Capacitors in Series

What the figure shows

The diagram presents two parallel-plate capacitors, labelled C₁ (left) and C₂ (right), placed side by side in a horizontal line. Each capacitor is drawn as a pair of vertical plates: the left plate of each capacitor is marked with a column of + signs, and the right plate with a column of − signs. Above the left plate of C₁ is the label Q, and above its right plate is −Q; similarly, above the left plate of C₂ is Q and above its right plate is −Q. A short connecting wire joins the right plate of C₁ to the left plate of C₂. Two external terminals are shown: one attached to the left plate of C₁ and the other to the right plate of C₂. The labels C₁ and C₂ appear below their respective capacitors.

The physical idea

The figure illustrates the series combination of two capacitors. The key insight is that when capacitors are connected in series, the charge on each capacitor is the same (magnitude $Q$). This happens because the connecting wire between C₁ and C₂ is initially neutral; if the charges on the inner plates were not equal and opposite, an electric field would exist in the wire, causing charge to flow until the net charge on each capacitor becomes zero. Consequently, the left plate of C₁ and the right plate of C₂ each carry $+Q$ and $-Q$ respectively, while the right plate of C₁ and the left plate of C₂ carry $-Q$ and $+Q$ respectively.

The total potential difference $V$ across the combination is the sum of the individual potential drops:

where $V_1 = \frac{Q}{C_1}$ and $V_2 = \frac{Q}{C_2}$.

Key formula derived from the figure

Using the above relations, the textbook obtains:

The effective capacitance $C$ of the series combination is defined by $C = \frac{Q}{V}$. Substituting $V$ gives:

For $n$ capacitors in series, this generalises to:

$\frac{1}{C} = \frac{1}{C_1} + \frac{1}{C_2} + \frac{1}{C_3} + \dots + \frac{1}{C_n}$ …

Figure 2.27 shows a series combination of $n$ capacitors. The diagram is a horizontal row of four parallel-plate capacitors labelled $C_1$, $C_2$, $C_3$, and $C_n$ from left to right. Each capacitor is drawn as a pair of vertical plates: the left plate has a $+Q$ charge and the right plate has a $-Q$ charge. The capacitors are connected end-to-end by wires, so that the right plate of one capacitor is directly wired to the left plate of the next. Between $C_3$ and $C_n$, a short dashed/broken wire segment indicates that many intermediate capacitors (the "…" up to $n$) have been omitted for clarity.

Physical idea: In a series combination, the same charge $Q$ appears on every plate. The left plate of $C_1$ gets $+Q$ from the battery, and the right plate of $C_n$ gets $-Q$. Because the connecting wires are conductors, charge redistributes until the net charge on each capacitor is zero — this forces the right plate of $C_1$ to have $-Q$ and the left plate of $C_2$ to have $+Q$, and so on. Thus, each capacitor stores the same magnitude of charge $Q$, but the potential difference across each is different (since $V_i = Q/C_i$).

Key formula derived from this figure: The total potential drop $V$ across the series combination is the sum of the individual drops:

The combination behaves like a single effective capacitor with charge $Q$ and potential difference $V$, so its effective capacitance $C$ satisfies $Q = C V$. Substituting gives:

Cancelling $Q$ (which is non-zero) yields the series capacitance formula:

Here: …