Electric Capacitance

Topic hub: Begin here for the capacitor model, then use the linked branch notes for networks, switching and the optional exponential derivation.

Enrichment outside active 9749

Electric capacitance and RC transients are not listed in the active 2026 H2 Physics 9749 learning outcomes. This dossier is retained as useful enrichment and preparation for courses that include capacitors. Do not treat it as required 9749 examination scope unless your teacher says otherwise.

Overview

A capacitor consists of two conductors separated by an insulating gap. Moving electrons from one conductor to the other creates charge separation, an electric field and a potential difference (p.d.) between them.

Capacitance tells us how much plate-charge magnitude is associated with each volt of p.d. Stored energy comes from the work required to build that charge separation.

For networks and time-dependent circuits, continue to Capacitor Networks and RC Switching.

Core Ideas

• Capacitance is defined by $C=Q/V$, where $Q$ is the magnitude of charge on either plate in the ideal two-conductor model.
• A capacitor stores charge separation and electric energy, not net charge of the whole initially neutral assembly.
• Dielectrics increase capacitance by polarising and reducing the p.d. for the same free plate-charge magnitude in an isolated fixed-$Q$ case.
• The stored energy is the area under the $V$-against-$q$ charging graph for an ideal capacitor.

Exam Relevance

This page is enrichment outside active 9749. If the material is taught locally, typical questions ask for careful definitions of $Q$, use of $E=\frac{1}{2}QV$, interpretation of dielectric insertion, and recognition of capacitor-switching limits.

Model and Definition

For an ideal linear capacitor whose geometry and insulating material remain fixed:

$$Q=CV,$$

so:

$$C=\frac{Q}{V}.$$

Here:

• $Q$ is the magnitude of charge on either plate in the standard two-conductor model;
• $V$ is the p.d. between the conductors;
• $C$ is capacitance, measured in farads.

$$1\mathrm{F}=1\mathrm{C}{\mathrm{V}}^{-1}.$$

Common submultiples are:

$$1\mu \mathrm{F}=10^{-6}\mathrm{F},1\mathrm{nF}=10^{-9}\mathrm{F},1\mathrm{pF}=10^{-12}\mathrm{F}.$$

The relation $Q=CV$ does not mean that capacitance increases whenever $Q$ increases. For an ideal linear capacitor with fixed construction, $C$ is constant and $Q$ changes in proportion to $V$.

Figure: In the ideal two-conductor model, transferring electrons from one plate to the other leaves plate charges $+Q$ and $-Q$. Their separation creates the p.d. $V$. The insulating gap may be vacuum, air or another dielectric; a solid dielectric is not required for the definition $Q=CV$.

Stored Plate Charge versus Net Charge

Suppose the two-conductor capacitor starts electrically neutral and charge is transferred internally from one plate to the other. The plates acquire $+Q$ and $-Q$, so:

$$Q_{\mathrm{net}}=+Q+(-Q)=0.$$

Nevertheless, $Q$ in $C=Q/V$ is nonzero: it is the magnitude on either plate. The capacitor stores charge separation and electric-field energy, not a newly created net charge.

This zero-net-charge statement belongs to the standard initially neutral ideal model. A real capacitor can also acquire additional common net charge relative to its surroundings; such stray/common-mode effects are outside this introductory treatment.

Figure: The charge ledger prevents two meanings of $Q$ from being confused. The algebraic charge of an initially neutral two-plate system remains zero, while the plate-charge magnitude used in $C=Q/V$ is $Q$.

Dielectrics

A dielectric is an insulating material that polarises in an electric field. Bound charges shift slightly within its molecules, producing a polarisation field that opposes part of the field due to the free plate charges.

For the same geometry, inserting a dielectric increases capacitance. What else changes depends on the constraint.

Isolated capacitor: fixed free charge

If the capacitor is disconnected so that plate charge cannot change appreciably:

• $Q$ is fixed;
• polarisation reduces the effective field and p.d.;
• $C=Q/V$ therefore increases;
• $E=Q^2/(2C)$ decreases.

