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It can be shown that for a parallel plate capacitor there are only two factors ( and ) that affect its capacitance . The capacitance of a parallel plate capacitor in equation form is given by
is the area of one plate in square meters, and is the distance between the plates in meters. The constant is the permittivity of free space; its numerical value in SI units is . The units of F/m are equivalent to . The small numerical value of 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.)
(a) What is the capacitance of a parallel plate capacitor with metal plates, each of area , separated by 1.00 mm? (b) What charge is stored in this capacitor if a voltage of is applied to it?
Strategy
Finding the capacitance is a straightforward application of the equation . Once is found, the charge stored can be found using the equation .
Solution for (a)
Entering the given values into the equation for the capacitance of a parallel plate capacitor yields
Discussion for (a)
This small value for the capacitance indicates how difficult it is to make a device with a large capacitance. Special techniques help, such as using very large area thin foils placed close together.
Solution for (b)
The charge stored in any capacitor is given by the equation . Entering the known values into this equation gives
Discussion for (b)
This charge is only slightly greater than those found in typical static electricity. Since air breaks down at about , more charge cannot be stored on this capacitor by increasing the voltage.
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 . This is due to the mainly negatively charged ions in the cell and the predominance of positively charged sodium ( ) ions outside. Things change when a nerve cell is stimulated. 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.
The previous example highlights the difficulty of storing a large amount of charge in capacitors. If is made smaller to produce a larger capacitance, then the maximum voltage must be reduced proportionally to avoid breakdown (since ). An important solution to this difficulty is to put an insulating material, called a dielectric , between the plates of a capacitor and allow to be as small as possible. Not only does the smaller make the capacitance greater, but many insulators can withstand greater electric fields than air before breaking down.
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