**Answer:**

**0.192 C**

**Explanation:**

The charge in a capacitor is given as,

Q = CtVt..................... Equation 1

Where Q = Charge, Ct = Effective Capacitance of the capacitors, Vt = Effective Voltage.

**Note: If the capacitors are connected to each other with the positive plate of each connected to the negative plate of the other means that the capacitors are connected in series**

The combined capacitor in series is given as,

1/Ct = 1/C1 + 1/C2

Where C1 = Capacitance of the first capacitor, C2 = Capacitance of the second capacitor.

Ct = C1C2/(C1+C2)...................... Equation 2

Given: C1 = 4.0 mF, C2 = 6.0 mF

Substitute into equation 2

Ct = (4×6)/(4+6)

Ct = 24/10

Ct = 2.4 mF.

Also,

Vt = V1 + V2................... Equation 4

Where V1 = Voltage in the first capacitor, V2 = Voltage in the second capacitor.

Given: V1 = 50 V, V2 = 30 V

Vt = 50+30

Vt = 80 V.

Substitute the value of Vt and Ct into equation 1

Q = 80(2.4)

Q = 192 mC

Q = 0.192 C.

Since both capacitors are in series, The same quantity of charge flows through them.

**Hence the final charge on the 6.0 mF capacitor = 0.192 C**