4. Ch. 2, part 1

Lecture 4 - Method of images

Recall that a general solution of the boundary value problem can be obtained if we know the Green function G and that we can write G(x, x′) = 1/ |x − x′| + F (x, x′), where F (x, x′) can be regarded as a potential due to charges external to the volume of interest, that is, in practical applications, to fictitious charges inside conductors. In a two-dimensional problem (nothing depends on z), one can similarly write G(x, x′) = −2 ln |x − x′| + F (x, x′).

In the cases a sphere (in 3-d) or a cylinder (in 2-d) one can attribute F to a single “image”. In the limit of infinite radius, the sphere reduces to a plane, the simplest and most useful case. Following Jackson, we find the potential for a generic point charge q; the Green function is obtained by setting q = 1.

4.1. The conducting plane

Placing a charge q at the point (0, 0, z_{o}) and the “image charge” −q at the mirror image point (0, 0, −z_{o}) assures that the potential

\frac{q}{\sqrt{x^{2} + y^{2} + (z − z_{o})^{2}}} − \frac{q}{\sqrt{x^{2} + y^{2} + (z + z_{o})^{2}}}  (4.1)

vanishes on the xy plane (z = 0). To be definite, we take z_{o} > 0. The actual potential of a charge q in front of a conducting plane is given by eq. (4.1) for z > 0, vanishes for z < 0. Green’s function is obtained for q = 1.

The force acting on q is simply

F_{z} = −\frac{q^{2}}{4z_{o}^{2}}  (4.2)

as if q and −q were two real charges at a distance 2z_{o}. The work needed to bring q from infinity to a distance z_{o} from the plane is then

W = −\frac{q^{2}}{4z_{o}}  (4.3)

If q and −q were two real charges at a distance 2z_{o}, the mutual energy would be −q^{2}/2z_{o}, twice as large as W. The difference arises because the image charge is not present to begin with.

The electric field component E_{z} on the xy plane (i.e. at the surface of the conductor) is obtained by taking

−\frac{∂}{∂z} [\frac{q}{\sqrt{x^{2} + y^{2} + (z − z_{o})^{2}}} − \frac{q}{\sqrt{x^{2} + y^{2} + (z + z_{o})^{2}}} ]

and setting z = 0. It is

E_{z} = −\frac{2qz_{o}}{(x^{2} + y^{2} + z_{o}^{2})^{3/2}}

The parallel field components E_{x} and E_{y} vanish, as they should.

The “induced” surface charge density is given by 4πσ = E_{z}. The total induced surface charge is −q, as expected, since:

\frac{1}{4π} ^{∫∫} \frac{2z_{o}}{(x^{2} + y^{2} + z_{o}^{2})^{3/2}} dx dy = ∫_{0}^{∞} \frac{z_{o}}{(ρ^{2} + z_{o}^{2})^{3/2}} ρ dρ = 1

The normal field at the surface due to q is \frac{1}{2}E_{z}. Thus the force exerted by q on the surface charge density is given by the integral of \frac{1}{2}σE_{z}:

F_{z} = \frac{1}{4π} ^{∫∫} \frac{2q^{2}z_{o}^{2}}{(x^{2} + y^{2} + z_{o}^{2})^{3}} dx dy = ∫_{0}^{∞} \frac{q^{2}z_{o}^{2}}{(ρ^{2} + z_{o}^{2})^{3}} ρ dρ

The integration gives F_{z} = q^{2}/4z_{o}^{2}, confirming the result (4.2) of the simpler argument.

It may also be desirable to confirm that the work (4.3) correctly represents the energy balance in the field. Consider then the quantities:

• W_{q}(∞) : the self energy of the charge q when it is at infinity (this is infinite, but never mind: it will cancel out)

• W_{q}(z_{o}) : the self energy of the charge q in its final position (also infinite, but will cancel)

• W_{σ} : the self-energy of the surface charge

• W_{qσ} : the interaction energy between the charge q in its final position and the surface charge

The energy of the final configuration is W_{q}(z_{o})+W_{σ} +W_{qσ}, that of the initial configuration is W_{q}(∞). One can easily show that the energy balance is given by W_{qσ}, because W_{q}(z_{o}) + W_{σ} = W_{q}(∞). But W_{qσ} is one half the interaction energy of the charges q and −q at distance 2z_{o}, because it is obtained by integrating the corresponding energy density over half space. Thus W_{qσ} = −\frac{1}{2}(q^{2}/2z_{o}), in agreement with 4.3.

