Can You Change the Capacitance of a Capacitor?

Yes. A capacitor’s capacitance can be changed, most directly by changing the arrangement of its conductors or the material between them. A common example is a variable capacitor, in which rotating plates changes how much their areas overlap.

The key distinction is that capacitance is not the same thing as charge or voltage. Changing the charge or voltage of an ideal capacitor with fixed construction does not, by itself, change its capacitance.

What determines capacitance?

Capacitance measures how much charge a capacitor stores per unit voltage difference:

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

Here, $Q$ is the magnitude of the charge on either conductor, and $\Delta V$ is the potential difference between them. For an ideal, linear capacitor, this ratio is constant for a given construction.

For a parallel-plate capacitor whose plates are large compared with their separation, the capacitance is approximately

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

where $A$ is the overlapping plate area, $d$ is the separation, and $\varepsilon$ is the permittivity of the material between the plates. This formula shows three ways to change capacitance:

• Increase the plate overlap area: capacitance increases.
• Decrease the plate separation: capacitance increases.
• Put in a material with greater permittivity: capacitance increases.

For a uniform dielectric with relative permittivity $\varepsilon_r$, $\varepsilon=\varepsilon_r\varepsilon_0$, where $\varepsilon_0$ is the permittivity of vacuum. Real capacitors can have more complicated geometry, but the same broad principle applies: capacitance depends on the conductors’ arrangement and the electrical properties of the surrounding materials.

A worked example

Consider an air-filled parallel-plate capacitor with plate area $A=0.010\ \text{m}^2$ and separation $d=1.0\ \text{mm}=1.0\times10^{-3}\ \text{m}$. Using $\varepsilon_0\approx8.85\times10^{-12}\ \text{F/m}$,

$$\begin{aligned} C&\approx\frac{\varepsilon_0 A}{d}\\ &=\frac{(8.85\times10^{-12})(0.010)}{1.0\times10^{-3}}\\ &=8.85\times10^{-11}\ \text{F}=88.5\ \text{pF}. \end{aligned}$$

If you halve the separation to $0.50\ \text{mm}$ without changing anything else, the capacitance doubles to about $177\ \text{pF}$. Alternatively, filling the original gap with an ideal dielectric whose relative permittivity is 4 would raise the capacitance to about $4(88.5)=354\ \text{pF}$.

In a practical variable capacitor, turning a shaft moves interleaved plates. The overlap area changes smoothly, which changes the capacitance. Other capacitors are adjusted by moving a dielectric, changing the gap, or using an electronic circuit that behaves like a variable capacitance.

What happens to charge and voltage?

The result depends on how the capacitor is connected.

Connected to a battery: The battery approximately holds $\Delta V$ fixed. If capacitance increases, the capacitor takes on more charge because $Q=C\Delta V$. For example, raising capacitance from $100\ \text{pF}$ to $300\ \text{pF}$ while connected to a $12\ \text{V}$ battery raises the charge from

$$Q_1=(100\ \text{pF})(12\ \text{V})=1.2\ \text{nC}$$

to

$$Q_2=(300\ \text{pF})(12\ \text{V})=3.6\ \text{nC}.$$

The extra charge flows from the battery.

Isolated from the battery: The capacitor’s charge is approximately fixed, since there is no conducting path for charge to enter or leave. In the same example, if it starts at $100\ \text{pF}$ and $12\ \text{V}$, its charge is $1.2\ \text{nC}$. Increasing its capacitance to $300\ \text{pF}$ then lowers its voltage to

$$\Delta V=\frac{Q}{C}=\frac{1.2\ \text{nC}}{300\ \text{pF}}=4\ \text{V}.$$

So changing capacitance can change voltage or charge, but which one changes depends on the circuit condition. If you physically move plates or a dielectric, mechanical work and energy transfer may also be involved.

Common mistakes

• Assuming voltage sets capacitance. Voltage and charge are related by $Q=C\Delta V$; for a linear capacitor with unchanged construction, varying one varies the other without changing $C$.
• Confusing a variable capacitor with a changing voltage. A variable capacitor changes its physical or electrical structure. A voltage change alone is not the same thing.
• Forgetting what is held fixed. A battery tends to fix voltage, while an isolated capacitor tends to keep its charge. These lead to different outcomes when capacitance changes.
• Treating every capacitor as perfectly linear. Some electronic devices have voltage-dependent capacitance. In that case, the effective capacitance can depend on operating voltage, and the simple constant-$C$ model needs qualification.

For an ordinary fixed capacitor in an introductory circuit problem, use its stated capacitance as a constant unless the problem says its geometry, dielectric, or voltage-dependent behavior changes.