A 3.65-pF capacitor is connected in series with an 8.55-pF capacitor and a 400-V potential difference is applied across the pair. What is the charge on the first (smaller) capacitor?

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.