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A normal electric dipole comprising $+q$ at height $h$ and $-q$ at height $h+d$, with $d\ll h$, induces the displayed leading surface charge density on a grounded plane. The image charge construction first gives ${\sigma }_q(a)=-qh/[2\pi (a^2+h^2{)}^{3/2}]$ for one point charge; subtraction at heights $h$ and $h+d$ gives $-d{\partial }_h{\sigma }_q$ to first order. The central negative region ends at $a=\sqrt{2}h$, outside which the induced surface charge density is positive. Integrating over a disk gives $Q(<a)=-qd{a}^2/(a^2+h^2{)}^{3/2}$ to this order, so the total induced electric charge is zero. At finite $d$, the exact sign-change radius obeys $a^2=h^{2/3}(h+d{)}^{2/3}[h^{2/3}+(h+d{)}^{2/3}]$.

For point electric charges at heights $h_j>0$ on one normal above an infinite grounded conducting plane, the method of images places opposite image charges at $-h_j$. At the point directly below them, tangential electric fields cancel and normal components add, giving the displayed one-sided field. The electric field inside the metal is zero instead. The boundary jump is supported by induced surface charge.

A real charge of height $z$ has a reflected image charge at $-z$. Its acceleration is $\ddot{z}=-\kappa /z^2$, with $\kappa =q^2/(16\pi {\epsilon }_{0}m)$. Released from rest at height $d$, its ideal-model arrival time isThe image follows the instantaneous reflected position. The induced-charge energy carries one-half the full real-image pair interaction factor, consistent with this force.

For a point electric charge $q$ in a homogeneous medium of permittivity ${ϵ}_1$ beside a plane interface with permittivity ${ϵ}_2$, its potential on the source side is reproduced by a mirrored image charge $q^{\prime }=q({ϵ}_1-{ϵ}_2)/({ϵ}_1+{ϵ}_2)$. The interface has no free surface charge; continuity of tangential electric field and normal electric displacement field fixes this coefficient.

For radii $a<b$ and a point charge $q$ at $r_0{\mathbf{e}}_z$, $a<r_0<b$, set $k=b/a$. The method of images gives the two families $(q{k}^n,r_0{k}^{2n}{\mathbf{e}}_z)$ and $(-qa{k}^n/r_0,a^2{k}^{2n}{\mathbf{e}}_z/r_0)$ for $n\in \mathbb{Z}$. The first family's $n=0$ term is the physical point charge. All other image charges lie outside the gap. Sum their Coulomb's law electric potentials with factor $1/(4\pi {ϵ}_0)$.

For a short proof, reflection of $(Q,z)$ in a grounded sphere of radius $R$ gives $(-RQ/z,R^2/z)$. On $r=a$ this pairs first-family index $n$ with second-family index $-n$; on $r=b$ it pairs $n$ with $1-n$. Each pair has zero boundary electric potential. The negative-index electric potential tails are $O(k^n)$ and the positive-index tails are $O(k^{-n})$, uniformly on the closed gap after excluding the physical singularity. These geometric series establish absolute convergence, boundary cancellation, and termwise differentiation away from the physical point charge. Uniqueness of Poisson equation identifies the solution. A spherical Gaussian surface just outside $a$ encloses the first family with $n\le -1$ and the second with $n\le 0$, so Gauss's law and the geometric series give $Q_a=-qa(b-r_0)/(r_0(b-a))$.

A real charge $q$ at $d>a$ outside a spherical conductor of radius $a$ has the usual grounded-sphere image charge $q^{\prime }=-aq/d$ at $x^{\prime }=a^2/d$. Adding a central image $q_0=-q^{\prime }$ makes the total image charge zero, representing an isolated neutral conductor. The off-centre pair has zero potential on $r=a$; the central image therefore sets the constant conductor potential to $q/(4\pi {\epsilon }_{0}d)$. Its induced electric dipole moment is $p_x=q^{\prime }{x}^{\prime }=-q{a}^3/d^2$. Keeping the applied central field ${\mathbf{E}}_0=-q\hat{\mathbf{x}}/(4\pi {\epsilon }_0{d}^2)$ fixed gives $\mathbf{p}=4\pi {\epsilon }_0{a}^3{\mathbf{E}}_0$. As $d\to \infty$, the external field becomes uniform near the sphere; its divergent constant potential can be removed without affecting the electric field.

For two grounded conducting half-planes meeting at angle $\pi /3$, place a point charge $q$ at radius $d$ on the bisector $\theta =\pi /6$. The method of images closes after six copies: positive copies at angles $\pi /6+2k\pi /3$ and negative copies at $-\pi /6+2k\pi /3$, for $k=0,1,2$. Only the first positive copy is real. Opposite-sign reflected pairs make the electric potential vanish on both faces; the uniqueness theorem for the electrostatic potential proves the construction. Summing the five image charge electric fields gives the electrostatic force $\mathbf{F}=-q^2(5/4-1/\sqrt{3}){\hat{\mathbf{e}}}_r/(4\pi {ϵ}_0{d}^2)$, towards the intersection. The coefficient follows from distances $d$, $\sqrt{3}d$, and $2d$ to the adjacent negative, nonadjacent positive, and opposite negative copies.

