The potential of the point $A$ is greater than that of $B$ by $19\, volt$. What is the potential difference in volts across the $3\,\mu F$ capacitor ?
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$\mathrm{V}_{\mathrm{A}}-\frac{\mathrm{q}}{2 \times 10^{-6}}-\frac{\mathrm{q}}{3 \times 10^{-6}}-8+15-\frac{\mathrm{q}}{4 \times 10^{-6}}=\mathrm{V}_{\mathrm{B}}$
$\left(\mathrm{V}_{\mathrm{A}}-\mathrm{V}_{\mathrm{B}}\right)+15-8=\frac{13}{12 \times 10^{-6}} \mathrm{q}$
$19+15-8=\frac{13}{12 \times 10^{-6}} \mathrm{q}$
$\mathrm{q}=24 \,\mu \mathrm{c}$
${{\rm{V}}_{3\mu F}} = \frac{{24 \times {{10}^{ - 6}}}}{{3 \times {{10}^{ - 6}}}} = 8\,{\rm{volt}}$
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