According to figure, 4µF,8µF,4µF are in parallel combination. So, C′=4+8+4=16µF Now, C′ and 5µF are in series combination. So, total capacitance in the circuit is, ‌
1
C‌total ‌
=‌
1
5
+‌
1
C′
From Eq. (i), C‌total ‌=‌
16×5
16+5
=‌
80
21
µF Total charge in the given circuit, q‌total ‌‌=C‌total ‌⋅V ‌=‌
80
21
×63=240µC Charge remains same in series combination. So, 5µF also have 240µC charge. The potential difference across 5µF capacitor is, ‌V′=‌