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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 m 2 , separated by 1.00 mm? (b) How much charge is stored in this capacitor if a voltage of 3.00 × 10 3 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 = ε 0 A d = ( 8.85 × 10 −12 F m ) 1.00 m 2 1.00 × 10 −3 m = 8.85 × 10 −9 F = 8.85 nF .

    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 = C V = ( 8.85 × 10 −9 F ) ( 3.00 × 10 3 V ) = 26.6 μ C .

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

A = C d ε 0 = ( 1.0 F ) ( 1.0 × 10 −3 m ) 8.85 × 10 −12 F / m = 1.1 × 10 8 m 2 .

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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Check Your Understanding The capacitance of a parallel-plate capacitor is 2.0 pF. If the area of each plate is 2.4 cm 2 , what is the plate separation?

1.1 × 10 −3 m

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Check Your Understanding Verify that σ / V and ε 0 / d have the same physical units.

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Spherical capacitor

A spherical capacitor is another set of conductors whose capacitance can be easily determined ( [link] ). It consists of two concentric conducting spherical shells of radii R 1 (inner shell) and R 2 (outer shell). The shells are given equal and opposite charges + Q and Q , respectively. From symmetry, the electrical field between the shells is directed radially outward. We can obtain the magnitude of the field by applying Gauss’s law over a spherical Gaussian surface of radius r concentric with the shells. The enclosed charge is + Q ; therefore we have

S E · n ^ d A = E ( 4 π r 2 ) = Q ε 0 .

Thus, the electrical field between the conductors is

E = 1 4 π ε 0 Q r 2 r ^ .

We substitute this E into [link] and integrate along a radial path between the shells:

V = R 1 R 2 E · d l = R 1 R 2 ( 1 4 π ε 0 Q r 2 r ^ ) · ( r ^ d r ) = Q 4 π ε 0 R 1 R 2 d r r 2 = Q 4 π ε 0 ( 1 R 1 1 R 2 ) .

In this equation, the potential difference between the plates is V = ( V 2 V 1 ) = V 1 V 2 . We substitute this result into [link] to find the capacitance of a spherical capacitor:

C = Q V = 4 π ε 0 R 1 R 2 R 2 R 1 .
The cross section of a spherical capacitor is shown in the form of two concentric circles. The radius of the smaller one is R subscript 1 and that of the larger one is R subscript 2. The smaller one has positive signs on it and the larger one has negative signs on it. Arrows radiate outwards from the inner circle to the outer circle. In between the two, is a third circle, with radius r, shown as a dotted line. This is labeled Gaussian surface.
A spherical capacitor consists of two concentric conducting spheres. Note that the charges on a conductor reside on its surface.

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 .

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

V = R 1 + E · d l = Q 4 π ε 0 R 1 + 1 r 2 r ^ · ( r ^ d r ) = Q 4 π ε 0 R 1 + d r r 2 = 1 4 π ε 0 Q R 1 .

The capacitance of an isolated sphere is therefore

C = Q V = Q 4 π ε 0 R 1 Q = 4 π ε 0 R 1 .

Significance

The same result can be obtained by taking the limit of [link] as R 2 . A single isolated sphere is therefore equivalent to a spherical capacitor whose outer shell has an infinitely large radius.

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Source:  OpenStax, University physics volume 2. OpenStax CNX. Oct 06, 2016 Download for free at http://cnx.org/content/col12074/1.3
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