Find voltage across capacitor `C_1`.

Explanation

Now from charge conservation at anode x and y for x `(x-4)C_(1)+(x-2)C_(2)+(x-y)C_(3)=0`
`rArr 2(x-4)+2(x-2)+(x-y)2=0`
`6x-2y-12=0 ...........(1)`
For y `(y-x)C_(3)+[y-(-4)]C_(4)+(y-0)C_(5)=0`
`rArr (y-x)2+(y+4)2+y2=0`
`=6y-2x+8=0 .............(2)`
eq. (1) & (2)
y=-3/4 x=7/4
So potential difference `=x-y=7/4-(-3/4)=5/2"volt"`