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The method of images replaces a conducting boundary by fictitious point charges placed outside the physical region being solved. Choose their positions and strengths so that the resulting electric potential has the required constant value on each conductor. The uniqueness theorem for the electrostatic potential then guarantees that this solution equals the physical solution in the allowed region, provided the other boundary conditions are also satisfied.

For a plane held at zero electric potential, a real charge $q$ a height $d$ above the plane has an image $-q$ the same distance below it. The pair's electric potential vanishes everywhere on the plane. Its tangential electric field consequently vanishes there, as required in electrostatic equilibrium. The image is a mathematical representation of the induced surface charge, not a physical charge placed inside the conductor. Forces on the real charge can be computed from the image field, but total physical electrostatic energy must account for the conductor's induced response rather than simply counting a full real-plus-image pair energy.

For the uniform space charge in a grounded spherical cavity, grounding sets the metal's electric potential to zero. With no external sources and zero electric potential at infinity, the exterior electric potential is also zero by the uniqueness theorem for the electrostatic potential. The original shell electric charge is no longer fixed: electric charge can flow to or from earth.

Inside the cavity, Gauss's law gives $\mathbf{E}=\rho r\hat{\mathbf{r}}/(3{ϵ}_0)$. Integrating the electric field inward from the zero boundary electric potential givesThe inner surface carries $-Q_{\rho }$, where $Q_{\rho }=4\pi \rho r_1^3/3$, and the outer surface carries zero net electric charge. Only the cavity contributes to the final electrostatic energy:The same result follows from $U=\frac{1}{2}\int \rho V{d}^{3}r$; the conductor's surface charge contributes zero because its electric potential is zero. This is the final stored electrostatic energy, not the change from the earlier central-charge configuration, whose ideal point charge has divergent self-energy.

Use the parallel-plate capacitor approximation, neglecting the fringing field of a capacitor. With plate area $A=\pi (0.10\mathrm{m}{)}^2$, the initial capacitance and opposite plate electric charges areAfter the electric battery is disconnected, the isolated plate electric charges stay fixed. Doubling the separation halves the capacitance. The quasistatic pulling work is the increase of electrostatic energy:The final electric potential difference is thereforeAn electron released at rest gains kinetic energy $e\Delta V_f$, where $e$ is the positive elementary charge. Neglecting radiation and the negligible perturbation caused by this one electron, conservation of energy givesThis is $0.0280c$, so the nonrelativistic kinetic energy is adequate. The circular plates have finite radii; the numerical results are leading parallel-plate capacitor values rather than exact finite-plate capacitances.

Assemble the spherically symmetric charge distribution by adding thin concentric shells, with the final charge density kept uniform. When the assembled radius is $s$, the enclosed electric charge and the next increment areThe electric potential of the already assembled material at that shell is $k_{e}q(s)/s$. Consequently its assembly work is $dU=k_{e}q(s)dq/s$, and the electrostatic energy of a uniformly charged solid sphere isThere is no extra factor of one half in this assembly integral: each pair of charge elements has already been counted exactly once, when the later element was brought in. A solid uniformly charged insulator has this value; a conducting sphere has its electric charge on the surface and instead has electrostatic energy $k_e{Q}^2/(2r)$.

Let ${\hat{\mathbf{e}}}_r$ point from the intersection to the real point charge. Reflection symmetry cancels the component of electrostatic force perpendicular to the wedge bisector. An image charge $q_j$ at angular separation $\Delta$ from the real point charge gives radial forceThe two adjacent negative image charges have $\Delta =\pm 60^{\circ }$, distance $d$, and combined radial force $-k{q}^2/d^2$. The two positive image charges have $\Delta =\pm 120^{\circ }$, distance $\sqrt{3}d$, and combined radial force $+k{q}^2/(\sqrt{3}{d}^2)$. The opposite negative image charge is at distance $2d$ and contributes $-k{q}^2/(4d^2)$. ThusThe electrostatic force is towards the intersection, independently of the sign of $q$. One sums the five image charge electric fields and multiplies by $q$; the factor $1/2$ used in induced-charge electrostatic energy must not be inserted into this direct force calculation.

