It’s due very soon and this is my last attempt

n (figure 1)
c
1
= 8.00
mf
,
c
2
= 7.00
mf
,
c
3
= 2.00
mf
, and
c
4
= 5.00
mf
. the capacitor network is connected to an applied potential difference
v
ab
. after the charges on the capacitors have reached their final values, the voltage across
c
3
is 40.0
v

85.9 V

Explanation:

The voltage across C₃ is 40.0 V, so the voltage between a and d must also be 40.0 V.

Find the equivalent capacitance of C₁, C₂, and C₃:

C₁₂₃ = 1 / (1/C₁ + 1/C₂) + C₃

C₁₂₃ = 1 / (1/8.00 + 1/7.00) + 2.00

C₁₂₃ = 5.73 mF

This is in series with C₄.  Capacitors in series have the same charge, so:

Vad C₁₂₃ = Vbd C₄

(40.0) (5.73) = Vbd (5.00)

Vbd = 45.9 V

The total voltage is therefore:

V = Vad + Vbd

V = 40.0 + 45.9

V = 85.9 V