LC Oscillations, Quantitatively — Question 6

Question 6

A rapid capacitance change at a charge turning point exchanges mechanical work

At q=Q and i=0, an ideal variable capacitor is changed rapidly from C to 4C. The charge is conserved during the change; L is unchanged. The actuator may exchange energy with the field.

1. Find the new energy and energy transferred to the actuator.

2. Find the subsequent charge amplitude, current amplitude, and period relative to the original values.

Question 6 - Solution

A rapid capacitance change at a charge turning point exchanges mechanical work

(a) Method and calculation

$$U_i=Q^2/(2C),\quad U_f=Q^2/(8C)=U_i/4.$$$$\boxed{W_{\mathrm{actuator,out}}=U_i-U_f=3U_i/4}.$$

(b) Interpretation and further calculation

$$Q_f=Q,\quad\omega_f=1/\sqrt{4LC}=\omega_i/2.$$$$\boxed{I_{\max,f}=I_{\max,i}/2,\quad T_f=2T_i}.$$

Physical check

Charge is fixed during the idealized change, but electromagnetic energy is not.