Capacitors in Parallel and in Series — Question 6

Question 6

Trapped junction charge breaks the usual equality of series charges

A capacitor C joins A to M and a capacitor 2C joins M to B. Set VA=V>0 and VB=0. The isolated junction M has prescribed net charge $Q_M=CV$.

1. Find VM from junction charge conservation.

2. Find the signed charges on the A plate and on the M plate facing B, and explain their unequal magnitudes.

Question 6 - Solution

Trapped junction charge breaks the usual equality of series charges

(a) Method and calculation

$$Q_M=C(V_M-V_A)+2C(V_M-V_B).$$$$CV=3CV_M-CV\Rightarrow V_M=2V/3.$$

(b) Interpretation and further calculation

$$q_A=C(V-V_M)=CV/3.$$$$q_{M\to B}=2CV_M=4CV/3,\quad(-CV/3)+(4CV/3)=Q_M.$$

Physical check

An isolated junction conserves its net charge; zero net charge is an additional assumption, not a universal rule.