Answer
(a) $B\lt A\lt C$
(b) $B\lt A=C$
Work Step by Step
(a) For circuit A
$Q_{upper}+Q_{lower}=2q+q=3q$
For circuit B
$V_{upper}=V_{lower}$
$\implies \frac{Q_{upper}}{3C}=\frac{Q_{lower}}{C}$
We know that
$Q_{upper}+Q_{lower}=-4q+0=-4q$
$\implies \frac{Q_{upper}}{3}+Q_{upper}=-4q$
$\implies Q_{upper}=-3q$
For circuit C
$Q_{lower}+Q_{upper}=6q$
$\implies \frac{Q_{upper}}{2}+Q_{upper}=6q$
$\implies Q_{upper}=4q$
Thus, for A: $B\lt A\lt C$
(b) We know that
For A: $Q_{lower}=2q$
For B: $Q_{lower}=-q$
For C: $Q_{lower}=2q$
Thus, we conclude that $B\lt A=C$
