in terms of $Q$ and $L$.

Solution

Position of $q_3$. For $q_3$ to be in equilibrium, it must lie between the two positive charges (only there can the two repulsions/attractions be opposite). Let $x$ be its distance from $q_1$; the distance from $q_2$ is $L-x$. Impose equal magnitude for the two forces:

$k\,\frac{Q\,|q_3|}{x^2} = k\,\frac{4Q\,|q_3|}{(L-x)^2} \;\Longrightarrow\; (L-x)^2 = 4x^2 \;\Longrightarrow\; L-x = 2x \;\Longrightarrow\; x = \tfrac{1}{3}L$

Sign and value of $q_3$. Now impose equilibrium of $q_1$: it is acted on by $q_2$ (repulsive) and $q_3$ (at distance $x=L/3$). For these to cancel, $q_3$ must attract $q_1$, so $q_3<0$. In magnitude:

$k\,\frac{Q\cdot 4Q}{L^2} = k\,\frac{Q\,|q_3|}{(L/3)^2} \;\Longrightarrow\; \frac{4Q^2}{L^2} = \frac{9\,Q\,|q_3|}{L^2} \;\Longrightarrow\; |q_3| = \tfrac{4}{9}Q$

One can check that with these values $q_2$ is also in equilibrium (the symmetry of the problem guarantees it). So:

$\ev{q_3=-\tfrac{4}{9}Q,\ x=\tfrac{1}{3}L}$

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Topics: Elettrostatica Concepts: Legge di Coulomb Skills: Principio di sovrapposizione · Impostazione simbolica