3. Ch.1, part 3

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Lecture 3: Capacitance and capacitors

3.1. A single conductor

If an isolated conductor is raised to the potential V, the potential everywhere is Φ(x) = V ϕ(x), where ϕ(x) = 1 on the conductor and vanishes at infinity. The charge on the conductor, as given by Gauss’s theorem, is

4πQ = −V ∫_{S} \frac{∂ϕ}{∂n} da

where S is (conveniently) the surface of the conductor. We see that Q is proportional to V. By definition, the capacitance is

C = \frac{Q}{V} = −\frac{1}{4π} ∫_{S} \frac{∂ϕ}{∂n} da  (3.1)

The energy of the charged conductor is

W = \frac{1}{2}QV = \frac{1}{2} CV^{2} = \frac{1}{2} \frac{Q^{2}}{C}

Another expression for W is the usual

W = \frac{1}{8π} ∫_{V} (V ∇ϕ)^{2} d^{3}x

where V is the volume external to the conductor. Comparing these two expressions for W, we get another formula for C

C = \frac{1}{4π} ∫_{V}(∇ϕ)^{2}d^{3}x

(The earlier formula is recovered from this with an integration by parts). We see that C is always positive. This follows also from the fact that ϕ(x) has its maximum, ϕ = 1, on the surface S, so that ∂ϕ/∂n is negative on S. ——————————

In esu, the unit of C the cm. In the MKSA system, it is the farad:

farad = \frac{coul}{volt} = \frac{3 · 10^{9}statcoul}{(1/300) statvolt} = 9 · 10^{11}cm

For a sphere of radius a, the (esu) capacitance is C = a.

The farad is a huge capacitance. In common use is the picofarad:

pF = μμF = 0.9 cm

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For a simple body, C is of the order of the size of the body. For a sphere of radius a, we have the basic result C = a (in gaussian units, of course; in SI units C = 4πϵ_{0}a). If body A can be inscribed in body B, it can be shown that C_{A} < C_{B} (Jackson’s problem 1.20). Thus the radii of the inscribed and circumscribed sphere give simple bounds on C. For a convex body without sharp edges, corners, or points, C is almost the same as the capacitance of a sphere having the same surface area S, or C < \sqrt{S/4π}.

Examples of capacitance for simple bodies:

• cube, side a (numerically from Van Bladen) and 2/π = .6366198

0.665 a < C < 0.6678 a

• prolate ellipsoid, axes a, b, b, focal distance c = \sqrt{a^{2} − b^{2}}:

C = \frac{2c}{ln \frac{a + c}{a − c}} = \frac{c}{ln \frac{a + c}{b}}

– needle (a >> b):

C = \frac{a}{ln \frac{2a}{b}}

• oblate ellipsoid, axes a, a, b, focal distance c = \sqrt{a^{2} − b^{2}}:

C = \frac{2c}{arctan \frac{c}{b}}

– disk (a >> b):

C = \frac{2a}{π}

• thin toroidal ring, ring radius a, cross-section radius b : C = \frac{πa}{ln \frac{8a}{b}}

• spherical bowl, radius a, subtended angle 2θ :

C = \frac{a}{π} (θ + sin θ)

C/a vs. x = θ/π

Note that the results for the ellipsoids and for the bowl reduce to C = a in the appropriate limit ( b = a for the ellipsoids, θ = π for the bowl). Also, for small θ the bowl reduces to a disk of radius aθ and its capacitance is C = 2aθ/π as expected.

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3.2. Many conductors

For two bodies, the relation between charges and voltages is

Q_{1} = C_{11}V_{1} + C_{12}V_{2}  (3.2)
Q_{2} = C_{21}V_{1} + C_{22}V_{2}

with C_{21} = C_{12}. It can be shown that C_{11} and C_{22} are positive, C_{12} is negative, and C_{11} ≥ |C_{12}| as well as C_{22} ≥ |C_{12}|.

To obtain these results, start from the two basic configurations where V_{1} = 1, V_{2} = 0 and V_{1} = 0, V_{2} = 1. Call the corresponding potentials ϕ_{1}(x) and ϕ_{2}(x). In the general configuration the potential is Φ(x) = V_{1}ϕ_{1}(x) + V_{2}ϕ_{2}(x) and Gauss’s theorem gives

4πQ_{1} = −V_{1} ∫_{S_{1}} \frac{∂ϕ_{1}}{∂n} da − V_{2} ∫_{S_{1}} \frac{∂ϕ_{2}}{∂n} da

4πQ_{2} = −V_{1} ∫_{S_{2}} \frac{∂ϕ_{1}}{∂n} da − V_{2} ∫_{S_{2}} \frac{∂ϕ_{2}}{∂n} da

This argument clearly can be generalized to any number of conductors.

