Ranking potentials

Problem

Ranking. Four point charges $+Q$ are placed at different distances $r$ from a point $P$:

• (A) $Q = 2\;\text{nC}$, $r = 10\;\text{cm}$.
• (B) $Q = 8\;\text{nC}$, $r = 10\;\text{cm}$.
• (C) $Q = 1\;\text{nC}$, $r = 10\;\text{cm}$.
• (D) $Q = 4\;\text{nC}$, $r = 20\;\text{cm}$.

Rank the electrostatic potentials $V = kQ/r$ at point $P$ from smallest to largest.

Solution

Formula. It suffices to compare the ratio $Q/r$ for each case, since $V = kQ/r$ and $k$ is common. It is convenient to measure $Q$ in nC and $r$ in cm (the common units simplify in the comparison): $V \propto \frac{Q}{r}$

Computing the ratios. $\begin{aligned} \text{A:}\ \frac{2}{10}=0{,}20 \qquad & \text{B:}\ \frac{8}{10}=0{,}80 \\ \text{C:}\ \frac{1}{10}=0{,}10 \qquad & \text{D:}\ \frac{4}{20}=0{,}20 \end{aligned}$

Ordering. Comparing the values $0{,}10 < 0{,}20 = 0{,}20 < 0{,}80$, we observe that A and D give the same potential (in D the doubled charge is compensated by the doubled distance): $\ev{\text{C} < \text{A} = \text{D} < \text{B}}$

The potential grows with the charge and falls with distance: doubling both leaves $V$ unchanged, as shown by the equality A = D.

Links

Topics: Electric field and potential Concepts: Electric potential · Electric charge