Energy Stored in an Electric Field — Question 8

Question 8

Minimum energy at fixed total node charge requires equal capacitor voltages

Two capacitors C1,C2 share a reference node. Their upper-node charges q1,q2 may redistribute while total upper charge Q=q1+q2 remains fixed. There are no other energy-storage terms.

1. Minimize the total stored energy with respect to q1.

2. Find the minimum energy and prove that any deviation from the minimizing allocation raises it.

Question 8 - Solution

Minimum energy at fixed total node charge requires equal capacitor voltages

(a) Method and calculation

$$U(q_1)=\frac{q_1^2}{2C_1}+\frac{(Q-q_1)^2}{2C_2},\quad U\prime=\frac{q_1}{C_1}-\frac{Q-q_1}{C_2}.$$$$q_{1,*}=\frac{C_1Q}{C_1+C_2},\quad q_{2,*}=\frac{C_2Q}{C_1+C_2},\quad V_1=V_2.$$

(b) Interpretation and further calculation

$$U_{\min}=\frac{Q^2}{2(C_1+C_2)}.$$$$q_1=q_{1,*}+\delta:\quad U-U_{\min}=\frac12\left(\frac1{C_1}+\frac1{C_2}\right)\delta^2>0\quad(\delta\ne0).$$

Physical check

The positive second derivative makes equal-voltage sharing a strict energy minimum for positive capacitances.