The figure shows a diagonal symmetric arrangement of capacitors and a battery. If the potential of $C$ is zero, then
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Now if $2 \mu F$ is connected between $B$ and $D$, charge will flow from $B$ to $D$ and finally $V_{B}>V_{D}$
$V_{A}=20 V$
$q_{2}=q_{1}+q_{3}$
or $\left(V_{A}-V_{B}\right) 4=\left(V_{B}-V_{D}\right) 2+\left(V_{B}-V_{C}\right) 2$
or $4\left(V_{A}-V_{B}\right)+2\left(V_{D}-V_{B}\right)=2 V_{B}$
$q_{2}=q_{1}+q_{3}$
or $4\left(V_{D}-V_{C}\right)+2\left(V_{A}-V_{D}\right)+2\left(V_{B}-V_{D}\right)$
or $2\left(V_{A}-V_{D}\right)+2\left(V_{B}-V_{D}\right)=4 V_{D}$
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