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To show that W_{q}(z_{o}) + W_{σ} = W_{q}(∞), note that all three quantities are obtained by integrating the same energy density, over all space for W_{q}(∞), over the half-space z > 0 for W_{q}(z_{o}), and (effectively) over the remaining half space for W_{σ}.

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4.2. The conducting sphere

In this case, a point charge and its image are not equal in magnitude. We put the charge q along the z axis at distance s from the center of the sphere. The image charge is then located on the z axis at

s′ = \frac{a^{2}}{s}

where a is the radius of the sphere, and has magnitude and sign given by

−q′ = q \frac{a}{s} = q \frac{s′}{a}

Strictly, the image method applies to the grounded sphere, and the result is extended to other cases by superposition of solutions.

4.2.1. Grounded sphere

It can be shown directly that the potential

Φ = \frac{q}{\sqrt{x^{2} + y^{2} + (z − s)^{2}}} + \frac{q′}{\sqrt{x^{2} + y^{2} + (z − s′)^{2}}}  (4.4)

vanishes on the sphere x^{2} +y^{2} +z^{2} = a^{2}. To be definite, we assume that s > a, but eq. (4.13) is valid for all s. One can also note that Φ is symmetric in q, s and q′, s′, so the solution of the “interior” problem (s < a) can be obtained from that of the “exterior” problem by interchanging q, s with q′, s′.

The force acting on q is simply

F_{z} = \frac{qq′}{(s − s′)^{2}} = −\frac{q^{2} \frac{a}{s}}{(s − \frac{a^{2}}{s} )^{2}} = −\frac{q^{2}as}{(s^{2} − a^{2})^{2}}  (4.5)

as if q and q′ were two real charges at a distance s − s′. This falls off as 1/s^{3} for large s, and for s near a reduces to

F_{z} = −\frac{q^{2}}{4 (s − a)^{2}}

in agreement with (4.2) for the plane. The work needed to bring q from infinity to a distance s from the center of the sphere is then

W = ∫_{∞}^{s} \frac{q^{2}az_{o}}{(r^{2} − a^{2})^{2}} dr = −\frac{1}{2} \frac{aq^{2}}{s^{2} − a^{2}}  (4.6)

If q and q′ were two real charges at a distance s−s′, the mutual energy would be qq′/ (s − s′), twice as large as W. The difference arises because the image charge is not present to begin with.

The electric field component E_{z} on the sphere is obtained by taking

−\frac{∂}{∂z} [\frac{q}{\sqrt{x^{2} + y^{2} + (z − s)^{2}}} + \frac{q′}{\sqrt{x^{2} + y^{2} + (z − s′)^{2}}} ]

which is

\frac{q(z − s)}{(\sqrt{x^{2} + y^{2} + (z − s)^{2}})^{3}} + \frac{q′(z − s′)}{(\sqrt{x^{2} + y^{2} + (z − s′)^{2}})^{3}}

and setting x^{2} + y^{2} + z^{2} = a^{2}. Keeping in mind that on the sphere

\frac{q}{\sqrt{x^{2} + y^{2} + (z − s)^{2}}} = −\frac{q′}{\sqrt{x^{2} + y^{2} + (z − s′)^{2}}}

one gets

E_{z} = \frac{q(z − s) − q′(q/q′)^{3}(z − s′)}{(a^{2} + s^{2} − 2sz)^{3/2}} =

q \frac{z [1 − (q/q′)^{2}] − s [1 − (q/q′)^{2}(s′/s)]}{(a^{2} + s^{2} − 2sz)^{3/2}}

