*Before you read this, I suggest you read* post 17.44.

A *capacitor* is a device for storing electrical charge (post 16.25); *capacitance* is the ability to store charge. But the amount of charge stored by a normal capacitor (its capacitance) is so small that the main application of capacitors is not to act as storage devices. Instead, they are used to change the wave shape (post 18.14) of time-dependent electrical signals or to change their phase (post 18.10). In this post, I will explain the first application. The second application will be described in a later post.

A capacitor consists of two sheets, or plates, of a material that can conduct electricity (black in the picture above) with an insulator (a material that can’t conduct electricity, shown in red) between them. (In practice, capacitors are usually made of thin sheets of material that can be rolled up to make the device as small as possible.) Now let’s suppose there is a potential difference, *V*, between the points A and B, in the picture (see post 17.44). This potential difference makes charges moves away from A towards B (post 17.44). (Remember that movement of charge is movement of electrons in the opposite direction – post 17.44).

However, the charges can’t move across the layer of insulator (also called the *dielectric*) so they remain at rest on the first conducting sheet/plate that they encounter. So, these charges are stored by the capacitor. If the charge stored is *Q*, the capacitance of the capacitor is defined by

*C* = *Q*/*V*.

This definition arises because the harder we push charge on to the capacitor (high value of *V*), the greater the charge (high *Q*) that can be stored. If *V* is increased sufficiently, the capacitor will cease to store the charge, because, at such a high potential difference, charges will move across an insulator – this is called *dielectric breakdown*. Since *Q* is measured in coulombs (post 16.25) and *V* is measured in volts, we might expect the unit of capacitance to be C.V^{-1} (post 16.12). Be careful – *C* is the symbol I have chosen to represent capacitance but C is the standard SI abbreviation (post 16.12) for the coulomb; *V* is the symbol I have chosen to represent potential difference but V is the standard SI abbreviation for the volt (post 17.44). But the unit C.V^{-1} is given a special name – the *farad* (abbreviated to F), named after the pioneer investigator of electricity and magnetism Michael Faraday. In practice, 1 F would be a huge capacitance and the capacitance of most capacitors in measured in μF or pF (see post 16.12 for the meaning of these prefixes).

Now let’s suppose that *V* depends on time as shown in the (red) graph above. When *V* has its peak value, the capacitor is being charged. But, the value drops and the capacitor is capable of storing less charge (according to the definition of capacitance). So, it loses its charge – we say that the capacitor *discharges*. The effect of this discharge is to remove some of the detail of the figure above which means loss of some of the high frequency components (post 18.14) of the potential difference across the capacitor. The effect is shown in the (blue) graph at the end of this post. The capacitor has changed the shape of the wave representing the electrical signal.

This effect is analysed in more detail below.

Suppose that the capacitor discharges through a resistor (post 17.44) whose resistance is *R*. From the definition of capacitance, the potential difference across a capacitor of capacitance *C*, that stores a charge *Q* is given by *V = Q/C*. This potential difference causes a current *I* = – *dQ*/*dt* (the current differentiated with respect to time, post 17.4) to flow through the resistor; the minus sign arises because the charge is leaving the capacitor. According to the definition of resistance (post 17.44) *V* = *IR* or

*Q/C* = – *RdQ/dt*.

This equation can be rearranged to give

*dQ/dt* = –*Q*/(*RC*)

which is identical to equation 4 of post 18.15 with *k* = -1/(*RC*). This describes exponential decay and, according to post 18.15, has the solution

*Q* = *Q*_{0}*e*^{–t/(RC)}.

We define the *half-life* of this decay process as the time *t*_{1/2} when *Q* has decayed to *Q*_{0}/2 (post 16.6).

Putting the definition of half-life into the previous equation gives the result that

*Q*_{o}/2 = *Q*_{o}exp(-*t*_{1/2}/[*RC*]) or ½ = exp(-*t*_{1/2}/[*RC*]).

This can be written as

2 = exp(-*t*_{1/2}/[*RC*])

(see post 18.2) or as

*t*_{1/2} = *RC*log* _{e}*2

(see post 18.3).

Let’s suppose that we have an electrical circuit in which a potential difference, *V*, that depends on time, can be represented by the graph above. This potential difference exists between the points labelled P and P’. The circuit is shown diagrammatically below, using standard symbols to represent a capacitor and a resistor.

The capacitor is charged by the applied potential difference and so adds an additional potential difference

*V* = *Q*/*C* = (*Q*_{0}/*C*)*e*^{–t/(RC)} = *V*_{0}*e*^{–t/(RC)}.

The graph below shows how *V* depends on time, *t*.

Now suppose that the time period (post 18.10), *T*, of the oscillating applied potential difference, is much less than *t*_{1/2} for the capacitor. The effect of combining the applied potential difference with the potential difference created by the charged capacitor is to produce a time-dependent potential difference that is shown in the (blue) graph below.

The capacitor has changed the shape of the red graph to give the blue graph.

__Related posts__

17.47 How can birds sit on high voltage electrical cables…

17.45 Electrical energy

17.44 Amps, volts and ohms

17.24 Fields and vectors

16.25 Electrical charge