Electric Potential — Question 5

Question 5

A responding source makes the finite-probe ratio misleading

A source responds to a probe so that its total insertion energy is $U(q)=qV_0+\alpha q^2/2$. Experiments give $U/q=20.0$ V for $q=1.00$ nC and 22.0 V for $q=2.00$ nC.

1. Find $V_0$, $\alpha$, and the zero-probe limit of U/q.

2. Find the incremental energy per added charge $dU/dq$ at 2.00 nC and compare it with U/q there.

Question 5 - Solution

A responding source makes the finite-probe ratio misleading

(a) Method and calculation

$$20=V_0+\tfrac12\alpha(1\,\mathrm{nC}),\quad22=V_0+\tfrac12\alpha(2\,\mathrm{nC}).$$$$\alpha=4.00\,\mathrm{V/nC},\quad V_0=18.0\,\mathrm V=\lim_{q\to0}U/q.$$

(b) Interpretation and further calculation

$$\frac{dU}{dq}=V_0+\alpha q=26.0\,\mathrm V\quad(q=2.00\,\mathrm{nC}).$$$$\frac{dU}{dq}-\frac Uq=\frac{\alpha q}{2}=4.00\,\mathrm V.$$

Physical check

The ratio is an average insertion cost; its slope correction matters when the source changes as charge is added.