"Pumping" Charges — Question 5

Question 5

A common potential shift does not alter the required pumping work

A two-terminal pump moves positive charge Q from A to B. Initially $V_A=-40$ V and $V_B=-25$ V. The reference is then changed by adding +100 V to both potentials.

1. Find the pump work for Q=6.00 mC before and after the shift.

2. Express the work using a line integral of the electrostatic field and show why the reference shift disappears.

Question 5 - Solution

A common potential shift does not alter the required pumping work

(a) Method and calculation

$$\Delta V=(-25)-(-40)=15.0\,\mathrm V,\quad W=Q\Delta V=.0900\,\mathrm J.$$$$\Delta V\prime=75-60=15.0\,\mathrm V,\quad W\prime=W.$$

(b) Interpretation and further calculation

$$\Delta V=-\int_A^B\vec E\cdot d\vec\ell,\quad W_{\mathrm{pump}}=-Q\int_A^B\vec E\cdot d\vec\ell.$$$$\vec E\prime=-\nabla(V+100\,\mathrm V)=-\nabla V=\vec E.$$

Physical check

Negative terminal potentials relative to an arbitrary reference do not prevent the source from raising charge energy.