Using

(\frac{q}{q′} )^{2} = (\frac{s}{a})^{2} = (\frac{a}{s′} )^{2} = \frac{s}{s′}

one finds finally

E_{z} = \frac{qz (1 − (s/a)^{2})}{(a^{2} + s^{2} − 2sz)^{3/2}}

Similarly:

E_{x} = \frac{qx (1 − (s/a)^{2})}{(a^{2} + s^{2} − 2sz)^{3/2}}

E_{y} = \frac{qy (1 − (s/a)^{2})}{(a^{2} + s^{2} − 2sz)^{3/2}}

We see that the vector E on the sphere is radially directed, as expected.

The induced surface charge density is then, for s > a,

σ = \frac{qa}{4π} \frac{1}{(a^{2} + s^{2} − 2sa cos γ)^{3/2}} \frac{a^{2} − s^{2}}{a^{2}}  (4.7)

where γ is the angle from the direction of the charge (see Figs. 2.3, 2.4). The total induced surface charge is q′ = −qa/s, as expected, since:

\frac{a}{4π} ∫_{0}^{π} \frac{2πa^{2} sin γ dγ}{(a^{2} + s^{2} − 2sa cos γ)^{3/2}} = \frac{a^{2}}{2s} (\frac{1}{s − a} − \frac{1}{s + a}) = \frac{a}{s} \frac{a^{2}}{s^{2} − a^{2}}

For s < a the sign of σ in (4.7) must be changed, because the normal is inwards. The total induced charge is −q, because in this case

\frac{a}{4π} ∫_{0}^{π} \frac{2πa^{2} sin γ dγ}{(a^{2} + s^{2} − 2sa cos γ)^{3/2}} = \frac{a^{2}}{2s} (\frac{1}{a − s} − \frac{1}{s + a}) = \frac{a^{2}}{a^{2} − s^{2}}

The field at the surface due to q has z component

\frac{q(z − s)}{(a^{2} + s^{2} − 2sz)^{3/2}}

Thus the force exerted by q on the surface charge density is given by

F_{z} = \frac{q^{2}}{2} [1 − (s/a)^{2}] ∫_{0}^{π} \frac{a (a cos γ − s)}{(a^{2} + s^{2} − 2sa cos γ)^{3}} a^{2} sin γ dγ

The integration gives F_{z} = q^{2}s/ (s^{2} − a^{2})^{2}, confirming the result (4.5) of the simpler argument.

It may also be desirable to confirm that the work (4.6) correctly represents the energy balance in the field, but we will skip it.

4.2.2. Insulated sphere

We can change the charge on the sphere, and its potential, by adding a fictitious charge at the origin. The potential

\frac{q}{\sqrt{x^{2} + (y − s)^{2} + z^{2}}} + \frac{q′}{\sqrt{x^{2} + (y − s′)^{2} + z^{2}}} + \frac{Q − q′}{\sqrt{x^{2} + y^{2} + z^{2}}}  (4.8)

describes a charge q at a distance s from the center of a sphere of radius a carrying a charge Q. If the sphere is insulated, Q stays constant as q moves about.

The force acting on q is simply

F_{z} = \frac{qq′}{(s − s′)^{2}} + \frac{q(Q − q′)}{s^{2}}

Using the previous result (4.5), this becomes

F_{z} = −\frac{q^{2}as}{(s^{2} − a^{2})^{2}} + \frac{qQ}{s^{2}} − \frac{q^{2}a}{s^{3}}  (4.9)

F_{z} is always attractive for an uncharged sphere, and is even attractive at small s − a when Q has the same sign as q. See Fig. 2.5. This explains why it is not so difficult to add charge to an already charged metal sphere, or a similar body. And why the charge stays on the body. However, the image force has nothing to do with the work function of metals. Jackson is wrong here.

4.2.3. Sphere at a fixed potential V

Just replace Q − q′ with V a in eq. (4.8) and the following. The force is now

F_{z} = −\frac{q^{2}ar}{(s^{2} − a^{2})^{2}} + \frac{qV a}{s^{2}}  (4.10)

It is still attractive for small s − a, even when qV is positive.

