19.5 Capacitors and dielectrics  (Page 3/13)

It can be shown that for a parallel plate capacitor there are only two factors ( $A$ and $d$ ) that affect its capacitance $C$ . The capacitance of a parallel plate capacitor in equation form is given by

$A$ is the area of one plate in square meters, and $d$ is the distance between the plates in meters. The constant ${\epsilon }_0$ is the permittivity of free space; its numerical value in SI units is ${\epsilon }_0=8.85\times {\text{10}}^{–\text{12}}\text{F/m}$ . The units of F/m are equivalent to ${\mathrm{C}}^2\text{/N}·{\text{m}}^2$ . The small numerical value of ${\epsilon }_0$ is related to the large size of the farad. A parallel plate capacitor must have a large area to have a capacitance approaching a farad. (Note that the above equation is valid when the parallel plates are separated by air or free space. When another material is placed between the plates, the equation is modified, as discussed below.)

Another interesting biological example dealing with electric potential is found in the cell’s plasma membrane. The membrane sets a cell off from its surroundings and also allows ions to selectively pass in and out of the cell. There is a potential difference across the membrane of about $\text{–70 mV}$ . This is due to the mainly negatively charged ions in the cell and the predominance of positively charged sodium ( ${\text{Na}}^{\text{+}}$ ) ions outside. Things change when a nerve cell is stimulated. ${\text{Na}}^{\text{+}}$ ions are allowed to pass through the membrane into the cell, producing a positive membrane potential—the nerve signal. The cell membrane is about 7 to 10 nm thick. An approximate value of the electric field across it is given by

This electric field is enough to cause a breakdown in air.

Dielectric

The previous example highlights the difficulty of storing a large amount of charge in capacitors. If $d$ is made smaller to produce a larger capacitance, then the maximum voltage must be reduced proportionally to avoid breakdown (since $E=V/d$ ). An important solution to this difficulty is to put an insulating material, called a dielectric    , between the plates of a capacitor and allow $d$ to be as small as possible. Not only does the smaller $d$ make the capacitance greater, but many insulators can withstand greater electric fields than air before breaking down.