Problem — Equilibrium on a line

Problem

Two charges $q_A = 4\cdot 10^{-3}\ \text{C}$ and $q_B = -10^{-3}\ \text{C}$ are placed at a distance $d = 3\ \text{m}$ on a line. Where must a charge $q_C$ be placed so that the net force on it is zero? Set up $k\,q_A q_C/x^2 = k\,|q_B| q_C/(d-x)^2$ and discuss which solution is physical.

Solution

The two charges have opposite sign. For a test charge $q_C$ placed between them, the force from $q_A$ and that from $q_B$ have the same direction (one repels, the other attracts, but both push it towards $q_B$): they cannot cancel. Equilibrium exists only outside the segment, on the side of the charge with smaller magnitude ($q_B$), where the two forces can oppose each other.

Measure $x$ from the position of $q_A$. In the region beyond $q_B$ ($x > d$) the distance from $q_B$ is $x-d$. Impose equal magnitude:

$k\,\frac{|q_A|\,q_C}{x^2} = k\,\frac{|q_B|\,q_C}{(x-d)^2} \;\Longrightarrow\; \frac{|q_A|}{x^2} = \frac{|q_B|}{(x-d)^2}$

With $|q_A| = 4\cdot 10^{-3}\ \text{C}$ and $|q_B| = 1\cdot 10^{-3}\ \text{C}$ the ratio is $|q_A|/|q_B| = 4$:

$\left(\frac{x}{x-d}\right)^2 = \frac{|q_A|}{|q_B|} = 4 \;\Longrightarrow\; \frac{x}{x-d} = \pm 2$

Root $+2$ (outer region, $x>d$):

$x = 2(x-d) \;\Longrightarrow\; x = 2x - 2d \;\Longrightarrow\; x = 2d = 6\ \text{m}$

Root $-2$ (inner region):

$x = -2(x-d) \;\Longrightarrow\; 3x = 2d \;\Longrightarrow\; x = \tfrac{2}{3}d = 2\ \text{m}$

The inner root $x = 2\ \text{m}$ must be discarded: there the forces are in the same direction (charges of opposite sign), so they do not actually cancel — it is a spurious algebraic solution of the equation of magnitudes. The physical solution is the outer one, $3\ \text{m}$ beyond $q_B$:

$\ev{x = 6\ \text{m from }q_A\ \ (\text{i.e. }3\ \text{m beyond }q_B,\ \text{on the side of the smaller charge})}$

Links

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