Electrostatics and Magnetostatics

Image charge method

The image charge method solves conductor boundary-value problems by replacing each conducting surface with fictitious charges placed outside the region of interest, chosen so that the required boundary condition holds. For a point charge +Q at height h above an infinite grounded conducting plane, the image is -Q at depth h below the plane, so the potential vanishes on the plane and the field above is the superposition of the two charges' fields. The attractive force on the real charge is the Coulomb force toward its image, F=\frac{Q^2}{4\pi\varepsilon_0(2h)^2}=\frac{Q^2}{16\pi\varepsilon_0 h^2}, directed toward the plane. The electrostatic energy of the system, counting the induced charge once, is U=-\frac{Q^2}{16\pi\varepsilon_0 h}, which is half the energy of the image pair; using the full pair energy -Q^2/(8\pi\varepsilon_0 h) is a common error. For a grounded conducting sphere of radius a, a charge q at distance d>a from the centre is imaged by -qa/d located at distance a^2/d from the centre on the same radial line, and the sum of the images fixes the zero potential of the sphere and its interior. In a symmetric two-charge arrangement, the net force on each charge is the mutual repulsion minus the attraction to the two images; the image attraction grows as a/d and can reverse the net force when the sphere is large enough. The induced surface density under a point charge above a grounded plane is \sigma=-\frac{qd}{2\pi(\rho^2+d^2)^{3/2}}, so it is not a simple power of the distance from the foot of the perpendicular. A useful distinction is that the potential of the real charge satisfies Poisson's equation at its own location, whereas each image term is harmonic in the region where its own source is absent.

Bound currents and magnetization

Magnetization \mathbf M, the magnetic dipole moment per unit volume, is equivalent to bound currents: a volume current \mathbf J_b=\nabla\times\mathbf M inside the material and a surface current \mathbf K_b=\mathbf M\times\hat n on its boundary, where \hat n is the outward normal. When \mathbf M is uniform the volume current vanishes and only the surface current remains; its magnitude is M\sin\theta, with \theta the angle between \mathbf M and \hat n, so it is largest where the normal is perpendicular to \mathbf M. For a uniformly magnetized sphere this is at the equator, where the surface current equals M, and the total dipole moment is \mathbf m=\frac{4}{3}\pi R^3\mathbf M. The auxiliary field is \mathbf H=\mathbf B/\mu_0-\mathbf M, so when \mathbf B=\frac{2}{3}\mu_0\mathbf M inside, \mathbf H=-\frac{1}{3}\mathbf M, and the far field is that of this dipole. For a magnetization frozen in along the axis of an infinitely long cylinder and depending only on the distance from the axis, \nabla\cdot\mathbf M=0 and there are no free currents, so \mathbf H=0 and \mathbf B=\mu_0\mathbf M inside; the bound volume current -\frac{dM_z}{ds}\hat\phi is exactly what \nabla\times\mathbf B requires. For a finite cylinder with uniform magnetization along its axis, the flat end faces have normals parallel to \mathbf M and carry no bound current, while the curved side carries the surface current M\hat\phi.

Solenoid magnetic field

For an ideal long solenoid with n turns per unit length carrying current I, the field is uniform inside, B=\mu_0 nI, and essentially zero outside. The vector potential is azimuthal, \mathbf A=A_\phi(r)\hat\phi, and Stokes' theorem gives 2\pi r A_\phi=\Phi(r), where \Phi(r) is the magnetic flux enclosed by a circle of radius r. Inside, \Phi=\pi r^2 B, so A_\phi=Br/2 rises linearly; outside, the flux is fixed at \pi R^2 B, so A_\phi=BR^2/(2r) falls as 1/r. The potential therefore peaks at the solenoid surface and is non-zero outside even where \mathbf B=0. The electromagnetic field also carries momentum with density \mathbf g=\varepsilon_0\mathbf E\times\mathbf B, which equals \mathbf S/c^2; where E and B are perpendicular its magnitude is \varepsilon_0 EB. If a capacitor's plates are parallel to a solenoid's axis, the capacitor field is perpendicular to the axial field, so the momentum density is non-zero and must be found from the cross product. As a quick check, n=1000 m^{-1} and I=2 A give B=\mu_0 nI\approx 2.5\times10^{-3} T.

Gauss's law flux integration

Gauss's law states \oint \mathbf E\cdot d\mathbf A=Q_{\text{enc}}/\varepsilon_0. For a purely radial field whose magnitude depends only on r, the flux through a sphere of radius R is simply E(R)\cdot 4\pi R^2, because E is constant over the surface and parallel to the normal. For example, a field E=\alpha e^{-r^2}/r gives flux 4\pi\alpha R e^{-R^2} through a sphere of radius R, so the flux must be evaluated on the given surface rather than assumed. When the field depends only on the cylindrical radius, as \mathbf E=k\boldsymbol{\rho}/\rho^2, the normal component on a sphere of radius R is not uniform in general; here \mathbf E\cdot\hat r=k/r, so the flux is 4\pi kR. Equivalently, the field is that of a line charge along the z-axis with charge per unit length 2\pi\varepsilon_0 k, and the flux is this line charge's enclosed length 2R divided by \varepsilon_0 times the line density, giving the same 4\pi kR.

