Energy Stored in an Electric Field — Question 7

Question 7

A nonlinear capacitor stores the integral of voltage against charge

A reversible device obeys $q(v)=C_0v+\beta v^3$ for v nonnegative, with positive C0 and beta. It begins at q=v=0.

1. Derive its stored energy at voltage V.

2. Compare with one half qV and find the ratio when $\beta V^2=C_0$.

Question 7 - Solution

A nonlinear capacitor stores the integral of voltage against charge

(a) Method and calculation

$$dq=(C_0+3\beta v^2)dv,\quad U=\int_0^Vv\frac{dq}{dv}dv.$$$$U=\frac12C_0V^2+\frac34\beta V^4.$$

(b) Interpretation and further calculation

$$\tfrac12qV=\frac12C_0V^2+\frac12\beta V^4,\quad U-\tfrac12qV=\frac14\beta V^4.$$$$\beta V^2=C_0:\quad U/(qV/2)=5/4.$$

Physical check

The familiar half-qV formula requires a linear charge-voltage law; the differential charging integral is the general relation here.