Energy Stored in an Electric Field — Question 1

Question 1

Integrating the charging voltage avoids double-counting stored energy

An initially uncharged linear capacitor C=8.00 nF is charged quasistatically to 50.0 V.

1. Derive its stored energy by integrating the incremental work.

2. Find its final charge and compare the stored energy with Q times the final voltage.

Question 1 - Solution

Integrating the charging voltage avoids double-counting stored energy

(a) Method and calculation

$$dU=v(q)\,dq=\frac qC\,dq,\quad U=\int_0^Q\frac qC\,dq=\frac{Q^2}{2C}=\frac12CV^2.$$$$U=\tfrac12(8.00\times10^{-9})(50.0)^2=10.0\,\mu\mathrm J.$$

(b) Interpretation and further calculation

$$Q=CV=400\,\mathrm{nC}.$$$$QV=20.0\,\mu\mathrm J=2U,\quad\overline v=V/2.$$

Physical check

The instantaneous charging voltage rises from zero; using the final voltage for every charge increment overestimates the work.