From molecular dipoles to macroscopic fields. A dielectric contains charges which remain bound to its atoms or molecules. An applied electric field displaces positive and negative charge centres and can orient permanent electric dipoles. The electric polarization $\mathbf{P}$ is electric dipole moment per unit volume, in $\mathrm{C}{\mathrm{m}}^{-2}$. Its associated bound charge isat the boundary of the polarized material, where $\hat{\mathbf{n}}$ points outward. A uniform electric polarization has no volume bound charge, but generally has surface bound charge. The electric field remains the force per unit test charge; it includes the fields of both free and bound charges.

Introducing the electric displacement field $\mathbf{D}={ϵ}_0\mathbf{E}+\mathbf{P}$ packages the free-charge source separately:For a homogeneous linear isotropic dielectric,The electric susceptibility ${\chi }_e$ is dimensionless; the permittivity $ϵ$ has units $\mathrm{F}{\mathrm{m}}^{-1}$. This is a constitutive approximation. An anisotropic dielectric requires a permittivity tensor, while a nonlinear dielectric need not have electric polarization proportional to electric field. The field in these equations is the macroscopic field, not necessarily the local field at an individual molecule.

Interfaces. Let $\hat{\mathbf{n}}$ point from medium 1 into medium 2. Integrating Gauss's law over a thin pillbox, and the electrostatic curl equation around a thin loop, gives the electrostatic interface boundary conditionsThus normal electric displacement is continuous when there is no free surface charge, whereas tangential electric field is continuous even if the permittivity changes. Normal electric field generally jumps because the interface carries bound charge. The electric potential is continuous for an ordinary interface without an ideal dipole sheet. These conditions, together with Laplace's equation in each homogeneous charge-free region, determine the field.

Energy and electrical sources. Reversibly assembling free charge requires incremental work $\delta W=\int \mathbf{E}\cdot \delta \mathbf{D}dV$. For a linear dielectric whose response can be reversibly integrated from zero field, the electrostatic energy isThis includes the material polarization response, rather than simply using ${ϵ}_0{E}^2/2$ everywhere. In a capacitor, $W=Q^2/(2C)=C{V}^2/2$. Inserting a dielectric increases capacitance. At fixed charge the stored energy decreases. At fixed voltage the stored field energy increases, but the voltage source supplies $VdQ$; the appropriate mechanical energy including that source changes by $-V^{2}dC/2$. Both cases predict attraction of dielectric into the capacitor. A force calculation must state which electrical quantity is fixed. A lossy or hysteretic dielectric also dissipates energy, so the reversible quadratic expression does not describe its full history.

Shape and depolarization. Suppose the dielectric is surrounded by vacuum and the distant imposed electric field is ${\mathbf{E}}_0$. Bound surface charge produces a depolarization field. For a uniformly polarized ellipsoid, along a principal axis,Here $N_d$ is the depolarization factor of a dielectric ellipsoid. This self-consistent field explains why the same material responds differently in different shapes.

An infinite slab with electric field normal to its faces has $N_d=1$, giving $E_{\mathrm{in}}=E_0/{ϵ}_r$; its opposite surface bound charges reduce the internal field. For electric field parallel to the faces, $N_d=0$, giving $E_{\mathrm{in}}=E_0$. End effects are neglected.

A very long circular rod in an axial electric field has negligible depolarization away from its ends, so $E_{\mathrm{in}}\simeq E_0$. In a transverse electric field the infinite-cylinder solution has $N_d=1/2$, giving $E_{\mathrm{in}}=2E_0/({ϵ}_r+1)$. To see the latter directly, take the applied field along $x$. The interior and exterior electric potentials can be written ${\varphi }_{\mathrm{in}}=-Br\cos \theta$ and ${\varphi }_{\mathrm{out}}=-E_{0}r\cos \theta +A\cos \theta /r$. Continuity of potential and normal electric displacement at $r=a$ gives $B=2E_0/({ϵ}_r+1)$.

For a sphere of radius $a$, $N_d=1/3$ andFor example, write ${\varphi }_{\mathrm{in}}=-Br\cos \theta$ and ${\varphi }_{\mathrm{out}}=-E_{0}r\cos \theta +C\cos \theta /r^2$. Continuity of potential gives $B=E_0-C/a^3$, and continuity of normal electric displacement gives $ϵB={ϵ}_0(E_0+2C/a^3)$. Solving yields $B=3E_0/({ϵ}_r+2)$ and $C=a^3{E}_0({ϵ}_r-1)/({ϵ}_r+2)$. The $1/r^2$ potential is that of the electric dipole moment above. For ${ϵ}_r=1$ all these induced effects vanish; as ${ϵ}_r$ becomes very large the normal-field sphere and slab approach the corresponding conductor field exclusion. The ideal infinitely long rod in an exactly axial imposed field has a different, singular end-free limit.

