Capacitors: Charge, Voltage and Stored Energy

1. Separate four quantities

We use an ideal, linear two-terminal capacitor model with equal and opposite plate charges. Its capacitance $C>0$ is constant while geometry and insulating material remain unchanged. We ignore leakage (unwanted loss of charge), breakdown (the insulator starts conducting when the electric field is too strong), and stray capacitances between nearby parts. The network results describe equilibrium: the final state after charging currents have stopped.

• Charge: $q_a$ is the signed charge on the plate connected to terminal a; the other plate carries $-q_a$. Its unit is the coulomb, $\mathrm C$.
• Voltage: $v=V_a-V_b$ is a directed potential difference between two terminals, in volts. Shifting the potential reference does not change this difference.
• Capacitance: $C$ relates charge to voltage. Its unit is the farad, $1\,\mathrm F=1\,\mathrm C/\mathrm V$.
• Stored energy: $U$ is the capacitor’s electrical energy, in joules. It is neither a charge nor a voltage.

$$q_a=Cv$$

For positive-polarity examples, write $Q=|q_a|$ and $V=|v|$, so $Q=CV$. A capacitor “charged to $Q$” has $+Q$ on one plate and $-Q$ on the other. Its net charge is zero; $Q$ is not the sum of the two charge magnitudes.

Common mistake: capacitance is not a maximum charge. In this linear model, doubling voltage doubles charge without changing $C$. A real component also has a voltage rating and operating limits; those are outside this lesson.

2. Connect energy with charge and voltage

When a fixed-capacitance capacitor is charged gradually from zero, its voltage rises from zero to $V$. The last small amounts of charge are moved through a higher voltage than the first. The average voltage with respect to transferred charge, not time, is $V/2$, giving the stored energy:

$$U=\frac12 QV$$

$$U=\frac12 CV^2=\frac{Q^2}{2C}$$

These three expressions describe the same state. Choose the form that uses the constrained quantity. Reversing polarity leaves energy positive: in signed form, $U=\tfrac12 Cv^2$.

Check the system: this formula gives the energy ultimately stored, not necessarily all the energy supplied by a source. A resistor may also receive energy during charging. Do not automatically equate battery work with capacitor energy.

3. Identify what the connection holds fixed

• Ideal voltage source still connected: the source fixes the potential difference. It can supply or receive charge. If $C$ changes, obtain the new charge from $Q=CV$.
• Both terminals disconnected and insulated: charge cannot enter or leave either plate. With no leakage or other connection, each plate’s charge stays fixed. If $C$ changes, obtain the new voltage from $V=Q/C$.
• Capacitors connected to one another without a source: connected plates can exchange charge. The conserved quantity is the total signed charge on each isolated conducting group, not necessarily the charge of each plate.

“Isolated” must refer to a specific system. An entire network can be isolated from the outside while charge redistributes within it. Likewise, having no battery does not guarantee conservation of stored energy: a resistor can heat up, and an external force can move plates.

4. Recognize parallel and series connections by nodes

A node groups points joined by ideal wires, at the same equilibrium potential. Redrawing a wire without changing its connections does not change the network. The positions of the symbols on the page do not decide “series” or “parallel.”

Parallel: the same two nodes

If both capacitors connect between the same nodes a and b, they share the same directed voltage $v$. Charges on the plates attached to node a add algebraically:

$$q_{a,\mathrm{tot}}=C_1v+C_2v$$

$$C_{\mathrm{eq}}=C_1+C_2$$

At the same voltage, the larger capacitance carries the larger charge magnitude and stores more energy. The equivalent capacitance exceeds each positive component capacitance.

Series: an isolated, neutral intermediate node

Consider a, then $C_1$, then b, then $C_2$, then c. Node b has no other branch. To use the “same charge” rule, the intermediate conductor, consisting of two plates and their joining wire, must also have zero initial net charge and no path to exchange charge with the outside.

For positive polarity from a to c, the inner plate of $C_1$ carries $-Q_1$ and the inner plate of $C_2$ carries $+Q_2$. Neutrality gives $-Q_1+Q_2=0$, hence $Q_1=Q_2=Q$. Voltages add:

$$V=\frac Q{C_1}+\frac Q{C_2}$$

$$\frac1{C_{\mathrm{eq}}}=\frac1{C_1}+\frac1{C_2}$$

The equivalent capacitance is smaller than the smallest component capacitance. At equal charge, the smaller capacitance has the larger voltage and stores more energy, because $V=Q/C$ and $U=Q^2/(2C)$.

Important limit: if the intermediate node has conserved net charge $q_b\ne0$, use the directed charges $q_1=C_1(V_a-V_b)$ and $q_2=C_2(V_b-V_c)$. The equation is then $-q_1+q_2=q_b$, and the directed charges are unequal. Keep that constraint instead of blindly applying the usual rule. Charge does not need to cross a capacitor’s insulating gap to redistribute on the intermediate conductor.

5. Reduce a network, then recover each capacitor

Worked example 2: a parallel block in a series connection

A source maintains $12\,\mathrm V$ between A and D. Capacitors $C_1=3\,\mu\mathrm F$ and $C_2=6\,\mu\mathrm F$ are in parallel between A and B. This block is in series with $C_3=18\,\mu\mathrm F$ between B and D. All capacitors start uncharged, and B has no other connection. Find the equilibrium voltages, charges and total stored energy.

