Answer:
2.56×10⁻¹⁰ C
Explanation:
Note: When capacitor's are connected in series, They act like resistance connected in parallel.
Therefore,
1/Ct = 1/C1+ 1/C2 .................................... Equation 1
Where Ct = Total Capacitance, C1 = Capacitance of the the first capacitor, C2 = Capacitance of the second capacitor.
Given: C1 = 3.65 pF, C2 = 8.55 pF.
Substitute into equation 1
1/Ct = 1/3.65 + 1/8.55
1/Ct = 0.274+0.117
1/Ct = 0.391
Ct = 2.56 pF.
Recall,
Q = CtV .................. Equation 2
Where,
Q = charge, V = potential difference.
Note: Since the capacitor are connected in series, the same charge flows through the first and the second capacitor.
Given: V = 400 V, Ct = 2.56 pF = 2.56×10⁻¹² F
Substitute into equation 2
Q = 2.56×10⁻¹²(100)
Q = 2.56×10⁻¹⁰ C.
