Electric Potential — Question 1

Question 1

A negative probe crosses a positive potential difference

A fixed source creates $V_B-V_A=80.0$ V. A test charge $q=-2.00$ nC moves from A to B without changing the source.

1. Calculate the change in electric energy and the work by the field.

2. Repeat for a positive probe of the same charge magnitude and find the quasistatic external work for each probe.

Question 1 - Solution

A negative probe crosses a positive potential difference

(a) Method and calculation

$$\Delta U=q\Delta V=(-2.00\times10^{-9})(80.0)=-160\,\mathrm{nJ}.$$$$W_e=-q\Delta V=+160\,\mathrm{nJ}.$$

(b) Interpretation and further calculation

$$q=+2.00\,\mathrm{nC}:\quad\Delta U=+160\,\mathrm{nJ},\quad W_e=-160\,\mathrm{nJ}.$$$$\Delta K=0:\quad W_{\mathrm{ext}}=\Delta U=\begin{cases}-160\,\mathrm{nJ}&q<0,\\+160\,\mathrm{nJ}&q>0.\end{cases}$$

Physical check

The potential difference belongs to the source configuration; reversing the probe changes energy and force signs, not the voltage.