For charge sharing between distant connected spheres, use the standard approximation that the spheres are sufficiently far apart for mutual electrostatic effects to be negligible, and that the wire adds negligible capacitance. The capacitance of an isolated conducting sphere is $C=4\pi {ϵ}_{0}r$. ThusThe initial charges are $Q_i=C_i{V}_i$. Connecting the spheres allows charge to flow until their potentials are equal. Charge conservation givesThe initial and final electrostatic energies are $U_i=\frac{1}{2}{C}_1{V}_1^2+\frac{1}{2}{C}_2{V}_2^2$ and $U_f=\frac{1}{2}(C_1+C_2)V_f^2$. Their difference becomes Joule heating in the resistive wire:Numerically,The resistance determines how fast this happens, not the final energy difference. Sufficient damping suppresses sustained charge-current oscillations; the calculation also neglects energy radiated away. At finite sphere separation the capacitance matrix would be needed, and the radii and initial potentials alone would not specify an exact correction.

The method of images solves certain electrostatics problems by replacing a conductor by fictitious electric charges outside the region where the solution is required. The replacement must satisfy the actual boundary values of the electric potential as well as the prescribed sources. The uniqueness of Poisson equation then establishes that the constructed electric potential equals the physical one in that region. The image locations are mathematical devices; they are not real charges inside the metal.

For a point charge $q$ at $(0,0,a)$ above an infinite grounded conducting plane $z=0$, put an image charge $-q$ at $(0,0,-a)$. In a cylindrical coordinate system with coordinates $(s,z)$ the potential in the physical region isThis vanishes at $z=0$, has the required singularity at the real charge, and tends to zero at infinity. With electric field $\mathbf{E}=-\nabla \varphi$, the induced surface charge density follows from the normal field immediately above the conductor:The force on $q$ is the field of the image evaluated at the real charge:The electrostatic energy relative to infinite separation is $U=\frac{1}{2}q{\varphi }_{\mathrm{image}}=-q^2/(16\pi {ϵ}_{0}a)$. The factor $1/2$ accounts for the fact that the induced charge changes as the real charge is assembled; treating the image as an independently fixed physical charge would give the wrong energy.

A second standard geometry is a grounded conducting sphere of radius $b$ with a charge $q$ a distance $d>b$ from its centre. Its exterior field is generated by an image $q^{\prime }=-qb/d$ at distance $b^2/d$ from the centre on the line towards the real charge. On the sphere the distances to the real and image charges are in the ratio $d/b$, making their potentials cancel. For an isolated neutral sphere, an additional central image $+qb/d$ makes the total induced charge zero and leaves the surface an equipotential. The construction is powerful for suitable symmetric boundaries, but arbitrary boundaries do not in general admit a finite image system.

The capacitor stores electrostatic energy, the inductor stores magnetic energy, and the resistor converts energy into heat. Their instantaneous values areThe sign convention for electric current does not change these expressions. With the initial voltage, $E_C(0)=C{V}_0^2/2=90\mu \mathrm{J}$ and $E_I(0)=P_R(0)=0$.

Two concentric conducting spherical shells of radii $a<b$ with charges $Q,-Q$ have an electric field only in the intervening vacuum: $E_r=Q/(4\pi {ϵ}_0{r}^2)$ by Gauss's law. Integrating gives a potential difference $Q(1/a-1/b)/(4\pi {ϵ}_0)$, and hence the displayed capacitance. The electrostatic energy is $Q^2/(2C)$, either by charging work or by integrating ${ϵ}_0|E{|}^2/2$ over the gap.

A spherical cavity of radius $a$ filled with fixed uniform charge density $\rho$ and surrounded by a grounded conductor has electric field $\rho r\hat{\mathbf{r}}/(3{ϵ}_0)$ inside and zero electric field outside the conductor when no external sources are present. Gauss's law supplies the interior electric field; integration from $V(a)=0$ gives the displayed electric potential. Its stored electrostatic energy is $2\pi {\rho }^2{a}^5/(45{ϵ}_0)$, obtained by integrating ${ϵ}_0|\mathbf{E}{|}^2/2$ over the cavity. Grounding permits the conductor's net electric charge to change.