Charge Is Conserved — Question 1

Question 1

Final charges do not uniquely determine every internal transfer

Three isolated components begin with charges $(Q_A,Q_B,Q_C)=(6,-2,1)$ nC and end with $(1,2,2)$ nC. Let $x,y,z$ be positive conventional charge transfers in nC along A to B, B to C, and C to A, respectively. No charge crosses the boundary of the three-component system.

1. Write the three balance equations and find the full family of nonnegative transfers consistent with the endpoint data.

2. If a separate sensor measures $z=2$ nC, determine $x$ and $y$. Explain what internal process remains invisible to endpoint charge measurements alone.

Question 1 - Solution

Final charges do not uniquely determine every internal transfer

(a) Method and calculation

$$6-x+z=1,\quad-2+x-y=2,\quad1+y-z=2.$$$$x-z=5,\quad x-y=4,\quad y-z=1.$$$$(x,y,z)=(5+t,1+t,t)\,\mathrm{nC},\quad t\ge0.$$

(b) Interpretation and further calculation

$$z=2\Rightarrow t=2,\quad(x,y,z)=(7,3,2)\,\mathrm{nC}.$$$$\Delta(Q_A+Q_B+Q_C)=0.$$An extra equal transfer around the closed loop leaves every endpoint charge unchanged. One balance equation is redundant because the total is conserved.

Physical check

The nonnegative-transfer condition restricts the free parameter but does not remove it.