Thank you for reporting, we will resolve it shortly
Q.
Find $V_{x}-V_{y}$ just after switch is closed?
Electrostatic Potential and Capacitance
Solution:
$Pd$. across $2\, \mu F$
(1) $V _{ A }- V _{ x }=\frac{30}{6} \times 4$
Pd. across $4 \mu F$
(2) $V _{ A }- V _{ y }=\frac{30}{6} \times 2$
by $(2)-(1)$
$\left( V _{ A }- V _{ y }\right)-\left( V _{ A }- V _{ x }\right)$
$=\frac{30}{6} \times 2-\frac{30}{6} \times 4 $
$V _{ x }- V _{ y }=-10$ volt