Integrate the surface charge density using the surface integral in spherical coordinates:Since $r_0>a$, the elementary integral isSubstitution yields the total induced electric chargeThe equality with the image charge also follows by applying Gauss's law to the exterior electric field: the physical and fictitious descriptions give the same electric flux through a sphere enclosing the conductor but excluding the physical point charge.

The image charge for a grounded conducting sphere is obtained by the Kelvin transform:It lies inside the electrical conductor, outside the region where it represents the physical electric field. The electric potential isAt the physical point charge, the electric potential has its usual Coulomb's law singularity. There is no physical singularity at the image charge: the interior expression is identically zero.

For $|\mathbf{r}|=a$, direct expansion of squared distances givesConsequently the two terms cancel exactly, proving the zero Dirichlet boundary condition on the sphere. At infinity each term is $O(r^{-1})$, and more explicitlyThus the Dirichlet boundary condition at infinity also holds. The candidate solves the Poisson equation with the one physical point charge; uniqueness of Poisson equation, with these Dirichlet boundary conditions, identifies it as the required electric potential.

Choose a spherical Gaussian surface of radius $s$ with $a<s<r_0$. Its physical enclosed electric charge is the total induced electric charge $Q_a$ on the inner sphere. The method of images produces the same electric field on that Gaussian surface, so Gauss's law identifies $Q_a$ with the sum of the enclosed image charges. These are the first family with $n\le -1$ and the second family with $n\le 0$:Summing these geometric series givesThe electric field in the empty central cavity vanishes by electrostatic shielding, so it contributes no additional inner-face surface charge density. The formula tends to $-q$ as $r_0\downarrow a$ and to zero as $r_0\uparrow b$, as expected when the point charge approaches one grounded boundary or the other.

Put $k=b/a>1$ and ${\mathbf{e}}_z=(0,0,1)$. The complete construction of image charges between concentric grounded spheres consists of two families, indexed by every integer:The term $(q_0,{\mathbf{r}}_0)$ is the physical point charge; every other term is an image charge. In the first family $n\le -1$ lies inside $r=a$ and $n\ge 1$ lies outside $r=b$. In the second family $n\le 0$ lies inside $r=a$ and $n\ge 1$ lies outside $r=b$. Thus no image charge lies in the physical gap.

To derive and verify these families, reflection of an axial point charge $(Q,z)$ in a grounded sphere of radius $R$ replaces it by the image charge $(-RQ/z,R^2/z)$. Reflection in $a$ sends $(q_n,{\mathbf{r}}_n)$ to $({\tilde{q}}_{-n},{\tilde{\mathbf{r}}}_{-n})$; reflection in $b$ sends it to $({\tilde{q}}_{1-n},{\tilde{\mathbf{r}}}_{1-n})$. Successive reflections therefore generate exactly the listed families. The method of images givesThe physical point charge itself is excluded when evaluating this formula. On $r=a$, pair the first-family term $n$ with the second-family term $-n$; their electric potentials cancel by the single-sphere distance identity. On $r=b$, pair $n$ with $1-n$ instead. These reindexings are justified by absolute convergence, as follows.

For $n\to -\infty$, both source positions tend to the origin while their electric charges have size $O(k^n)$. On the closed gap their distances are eventually at least $a/2$, so both electric potential terms are $O(k^n)$. For $n\to +\infty$, both distances are at least a positive constant times $k^{2n}$ and the electric charges are $O(k^n)$, giving $O(k^{-n})$ terms. Both tails are bounded by convergent geometric series. This proves absolute convergence and uniform convergence of the tails on the closed gap. On any compact set avoiding the physical point charge, the differentiated tails obey the same type of bounds: for example, the large positive tail of the electric field is $O(k^{-3n})$. Thus termwise differentiation is valid and the sum is a harmonic function except at the physical point charge, where it has the correct Coulomb's law singularity. The boundary cancellations and uniqueness of Poisson equation complete the construction.

The charge and its proposed image charge give the exterior electric potentialAt $z=0$ the terms cancel, so $\Phi =0$ everywhere on the plane and its tangential derivatives vanish. Thus the image construction satisfies the earthed-conductor boundary conditions. It has the correct point-charge singularity, is harmonic elsewhere in $z>0$, and decays at infinity; electrostatic uniqueness identifies it with the exterior solution.

At instantaneous height $z$, the image charge is a distance $2z$ below the real charge, giving force $m\ddot{z}=-K/z^2$ with $K=q^2/(16\pi {\epsilon }_0)$. Multiplying by $\dot{z}$ and using rest at $z=d$ givesThe induced-conductor potential energy is $-K/z$, not the full energy of a pair of independently movable charges. On the falling branch, integrate the reciprocal speed:The last integral is $\pi /2$ by $s={\sin }^2\theta$. Hence the fall time of a charge toward a grounded plane in the electrostatic Newtonian model isFor $q=0$ there is no force and the particle remains at rest.