Step A: reduce the inner block first

$$C_{12}=(3+6)\,\mu\mathrm F$$

$$C_{12}=9\,\mu\mathrm F$$

$$C_{\mathrm{eq}}=\frac{9\times18}{9+18}\,\mu\mathrm F$$

$$C_{\mathrm{eq}}=6\,\mu\mathrm F$$

Check the direction of the result: $C_{12}>6\,\mu\mathrm F$ for the parallel step, then $C_{\mathrm{eq}}<9\,\mu\mathrm F$ for the series step.

Step B: work back from the network to its branches

$$Q=C_{\mathrm{eq}}V=72\,\mu\mathrm C$$

$$V_{AB}=\frac{72}{9}\,\mathrm V=8\,\mathrm V$$

$$V_{BD}=\frac{72}{18}\,\mathrm V=4\,\mathrm V$$

The $C_{12}$ block and $C_3$ have equal charge magnitudes. Inside the parallel block, the two capacitors instead have equal voltages:

$$Q_1=(3\times8)\,\mu\mathrm C$$

$$Q_1=24\,\mu\mathrm C$$

$$Q_2=(6\times8)\,\mu\mathrm C$$

$$Q_2=48\,\mu\mathrm C$$

$$Q_3=72\,\mu\mathrm C$$

Checks: $(8+4)\,\mathrm V=12\,\mathrm V$ and $Q_1+Q_2=Q_3$. At node B, the signed charges are $(-24-48+72)\,\mu\mathrm C=0$. Neutrality is satisfied. Saying “all three are in series, so each carries 72 microcoulombs” would miss the parallel split.

Step C: check energy in two ways

$$U_1=96\,\mu\mathrm J$$

$$U_2=192\,\mu\mathrm J$$

$$U_3=144\,\mu\mathrm J$$

$$U_{\mathrm{tot}}=432\,\mu\mathrm J$$

$$\frac12 C_{\mathrm{eq}}V^2=432\,\mu\mathrm J$$

The individual energies sum to the equivalent capacitor’s energy. Adding charge magnitudes over every plate would give neither the charge delivered to one network terminal nor its energy.

6. Move the plates apart: one change, two constraints

For large parallel plates in vacuum, with separation $d$ small compared with their lateral dimensions, neglect edge effects:

$$C=\frac{\varepsilon_0 A}{d}$$

Here $\varepsilon_0$ is the vacuum permittivity, approximately $8.85\times10^{-12}\,\mathrm{F/m}$. Area $A$ belongs to one plate. Doubling $d$ at fixed area halves the capacitance, provided the approximation remains valid. Geometry determines the capacitance change; the connection determines what happens to $Q$, $V$ and $U$.

7. Join two capacitors: conserve charge, track energy

Two fixed-capacitance capacitors are disconnected from every source. Join their a terminals through a finite positive resistance and their b terminals together, then wait for equilibrium. Neglect external leakage, inductance, radiation and plate motion. The resistance allows transient current to decay while dissipating energy.

Initial voltages $v_1$ and $v_2$ both use a minus b as their sign convention. The a-plate group conserves its total signed charge. At equilibrium the common voltage is $v_f$:

$$q_{a,\mathrm{tot}}=C_1v_1+C_2v_2$$

$$q_{a,\mathrm{tot}}=(C_1+C_2)v_f$$

$$v_f=\frac{C_1v_1+C_2v_2}{C_1+C_2}$$

The final voltage is a capacitance-weighted average; it is the arithmetic mean only when the capacitances are equal. The b-plate group carries the opposite charge. Merely saying “the total charge of both capacitors is zero” is not enough to determine the final voltage.

8. Try six graduated questions

Draw your own connection sketch and name the constraint first. Open the hint only if needed, then compare the solution’s reasoning.

9. Check the method before the arithmetic

• Mark nodes and voltage polarity. Do not identify connections by appearance alone.
• Write “source fixes voltage,” “each plate’s charge is fixed,” or “isolated group charge is conserved.” Specify the group.
• For series connections, justify intermediate-node neutrality before equating charges. For parallel connections, verify both common nodes.
• Reduce the network, then work back to each component. A parallel block’s total charge differs from its individual branch charges.
• Check units, summed voltages and signed charges on isolated nodes. Compare two energy calculations when possible.
• Do not impose conservation of capacitor energy alone when a source, resistor or mechanical force exchanges energy with the capacitors.

To revisit the distinction between potential and energy, see Electric Field and Potential. For voltage signs in steady resistive circuits, continue with Kirchhoff’s Laws. Solving the time dependence of RC circuits is a separate next step and is not developed here.

10. References and course connection

Course: Concordia PHYS 205 — Electricity and Magnetism.

For further explanations of the principles, see OpenStax, University Physics, Volume 2. This guide’s problem statements, numerical choices and figures were created for Nablio; they do not reproduce course assignments or examinations.

• 8.1 — Capacitors and Capacitance
• 8.2 — Capacitors in Series and in Parallel
• 8.3 — Energy Stored in a Capacitor