The situation I'm referring can be seen in the following link (ignore the numerical value, let's just call it $V$)


I want to find the charge, $Q$, on plate B. Im pretty sure I can just view it as two parallel capacitors and this results in $Q=(C_1+C_2)V$, where $C_i$ can be calculated with the usual formula for parallel capacitors (just suppose that we know the area and the two distances $d_i$ between $AB$ and $BC$). The problem arises when I try to intuitively picture this as charges moving from one plate to another, by means of the generator. The generator will move a charge equal to $Q_1=C_1V$ from plate $A$ to plate $B$ and a charge $Q_2=C_2V$ from plate $C$ to plate $A$. This will result in plate B having a negative charge equal to $-Q_2-Q_1$ and plates $A$ and $C$ having, respectively, charges $Q_1$ and $Q_2$. This just does not feel right to me, because capacitors usually have net charge $=0$, but if I consider capacitor $AB$ the net charge is $-Q_2$ and similarly for capacitor $BC$. Where is the flaw in my reasoning? Thanks a lot


1 Answer 1


Imagine Three boxes, enclosing respectively $A$, $B$ and $C$ (each box enclosing only one plate and having a top and a bottom side of area S).

Applying Gauss’s theorem on the Box $Ba$ surrounding $A$

$Eab.S = Qa$

Ditto for $B$ and $C$:

$ (Eab + Ebc).S = Qb$


$Ebc.S = Qc$

Since $Eab = Ebc = V$,

$Qa = Qb = (1/2) Qbc$


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