Answer:
Q₁ = 2.4 10⁻⁴ C
Explanation:
We have a circuit with several capacitors, let's find the equivalent capacitor of the parallel
[tex]C_{eq1}[/tex] = C₂ + C₃
C_{eq1} = (10 +30) 10⁻⁶
C_{eq1} = 40 10⁻⁶ F
There remains a series system between C₁ and C_{eq1}, let's find the equivalent capacitor
1/C_{eq2} = 1 / C₁ + 1 / C_{eq1}
1 /C_{eq2} = 1 / 20 10⁻⁶ + 1/40 10⁻⁶
1 / C_{eq2} = 0.075 10⁶
C_{eq2} = 13.33 10⁻⁶ F
let's use the relationship
V = Q / C_{eq2}
Q = V C_{eq2}
Q = 18 13.33 10⁻⁶
Q = 2.4 10⁻⁴ C
In a combination of capacitors in series the charge is constant, so the charge on C₁ is the same
Q₁ = 2.4 10⁻⁴ C