From the graph to the sign of the charges

Problem

From the graph to the sign of the charges. Two point charges are fixed on the $x$ axis at points $A=(0,0)$ and $B=(d,0)$. The graph shows the qualitative behaviour of the electric potential $V(x)$ along the line joining them, with $V\to 0$ at infinity and a single point where $V=0$ between the charges.

(a) Which axis of the graph corresponds to the line on which the charges lie? (b) What sign do $q_1$ (at $A$) and $q_2$ (at $B$) have? (c) Compare $|q_1|$ and $|q_2|$, justifying it from the heights of the peaks.

Solution

(a) Axis of the charges. The line joining the charges, on which they lie, is the line along which the position varies: it is the horizontal axis ($x$). The vertical axis shows the value of the potential $V$, not a position. $\ev{\text{the line of the charges is the } x \text{ axis}}$

(b) Signs of the charges. Near each charge the potential diverges: $V\to +\infty$ near a positive charge, $V\to -\infty$ near a negative charge. The graph shows a positive peak at $A$ and a negative peak at $B$: $\ev{q_1 > 0 \ (\text{at } A), \qquad q_2 < 0 \ (\text{at } B)}$

(c) Comparing the magnitudes. The total potential is the sum $V(x) = k q_1/|x| + k q_2/|x-d|$. The fact that the point where $V=0$ falls between the charges but shifted towards $A$ (the negative peak at $B$ is more pronounced and already “dominates” in the intermediate region) indicates that the negative charge is larger in magnitude. More formally, where $V=0$ we have $|q_1|/r_1 = |q_2|/r_2$; since that point is closer to $A$ ($r_1 < r_2$), it must be $|q_1| < |q_2|$. $\ev{|q_1| < |q_2|}$

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Topics: Electric field and potential Concepts: Electric potential · Electric charge Skills: Superposition principle