Capacitance

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Prerequisite Materials:

Electricity

Exponential Function

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Capacitors consist of two conducting plates separated by an insulator, called a dielectric, or a vacuum. Capacitors store charge, positive on one plate $\mathrm{+Q}$ and negative $\mathrm{-Q}$ on the other, so the total charge is always zero.

$$Q=CV$$

• where
• $Q$ is the charge in coulombs, $C$;
• $C$ capacitance in farads, $F$, usually measured in $\mathit{\mu F}$ or smaller;
• and
• $V$ is the potential difference in volts, $V$;

The graph above shows that capacitor $A$ has a greater capacitance than capacitor $B$. However, it doesn't clarify how their capacitances differ. To understand the underlying factors, it's helpful to consider the principles of parallel plate capacitors, which highlight how plate area, separation distance, and the dielectric material influence capacitance.

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Parallel Plate Capacitor

$$C=\frac{{\epsilon }_{\text{r}}{\epsilon }_{0}A}{d}$$

• where
• $C$ capacitance in farads, $F$;
• ${\epsilon }_0$ is the permittivity of free space;
• ${\epsilon }_{\text{r}}$ is the relative permittivity and is dimensionless;
• and
• $d$ is the separation of the plates in metres.

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Combining Capacitors

For capacitors in parallel the total capacitance is given by

$$C_{\text{Total}}=C_1+C_2+C_3+...$$

For capacitors in series the total capacitance is given by

$$\frac{1}{C_{\text{Total}}}=\frac{1}{C_1}+\frac{1}{C_2}+\frac{1}{C_3}+...$$

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Capacitors as Energy Stores

A capacitor can store energy which can be determined by

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

To aid understanding and remembering note the similarity to energy store in a spring.

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Capacitor Discharge

When a capacitor discharges through a resistor, the charge stored on it decreases exponentially. The amount of charge remaining on the capacitor (Q) after some elapsed time (t) is governed by the exponential decay equation:

Tha graph of voltage and current against time on discharge has the same shape.

$$Q=Q_0{\mathrm{e}}^{\frac{\mathrm{-t}}{RC}}$$

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Capacitance Time Constant, $RC$

$$\text{time constant, }s=\text{resistance, }\Omega \times \text{capacitance, }F$$

$$\tau =RC$$

Substituting $t=RC$ into $Q=Q_0{\mathrm{e}}^{\frac{\mathrm{-t}}{RC}}$ we get,

$$Q=Q_0{\mathrm{e}}^{-1}$$

$$\frac{Q}{Q_0}={\mathrm{e}}^{-1}\approx 0.37$$

So, after a time equal to the time constant the proportion of charge remaining on the plates with about 37% of the starting charge. This is the same for the potential difference across the plates and for the current through the capacitor.

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Questions:

Questions and Revision
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Questions
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Structured Questions
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Capacitance of a Sphere

Charge can be stored on the surface of a sphere. An isolated, charged sphere of radius $R$ must therefore have a capacitance. To calculate the capacitance of this unusual single plate capacitor is the ratio of the charge, $Q$ and the potential, $V$ at the spheres surface.

$$C=\frac{Q}{V}=\frac{Q}{\frac{Q}{4\mathrm{\pi }{\epsilon }_{0}R}}=4\mathrm{\pi }{\epsilon }_{0}R$$

By discribing this sphere as isolated, we mean that it is very far from any other charged objects.

More on Capacitance of an Isolated Sphere
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