Two capacitors C₁ = 3 μF and C₂ = 6 μF are connected in series and charged by a 9 V battery. What is the charge stored on C₂?
A54 μC — C₂ sees the full battery voltage, so Q₂ = C₂ × V = 6 × 9
B36 μC — in series, charge distributes proportionally to capacitance
C18 μC — in series all capacitors carry the same charge: C_eq = 2 μF, so Q = C_eq × V = 18 μC
D9 μC — the voltage splits equally between the two capacitors
In series, all capacitors carry the same charge Q regardless of their individual capacitances. C_eq = 1/(1/3 + 1/6) = 1/(1/2) = 2 μF. Q = C_eq × V = 2 × 9 = 18 μC — and this same 18 μC appears on both C₁ and C₂. Option A is the most tempting error: applying Q = CV to C₂ with the full 9 V, as if C₂ were connected directly across the battery. In reality, the voltage across C₂ is V₂ = Q/C₂ = 18/6 = 3 V, not 9 V. The voltage is split, not the charge.
Question 2 Multiple Choice
Adding a capacitor in series always decreases C_eq below the value of either individual capacitor. What is the physical reason?
ASeries capacitors share the battery voltage, leaving less energy available for each
BSeries connection is analogous to increasing the total plate separation — greater separation reduces capacitance, just as in a single capacitor with wider plates
CThe inner plates of series capacitors carry opposite charges that partially cancel
DThe equivalent series resistance increases, reducing the effective capacitance
Capacitance C = ε₀A/d decreases when plate separation d increases. Connecting capacitors in series is physically equivalent to building one capacitor with the sum of their internal gaps — the total effective plate separation is the sum of the individual gaps. More separation → less capacitance → C_eq is smaller than either individual capacitor. Parallel connection does the opposite: it adds plate area (A increases → C increases → C_eq larger than either alone). This physical picture explains why the combination rules for C are the opposite of those for R.
Question 3 True / False
Capacitors in parallel combine the same way resistors in series do: the equivalent value equals the direct sum of the individual values.
TTrue
FFalse
Answer: True
Capacitors in parallel: C_eq = C₁ + C₂ + ... (direct sum). Resistors in series: R_eq = R₁ + R₂ + ... (direct sum). Both exhibit direct addition because the relevant quantity accumulates in the same direction: parallel capacitors add effective plate area, series resistors add path length for current. This is part of the broader duality between C and R: the combination rules for capacitors are exactly swapped relative to resistors, which the topic explicitly notes as a useful memory anchor.
Question 4 True / False
When capacitors are connected in series, they most have the same voltage across their terminals.
TTrue
FFalse
Answer: False
In series, all capacitors carry the same charge Q — not the same voltage. The voltage across each capacitor is V = Q/C, so capacitors with different capacitances will have different voltages. Specifically, smaller capacitors carry larger voltage drops (for the same Q, smaller C means larger V = Q/C). Equal voltage is the defining property of parallel connection, not series. Confusing which quantity is shared (charge in series, voltage in parallel) is the most common error in capacitor network problems.
Question 5 Short Answer
Why do all capacitors in a series combination end up with the same charge, regardless of their individual capacitance values?
Think about your answer, then reveal below.
Model answer: The inner plates of adjacent capacitors in a series combination are electrically isolated from the external circuit — no charge can flow onto or off them from outside. When the combination charges, charge accumulates on the outermost plates only. The electric field from those plates induces equal and opposite charges on the adjacent inner plates by polarization. Since each isolated inner conductor must have zero net charge (charge cannot build up on an isolated conductor), whatever charge appears on one face of an inner plate must be exactly balanced on the other face — forcing the same charge Q through every capacitor in the chain.
This is charge conservation applied to isolated internal conductors. If C₁ develops charge +Q on its outer plate, the field induces −Q on the inner face of C₁, which forces +Q on the near face of C₂, which induces −Q on C₂'s outer plate — and so on. Each capacitor receives the same Q regardless of its capacitance. The individual capacitances only determine how much voltage each capacitor 'needs' to accommodate that charge: V = Q/C. This is why a smaller capacitor in a series string has a larger voltage drop than a larger one.