The energy decrease corresponds to mechanical work/energy transfer during dielectric insertion; it does not disappear.

Battery-connected capacitor: fixed p.d.

If an ideal battery remains connected:

• $V$ is fixed;
• increased capacitance causes extra charge to flow from the source, $Q=CV$;
• stored energy $E=\frac{1}{2}C{V}^2$ increases;
• the source and any mechanical work must be included in the full energy account.

Figure: The main panel shows the isolated fixed-$Q$ case: bound surface charges create an opposing polarisation field, reducing $V$ and increasing $C$. The comparison strip distinguishes this from the battery-connected fixed-$V$ case, where extra free charge flows onto the plates.

Dielectric breakdown

If the field becomes too large, an insulating material can become conducting. Charge may cross the gap suddenly, causing a spark, heating or permanent damage. This is dielectric breakdown.

Quantitative permittivity and dielectric-strength calculations are further enrichment and are not developed here.

Energy Stored in a Capacitor

For a small increase $dq$ in separated charge, the work transferred into the capacitor is:

$$dE=Vdq.$$

For an ideal linear capacitor, $V=q/C$. Charging quasi-statically from $q=0$ to $q=Q$ gives:

$$E={\int }_0^Q\frac{q}{C}dq=\frac{Q^2}{2C}.$$

Using $Q=CV$:

$$E=\frac{1}{2}QV=\frac{1}{2}C{V}^2=\frac{Q^2}{2C}.$$

Figure: The straight line is the exact ideal-capacitor relation $V(q)=q/C$. The stored energy is the area under the $V$-against-$q$ graph from $0$ to $Q$, giving the triangular area $\frac{1}{2}QV$.

Stored energy is not always the source energy

If an initially uncharged capacitor is charged from an ideal constant-voltage source $V$ through resistance until its final charge is $Q=CV$, the source transfers:

$$E_{\mathrm{source}}=QV=C{V}^2.$$

Only:

$$E_{\mathrm{stored}}=\frac{1}{2}C{V}^2$$

remains in the capacitor. In the simple resistive charging process, the other half is dissipated as internal energy in the resistance. This distinction depends on the charging process; $\frac{1}{2}C{V}^2$ is the final capacitor energy, not automatically the total energy supplied by every source arrangement.

Choosing the Energy Formula

Known or fixed quantities	Useful form	Interpretation
$Q$ and $V$	$E=\frac{1}{2}QV$	triangular graph area
fixed $C$ and known $V$	$E=\frac{1}{2}C{V}^2$	battery-connected/final-voltage problems
fixed $Q$ and known $C$	$E=Q^2/(2C)$	isolated capacitor problems

Always identify what remains fixed before predicting how energy changes.

Worked Example

An ideal $4.0\mu \mathrm{F}$ capacitor is charged to $12\mathrm{V}$.

Charge magnitude:

$$Q=CV=(4.0\times 10^{-6})(12)=48\mu \mathrm{C}.$$

Stored energy:

$$E=\frac{1}{2}C{V}^2=\frac{1}{2}(4.0\times 10^{-6})(12{)}^2=2.88\times 10^{-4}\mathrm{J}.$$

If it was charged from zero through resistance by an ideal $12\mathrm{V}$ source, the source supplied $QV=5.76\times 10^{-4}\mathrm{J}$; half became stored energy and half was dissipated in the resistance.

Common Misconceptions

• “A capacitor creates charge.” It separates existing charge.
• “The whole capacitor must have net charge $Q$.” In the standard initially neutral model, its two plate charges sum to zero.
• “$C$ increases because $Q$ increased.” For fixed construction, $C$ is a device parameter and $Q=CV$.
• “A dielectric always keeps $Q$ fixed.” Only an isolated capacitor fixes free charge; a connected ideal battery fixes $V$.
• “The source always supplies $\frac{1}{2}QV$.” That is the final stored energy, not necessarily total source transfer.

Links

• Capacitor Networks and RC Switching
• Why RC Curves Are Exponential
• Core prerequisite: Current Electricity Fundamentals
• Core prerequisite: D.C. Circuits
• Related core topic: Electric Fields