LC Oscillations, Qualitatively — Question 7

Question 7

A vanishing current marks maximum capacitor energy rather than no dynamics

At an instant in an ideal LC loop, $i=0$ and $q=-Q$. Use $L\dot i=-q/C$.

1. Find the sign of the current derivative and the next direction of current.

2. Find the leading small-time changes in current and charge, measuring time from that instant.

Question 7 - Solution

A vanishing current marks maximum capacitor energy rather than no dynamics

(a) Method and calculation

$$\dot i(0)=Q/(LC)>0\Rightarrow\boxed{i(dt)>0}.$$$$i(dt)=\frac{Q}{LC}dt+O(dt^3).$$

(b) Interpretation and further calculation

$$q(dt)=q(0)+\int_0^{dt}i(t)dt.$$$$\boxed{q(dt)=-Q+\frac{Q}{2LC}dt^2+O(dt^4)}.$$

Physical check

The restoring capacitor voltage is largest when current is momentarily zero.