Energy Stored in an Electric Field — Question 4

Question 4

Parallel charge sharing loses an amount set by the initial voltage difference

Two linear capacitors C1,C2 initially have signed voltages V1,V2. Their corresponding terminals are joined without a source. Assume the transient settles through arbitrarily small dissipation or radiation.

1. Find the common final voltage and stored energy.

2. Derive the energy missing from the final electrostatic field and identify when it vanishes.

Question 4 - Solution

Parallel charge sharing loses an amount set by the initial voltage difference

(a) Method and calculation

$$V_f=\frac{C_1V_1+C_2V_2}{C_1+C_2}.$$$$U_f=\frac12(C_1+C_2)V_f^2=\frac{(C_1V_1+C_2V_2)^2}{2(C_1+C_2)}.$$

(b) Interpretation and further calculation

$$U_i-U_f=\frac12(C_1V_1^2+C_2V_2^2)-\frac{(C_1V_1+C_2V_2)^2}{2(C_1+C_2)}.$$$$U_{\mathrm{lost}}=\frac{C_1C_2}{2(C_1+C_2)}(V_1-V_2)^2\ge0,$$with equality only for equal initial voltages.

Physical check

Charge is conserved, but final electrostatic energy need not be; an exactly lossless dynamical circuit would not simply settle.