"Pumping" Charges — Question 1

Question 1

A source raises positive charge in potential without raising its kinetic energy

Inside an ideal source, a positive charge $q=4.00$ mC moves quasistatically from terminal A at 2.00 V to terminal B at 14.0 V. Neglect internal dissipation.

1. Find the electric force work and the non-electrostatic work.

2. Find the energy transferred to an external load if the same charge returns from B to A, and explain why the source is necessary.

Question 1 - Solution

A source raises positive charge in potential without raising its kinetic energy

(a) Method and calculation

$$\Delta U=q(V_B-V_A)=(.00400)(12.0)=.0480\,\mathrm J.$$$$W_E=-\Delta U=-.0480\,\mathrm J,\quad W_{\mathrm{pump}}=+.0480\,\mathrm J.$$

(b) Interpretation and further calculation

$$W_{E,\mathrm{return}}=q(V_B-V_A)=.0480\,\mathrm J.$$$$W_{E,\mathrm{closed}}=W_E+W_{E,\mathrm{return}}=0.$$The source supplies the energy dissipated by the load, while the electrostatic field has zero net loop work.

Physical check

The name charge pump describes energy conversion, not creation of net electric charge.