Bound polarization charge in dielectrics

Polarization \mathbf P, the dipole moment per unit volume, produces a bound volume charge \rho_b=-\nabla\cdot\mathbf P inside the material and a bound surface charge \sigma_b=\mathbf P\cdot\hat n on its boundary. A uniformly polarized body therefore has no volume bound charge, and its bound charge resides only on the surface with net charge zero. The dipole moment of a neutral distribution does not depend on the choice of origin, and for a uniformly polarized body it equals \mathbf p=\int\mathbf P\,dV, so translating the body leaves it unchanged. For a point charge q embedded in a linear dielectric of relative permittivity \varepsilon_r, \mathbf D depends only on the free charge, so E=\frac{q}{4\pi\varepsilon_0\varepsilon_r r^2} and P=\varepsilon_0(\varepsilon_r-1)E, which falls as 1/r^2. The total bound (screening) charge is -q\left(1-\frac{1}{\varepsilon_r}\right), whose magnitude is less than q for \varepsilon_r>1 and zero in vacuum.

Dielectric-filled spherical capacitor

When a spherical capacitor is partly filled with a dielectric, the free charge Q on the inner sphere fixes \mathbf D through Gauss's law, \oint\mathbf D\cdot d\mathbf A=Q, but the field distribution depends on how the dielectrics are arranged. If the interface between two dielectrics is a radial plane, such as a hemispherical shell filling half the gap, the interface normal is azimuthal and the radial field is tangential to it. Tangential \mathbf E is continuous across any interface, so E has the same value in both halves, while D differs by the permittivity ratio. Applying Gauss's law to each half gives (\varepsilon_0+\varepsilon_1)E\,2\pi r^2=Q, so the field is E=\frac{Q}{2\pi(\varepsilon_0+\varepsilon_1)r^2} radially everywhere between the shells. The factor 2\pi rather than 4\pi reflects that only a hemisphere is considered in each region.

Magnetic circuit with air gap

A magnetic circuit is analysed in analogy with an electric circuit using Ampère's circuital law, \oint\mathbf H\cdot d\mathbf l=NI for a coil of N turns carrying current I. Each part of the core contributes H_k l_k, and the magnetomotive force is shared between iron and air gap, NI=H_{\text{iron}}l_{\text{iron}}+H_g l_g. For a uniform cross-section the flux is continuous, so B is the same in the iron and the gap. Iron with a very large relative permeability needs only a tiny H, so its drop is negligible, and B\approx \mu_0 NI/l_g. For instance, a 1 cm gap with NI=1000 A gives B\approx 0.126 T. Keeping the iron drop negligible is valid when l_{\text{iron}}/\mu_r is small compared with l_g.

Laplace equation mean-value property

A potential that satisfies Laplace's equation in a charge-free region is harmonic, and harmonic functions obey the mean-value property: the value at the centre of a sphere lying entirely inside that region equals the average of the values over the sphere's surface. On a spherical shell the average is weighted by area, since equal solid angles carry equal weight. If the surface potential takes constant values on fractions of the area, the centre value is the area-weighted sum of those values. For example, if half of the surface is at 0 V and half at 6 V, the centre potential is 3 V. The same property implies harmonic functions have no interior maxima or minima.

Charged disc axial field

A uniformly charged disc of radius R and surface charge density \sigma produces on its axis, at distance d from the centre, the field E=\frac{\sigma}{2\varepsilon_0}\left(1-\frac{d}{\sqrt{d^2+R^2}}\right), directed away from the disc for positive \sigma. As R\to\infty or d\ll R, the bracket tends to 1 and the field approaches that of an infinite plane, \sigma/2\varepsilon_0. The fractional deviation from the plane value is therefore \frac{d}{\sqrt{d^2+R^2}}. If the deviation is a fraction f, then d=\frac{fR}{\sqrt{1-f^2}}; for a 5% deviation, d\approx 0.050R.

Capacitor energy at constant voltage

For a capacitor held at a fixed voltage V by a source, the charge is Q=CV, the stored energy is U=\frac{1}{2}CV^2, and the force on a plate is F=\frac{1}{2}V^2\left|\frac{dC}{dx}\right|, with dC/dx taken at constant V. For a parallel plate capacitor with gap x, C=\varepsilon_0A/x, which gives an attractive force \varepsilon_0AV^2/(2x^2). When the plates are pulled apart, the source must absorb charge and receives energy V^2|dC|, of which half becomes mechanical work and half is the decrease of the capacitor's stored energy. The energy bookkeeping is therefore V\,|\Delta Q|=W_{\text{mech}}+|\Delta U_C|, with W_{\text{mech}}=|\Delta U_C| at constant voltage.