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Homework Assignment

Analogously to the one-conductor case, give explicit expressions for C_{ij} and use energy arguments, or flux arguments, to show that C_{ii} is positive, C_{ij} negative for i ≠ j, and ∑_{j} C_{ij} is non-negative. Use Green’s theorem to show that C_{ij} = C_{ji}.

—————————— 3.2.1. Capacitors

Of practical interest is the configuration with zero net charge, Q_{1} = Q and Q_{2} = −Q.

• In an ideal capacitor C_{12} = −C_{11} (so that no flux escapes): then eq. (3.2) gives Q = C_{11}ΔV, where ΔV = V_{1} − V_{2} (note that C_{11} is not the same as the capacitance of the isolated conductor 1).

• In the case of a symmetrical capacitor, V_{1} = ΔV/2 and V_{2} = −ΔV/2.

• Quite generally we can solve for V_{1} and V_{2} to obtain ΔV = Q/C, with the internal capacitance now given by

C = \frac{C_{11}C_{22} − C_{12}C_{21}}{C_{11} + C_{22} + C_{12} + C_{21}}  (3.3)

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Homework Assignment

Obtain eq. (3.3) and its simpler forms that apply when:

• The two conductors are equal

• Conductor 2 surrounds conductor 1

A spherical capacitor consists of a metal sphere of radius a and a metal shell of inner radius b_{1}, outer radius b_{2}. Find C and all the C_{ij} elements for this capacitor. When the capacitor is charged (Q on the inner sphere, −Q on the outer spherical shell), what are the potentials V_{1} and V_{2}? How is the charge distributed on the outer shell (how much is on the inner surface, how much on the outer surface?

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Here is a list of C for various capacitors.

• Two closely spaced parallel plates of area A, separation d (A >> d^{2}):

C = \frac{A}{4πd}

• Two concentric cylinders, radii a and b with a > b, length L >> (a − b):

C = \frac{L}{2 ln \frac{a}{b}}

• Two parallel cylinders, both of radius a, separation D > 2a, length L >> D :

C = \frac{L}{4 ln (\frac{D}{2a} + \sqrt{(\frac{D}{2a})^{2} − 1})}

– for D >> a, this reduces to

C = \frac{L}{4 ln \frac{D}{a}}

• Any two parallel cylinders, radii a, b, separation D, length L :

C = \frac{L}{2 ln (x + \sqrt{x^{2} − 1})}

where

x = |\frac{D^{2} − a^{2} − b^{2}}{2ab} |

This formula applies when the cylinders are external to each other (D > a + b) and L >> D, as well as when they are inside each other (D < |a − b|) and L >> |a − b| . The case of two intersecting cylinders is different, since they must be at the same potential.

• Two concentric spheres, radii a and b, with a > b :

C = \frac{1}{\frac{1}{b} − \frac{1}{a}}

3.2.2. Forces on charged conductors

Forces can be obtained simply by differentiating at constant Q_{i} the energy formula

W = \frac{1}{2} ∑_{i} V_{i}Q_{i} = \frac{1}{2} ∑_{ij} (ℂ^{−1})_{ij} Q_{i}Q_{j}

where ℂ^{−1} is the inverse of the capacitance matrix. The force in the general direction of the parameter ξ is

F_{ξ} = − (\frac{∂W}{∂ξ} )_{Q} = −\frac{1}{2} ∑_{ij} Q_{i}Q_{j} \frac{∂}{∂ξ} (ℂ^{−1})_{ij}

• For a single conductor

W = \frac{Q^{2}}{2C}

and the force in the general direction of the parameter ξ is

F_{ξ} = − (\frac{∂W}{∂ξ} )_{Q} = −\frac{Q^{2}}{2} \frac{∂(1/C)}{∂ξ} = \frac{Q^{2}}{2C^{2}} \frac{∂C}{∂ξ}  (3.4)

For example, a charged sphere of radius r experiences a radial force

F_{r} = −\frac{∂W}{∂r} = \frac{Q^{2}}{2r^{2}}

As expected, this force pushes outward, expanding the sphere.

• For a capacitor the force between the plates is also given by eq. (3.4) where C is now the internal capacitance. For example, the mutual force between the plates of a thin capacitor (C = A/4πx) is

F_{x} = −Q^{2} \frac{∂}{∂x} (\frac{1}{2} \frac{4πx}{A}) = −\frac{2πQ^{2}}{A}

As expected, this force is attractive and its magnitude per unit area is 2πσ^{2}, where σ = Q/A is the surface charge density.

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If we want to obtain the forces by differentiating at constant V, we must use the formula (with the seemingly “wrong” sign)

F_{x} = (\frac{∂W}{∂x} )_{V} = \frac{∂}{∂x} (\frac{1}{2}CV^{2}) = \frac{V^{2}}{2} \frac{∂C}{∂x}  (3.5)

The reason for the ”wrong” sign is that the virtual work at constant V is δW − V δQ, where the extra term is the work (typically done by batteries) to restore the charge. Since δW = \frac{1}{2}V δQ, we find that δW − V δQ = −δW.