116-117 - Capacitance and Energy Stored

Capacitance As Charge Per Volt

A capacitor is a component that stores charge by separating equal and opposite charges on two conducting plates. The whole component remains neutral overall, but one plate has charge $+Q$ and the other has charge $-Q$. The stored charge $Q$ usually means the magnitude of the charge on either plate.

When a capacitor is connected to a supply, a potential difference is built up across its plates. A capacitor with a larger capacitance stores more charge for the same potential difference. That is what capacitance measures:

The unit follows from the equation:

So a $1\,\mathrm{F}$ capacitor stores $1\,\mathrm{C}$ of charge for each volt across it. In school and exam circuits, capacitances are usually much smaller, such as $\mathrm{\mu F}$, $\mathrm{nF}$ or $\mathrm{pF}$.

Worked example:

A capacitor stores $1.8 \times 10^{-3}\,\mathrm{C}$ when the potential difference across it is $12\,\mathrm{V}$. Calculate its capacitance.

The question gives charge and potential difference, and asks for charge stored per volt, so use the definition:

Substitute with units:

This is often clearer as:

Interpretation: each volt across this capacitor is associated with $1.5 \times 10^{-4}\,\mathrm{C}$ of stored charge.

Misconception contrast:

For an Edexcel “calculate” item, show the formula, substitute SI units, and include the final unit. If you give the answer in $\mathrm{\mu F}$, make the power-of-ten conversion visible.

Why Energy Is Half QV

Potential difference is energy transferred per unit charge. If a fixed charge moved through a fixed potential difference, the energy transfer would be $QV$.

A capacitor is different while it is being charged. At the start, the capacitor has almost no charge on its plates, so its potential difference is almost zero. Later, as more charge has already been separated, the potential difference is larger. The final bit of charge is moved through nearly the final p.d.; the first bit was not.

For a capacitor with constant capacitance:

so:

That means $V$ is directly proportional to $Q$. On a graph of potential difference $V$ on the vertical axis against charge stored $Q$ on the horizontal axis, the line is straight and passes through the origin.

The work done in adding a small amount of charge $\Delta Q$ is:

On a $V$ against $Q$ graph, each small piece of work is a thin strip of area. Therefore:

For a straight line from $(0, 0)$ to final charge $Q$ and final p.d. $V$, the area is a triangle:

The unit check is useful:

So the graph area has the correct unit for energy.

This is why the answer is not $QV$. Multiplying final $Q$ by final $V$ treats every bit of charge as if it crossed the final p.d. In the linear capacitor model, the average p.d. during charging is $V/2$, so the stored energy is $Q(V/2)$.

[DIAGRAM: asset_name: Linear capacitor voltage-charge graph with triangular stored-energy area and reciprocal-capacitance gradient; asset_slug: fields_capacitor_energy_graph; recommended_method: matplotlib; description: Linear capacitor voltage-charge graph with triangular stored-energy area and reciprocal-capacitance gradient]

Three Energy Forms

Edexcel gives the energy stored in a capacitor in three equivalent forms:

They are not three separate ideas. They are the same graph-area result written for different information.

Start with:

If the question gives $C$ and $V$, use $Q = CV$:

If the question gives $Q$ and $C$, use $V = Q/C$:

Worked example:

A $470\,\mathrm{\mu F}$ capacitor is charged to a potential difference of $9.0\,\mathrm{V}$. Calculate the charge stored and the energy stored.

Decision point 1: convert the capacitance into farads.

Decision point 2: charge comes from $Q = CV$.

Decision point 3: for energy, $C$ and $V$ are already known, so $W = \frac{1}{2} CV^{2}$ is the direct route.

The result is small in joules, but it is not small because charge has disappeared. It is small because the capacitance and p.d. are modest. The energy is stored in the electric field associated with the separated charge.