For the network of capacirors as shown in figure.
.
a. Find the potential of junction `B`.
b. Find the potential of function D,
c. Find the charge on `2 muF` capacitor.

Explanation

`V_(C)=0`
Loop `ADBMA: q_(2)/6+(q_(1)-q_(3))/3=10` (i)
Loop `ACBMA: (q_(2))/2+(q_(2)+q_(3))/1=10` (ii)
Loop `CDAC: -q_(2)/2+q_(3)/1+q_(1)/6=0` (iii)

Solving, we get `q_(1) = 20 mu C rArr q_(2) = 20//3 mu C, q_(3) = 0`
a. `V_(C)-V_(B)=(q_(2)+q_(3))/1` or `0-V_(B)=(20)/3+0`
`V_(B)=-(20)/3V`
b. `V_(C)-V_(D)=(-q_(3))/1` or `V_(D)=0`
c. `q_(2)=(20)/3 muC`.