"Pumping" Charges — Question 4

Question 4

An open external path allows only finite terminal charging

A charge pump transfers positive charge from A to B. The isolated terminals have capacitance C, initially zero voltage, and acquire $q_B=+Q$, $q_A=-Q$. The pump stops when $V_B-V_A=V_s$. Ignore leakage.

1. Find the final separated charge and stored electric energy.

2. For C=20.0 pF and Vs=15.0 V, calculate both quantities and explain why a steady external current cannot persist.

Question 4 - Solution

An open external path allows only finite terminal charging

(a) Method and calculation

$$V_B-V_A=Q/C\Rightarrow Q_f=CV_s.$$$$U_f=\int_0^{Q_f}\frac Q C\,dQ=\frac{Q_f^2}{2C}=\tfrac12CV_s^2.$$

(b) Interpretation and further calculation

$$Q_f=(20.0\times10^{-12})(15.0)=300\,\mathrm{pC}.$$$$U_f=\tfrac12(20.0\times10^{-12})(15.0)^2=2.25\,\mathrm{nJ}.$$After the finite charging transient, an open path carries no steady current.

Physical check

An open-circuit source can maintain voltage without delivering continuous load power.