As the real charge descends to height $z(t)$, its instantaneous image charge is at $-z(t)$. The force is thereforeThe image is not kept at its initial position. Multiplying by $\dot{z}$ and using $z(0)=d$, $\dot{z}(0)=0$ givesFor the falling charge above a grounded conducting plane, the arrival time isPut $z=d{\sin }^2\theta$; the integral becomes $2d^{3/2}{\int }_0^{\pi /2}{\sin }^2\theta d\theta =\pi d^{3/2}/2$. HenceThis is the finite arrival time in the specified ideal force model, for $q\ne 0$. The field-force derivation also avoids incorrectly using the full real-image pair interaction energy: the conductor's effective energy is $-q^2/(16\pi {\epsilon }_{0}z)$, with the factor one-half appropriate to induced charge.

The method of images places an image charge $-q$ at the reflected position below the plane. The candidate potential in the physical upper half-space isAt $z=0$ the two terms cancel, so $\Phi =0$ and $E_x=E_y=0$. Direct differentiation givesThus the upper-side field satisfies exactly the conductor boundary conditions from part (i). The fictitious pair does have a field below the plane, but that part of its solution is not used for the physical conductor.

In $z>0$, the potential has the correct real point source, the correct zero Dirichlet boundary condition, and the required decay at infinity. By Uniqueness of the Dirichlet problem, it is the physical potential there. This is why the artificial charge can represent the induced conductor response without being a second physical charge.

The method of images uses the real line charge at the origin and the opposite image charge at $(2d,0)$. LetSuperposing their logarithmic electrostatic potentials gives, in the physical region $x<d$,At $x=d$, $r_1=r_2$, so $\varphi =0$. Its tangential derivatives also vanish on that plane, giving $E_y=E_z=0$, exactly the electrostatic boundary conditions at a conductor. The image lies outside the physical domain, so it creates no extra physical source there. Together with the appropriate behaviour at infinity, uniqueness of Poisson equation identifies this field with the grounded-conductor solution.

The force on the real line charge is due to the induced conductor field, represented by the image charge; its own singular field must be omitted. The opposite image at distance $2d$ produces at the origin a field toward the conductor,Multiplying by the charge per unit length gives the attractive force per unit length:The attraction holds for either sign of $\mu$. The geometry assumes $d>0$, so the real line lies outside the conductor.

Differentiating the image charge potential givesAt the conducting plane, $E_x(d^{-})=\mu d/[\pi {\epsilon }_0(d^2+y^2)]$. With outward conductor normal $-{\mathbf{e}}_x$, the surface charge density is thereforeIt is independent of $z$ by translational symmetry. As a check, integrating across the plane gives ${\int }_{-\infty }^{\infty }\sigma (y,z)dy=-\mu$, the total induced charge per unit length, opposite to the real line charge.

Write $R=|\mathbf{p}|>a$. In SI units, the exterior electrostatic potential solveswith the point-charge singularity $q/(4\pi {\epsilon }_0|\mathbf{x}-\mathbf{p}|)$. The image charge for a grounded conducting sphere choiceslie inside the sphere. On $|\mathbf{x}|=a$, direct squaring gives $|\mathbf{x}-{\mathbf{p}}^{\prime }|=(a/R)|\mathbf{x}-\mathbf{p}|$. Thereforesatisfies the source equation, grounded boundary value and decay at infinity. The Uniqueness of the Dirichlet problem, applied to the harmonic difference with its singularity removed, identifies this with the physical exterior potential. Thus its gradient gives the required exterior electric field.

The charge feels the induced field, equivalently the field of the image charge, excluding its own singular field. Their separation is $(R^2-a^2)/R$, so Coulomb's law givesIts magnitude is $q^{2}aR/[4\pi {\epsilon }_0(R^2-a^2{)}^2]$ and its direction is toward the sphere's center, independent of the sign of $q$. No factor one-half belongs in this force; that factor occurs in induced electrostatic energy instead.

The method of images replaces a conductor's induced surface charge by suitable fictitious image charges located outside the physical region in which the electric field is sought. Choose them so that the combined electric potential satisfies the specified conductor boundary values, has the required behaviour at infinity, and has the correct physical sources. The uniqueness theorem for the electrostatic potential then identifies this constructed electric potential with the physical solution in that region.

For an infinite grounded conducting plane at $z=0$, a point charge $q$ at $(x,y,z)$ with $z>0$ has an image charge $-q$ at $(x,y,-z)$. Distances to the pair are equal on the plane, so their electric potentials cancel there. This construction determines the electric field above the plane; below it the actual conductor has constant electric potential and zero electric field, rather than the fictitious image's continuation.