Find voltage across capacitor `C_1`

Explanation

Now from charge conservation at node x and y for x
` (x-4) C_1 +(x-2) C_2 +( x- y) C_1=0`
` rArr 2(x-4) +2( x-2) +(x-y) 2=0 `
` 6x - 2y -12=0 `
For y
` (y-x) C_2 + [y-(-4) ]C_4 +( y-0) C_5 =0 `
`rArr " "( y-x) 2+ (y+4) +y 2=0 `
`=6y -2x +8=0 `
eq. (1) & (2)
` y=- 3//4" " x= 7//4 `
so potential difference = x-y ` = (7)/(4) -((-3)/(4)) =(5)/(3)` volt.