4.2.4. Sphere in a uniform field

The sphere could be either grounded or uncharged and insulated, since an applied uniform field does not induce a net charge. The trick is to generate the uniform field E_{z} by putting a charge Q at z = −R and a charge −Q at z = R, then letting Q and R go to infinity (Fig. 2.6). The potential of these charges

\frac{Q}{\sqrt{x^{2} + y^{2} + (z + R)^{2}}} − \frac{Q}{\sqrt{x^{2} + y^{2} + (z − R)^{2}}}  (4.11)

reduces to

\frac{Q}{R + z} − \frac{Q}{R − z} = −\frac{2Qz}{R^{2}} = −E_{0}z  (4.12)

where E_{0} = 2Q/R^{2} is the applied field. The image of Q is a charge −Qa/R at z = −a^{2}/R; the image of −Q is a charge Qa/R at z = a^{2}/R. In the limit of large R/a these two charges are a dipole of dipole moment

D = \frac{Qa}{R} \frac{2a^{2}}{R} = a^{3}E_{0}

and generate the dipolar potential

D\frac{z}{r^{3}} = a^{3}E_{0} \frac{z}{r^{3}} = a^{3}E_{0} \frac{cos θ}{r^{2}}

The total potential for r > a is then

Φ = −E_{0} (r − \frac{a^{3}}{r^{2}} ) cos θ

The induced surface charge density is

σ = −\frac{1}{4π} \frac{∂Φ}{∂r}|_{r=a} = \frac{3}{4π} E_{0} cos θ

It is worth remembering that a cos θ charge distribution on a spherical shell gives a uniform field inside the shell, and a dipole fileld outside. It is also notable that the polarizability of a conducting sphere of radius a is simply

\frac{D}{E_{0}} = a^{3}

4.2.5. Green’s function

We simply rewrite the point-charge solution,

\frac{q}{\sqrt{x^{2} + y^{2} + (z − s)^{2}}} − \frac{q′}{\sqrt{x^{2} + y^{2} + (z − s′)^{2}}}  (4.13)

putting q = 1 and placing the charge at the generic point r′ = (x′, y′, z′) instead of (0, 0, s). The image charge then goes at (a/ |r′|)^{2} (x′, y′, z′) and we obtain

G(r, r′) = \frac{1}{|r − r′|} − \frac{a/ |r′|}{|r−a^{2}r′/ |r′|^{2}|}

which can be rewritten

G(r, r′) = \frac{1}{|r − r′|} − \frac{a}{\sqrt{|r|^{2} |r′|^{2} + a^{4} − 2a^{2}r · r′}}

For boundary-value problems, we need ∂G/∂n′ on the surface of the sphere, |r′| = a. This is given by Jackson, which then goes on to use the result to compute the field of a sphere with hemispheres at different potentials, an uninteresting problem that we will skip. However, it may be useful as guidance for problem 2.3.

4.3. The conducting cylinder

The treatment is similar to that for the sphere and is left by Jackson as problem 2.7. Problem 2.11 actually gives part of the solution, and problem 2.4 is closely related. The following points are worth noting:

• The image charges are q and −q (unlike the case of the sphere)

• You cannot make the potential vanish both on the cylinder and at ρ = ∞. With the image charge alone, the potential on the cylinder is not zero (unlike the case of the sphere). You can add a charge to the cylinder (as in the case of the sphere), but then the potential will not go to zero at ρ = ∞.

• All the equipotentials are cylinders. Thus you can solve the problem 2.4 of two oppositely charged cylinders, external to each other or inside each other, in general not coaxial. One result of the problem is the capacitance

C = \frac{L}{2 cosh^{−1} |\frac{D^{2} − a^{2} − b^{2}}{2ab} |}

This is the result quoted in lecture 3, with the same symbols, since

cosh^{−1} x = ln (x + \sqrt{x^{2} − 1})