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Applying Kirchhoff loop law in loop $I$ :

$1q_{1} +3q_{1}+q_{2} −190=0$

$⟹4q_{1}+q_{2}=190×3$

Applying Kirchhoff loop law in loop $II$ :

$3q_{1}+q_{2} +3q_{2} +1q_{2} =0$

$⟹q_{1}=−5q_{2}$

Putting this in equation 1, we get $4(−5q_{2})+q_{2}=190×3$

$⟹q_{2}=30μC$

Thus potential difference $V_{AB}=3μFq_{2} =3μF30μC =10V$

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