Answer:
0.192 C
Explanation:
The charge in a capacitor can be expressed using the formula:
Q = CtVt..................... Equation 1
Where Q represents Charge, Ct = Effective Capacitance, and Vt = Effective Voltage.
Note: When the positive plates of two capacitors are joined to the negative plates of each other, this signifies they are in series
The formula for combined capacitance in series is given by:
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
Here, we have: C1 = 4.0 mF, C2 = 6.0 mF
Substituting into equation 2:
Ct = (4×6)/(4+6)
Ct = 24/10
Ct = 2.4 mF.
Further,
Vt = V1 + V2................... Equation 4
Where V1 is the voltage of the first capacitor, and V2 is the voltage of the second.
Given: V1 = 50 V, V2 = 30 V
Vt = 50 + 30
Vt = 80 V.
Now substituting the values of Vt and Ct into equation 1 gives:
Q = 80(2.4)
Q = 192 mC
Q = 0.192 C.
As the capacitors are in series, the same charge flows through each.
Therefore, the final charge on the 6.0 mF capacitor is 0.192 C.