8.1 Capacitors and capacitance  (Page 3/9)

Capacitance and charge stored in a parallel-plate capacitor

(a) What is the capacitance of an empty parallel-plate capacitor with metal plates that each have an area of $1.00{\text{m}}^2$ , separated by 1.00 mm? (b) How much charge is stored in this capacitor if a voltage of $3.00\times {10}^3\text{V}$ is applied to it?

Strategy

Finding the capacitance C is a straightforward application of [link] . Once we find C , we can find the charge stored by using [link] .

Solution

1. Entering the given values into [link] yields
   
   $C={\epsilon }_0\frac{A}{d}=\left(8.85\times {10}^{\mathrm{-12}}\frac{\text{F}}{\text{m}}\right)\frac{1.00{\text{m}}^2}{1.00\times {10}^{\mathrm{-3}}\text{m}}=8.85\times {10}^{\mathrm{-9}}\text{F}=8.85\text{nF}\text{.}$
   
   This small capacitance value indicates how difficult it is to make a device with a large capacitance.
2. Inverting [link] and entering the known values into this equation gives
   
   $Q=CV=(8.85\times {10}^{\mathrm{-9}}\text{F})(3.00\times {10}^3\text{V})=26.6\mu \text{C}\text{.}$

Significance

This charge is only slightly greater than those found in typical static electricity applications. Since air breaks down (becomes conductive) at an electrical field strength of about 3.0 MV/m, no more charge can be stored on this capacitor by increasing the voltage.

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A 1-f parallel-plate capacitor

Suppose you wish to construct a parallel-plate capacitor with a capacitance of 1.0 F. What area must you use for each plate if the plates are separated by 1.0 mm?

Solution

Rearranging [link] , we obtain

Each square plate would have to be 10 km across. It used to be a common prank to ask a student to go to the laboratory stockroom and request a 1-F parallel-plate capacitor, until stockroom attendants got tired of the joke.

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Capacitance of an isolated sphere

Calculate the capacitance of a single isolated conducting sphere of radius $R_1$ and compare it with [link] in the limit as $R_2\to \infty$ .

Strategy

We assume that the charge on the sphere is Q , and so we follow the four steps outlined earlier. We also assume the other conductor to be a concentric hollow sphere of infinite radius.

Solution

On the outside of an isolated conducting sphere, the electrical field is given by [link] . The magnitude of the potential difference between the surface of an isolated sphere and infinity is

The capacitance of an isolated sphere is therefore

Significance

The same result can be obtained by taking the limit of [link] as $R_2\to \infty$ . A single isolated sphere is therefore equivalent to a spherical capacitor whose outer shell has an infinitely large radius.

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