Four capacitors `C_(1)=8 muF`,
`C_(2)=2 muf, C_(3)=6 muF`, and `C_(4)=6, muF` are arranged as shown in. Find the charge on all the capacitors in the circuit.
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Explanation

The figure shows an isolated system that includes all the capacitor plates isolted from other parts of circuit. We can conclude net sum of charge should be zero.
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`2(x+10)+4(x+20)=6(0-x)+4(x-20)=0`
or `4x+20=0` or `x=-5V`
Hence `q_(1)=4[20-(5)]=100 muC`
`q_(1)=2[((-5)+10)-0)]=10 muC`
`q_(3)=6[0-(-5)]=30 muC`
`q_(4)=4[(-5+20)-0]=60 muC`.