Image charge

An image charge is a fictitious point charge placed outside the physical solution region so that its potential, combined with the real sources, enforces the required conductor boundary values. The Uniqueness of the Dirichlet problem then identifies the constructed potential with the physical one. The image need not describe the actual induced surface-charge distribution.

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 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 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 $Q_a=-qa(b-r_0)/(r_0(b-a))$.

For conducting sheets $y=\pm x$ at common zero potential and a charge in $y>|x|$, successive reflections give two negative images and one positive double-reflection image. On the bisector, the negative images lie at $(\pm y,0)$ and the positive image at $(0,-y)$. Summing their fields gives the displayed downward force. Image forces exclude the real charge's self-field; replacing the induced charge by images does not make the images physical particles.

The electrostatic bisector force is laterally anti-restoring, with the positive stiffness coefficient on the right. A spring of tension $T$ and length $\ell$ adds transverse restoring stiffness $T/\ell$ when its fixed support is directly above the charge. The displayed inequality therefore distinguishes actual transverse stability from the saddle of a model with only a vertical spring force. Axial spring stiffness alone does not specify the transverse verdict. At fixed finite support geometry and positive gravitational load, sufficiently small electric charge preserves lateral stability: the destabilizing coefficient tends to zero as $q^2$, while the tension term approaches $mg/\ell >0$. The vertical equilibrium must separately satisfy its own positive-stiffness condition.

For an attractive force $-\beta /y^2$ and an otherwise linear vertical spring, differentiation about the charged equilibrium $y_e$ reduces the spring stiffness by $2\beta /y_e^3$. If the uncharged height is $y_0$ and the shift is small, the fractional frequency change is $-\beta /(k{y}_0^3)$. Using only the force value, without its derivative, would miss this frequency shift.

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 a real charge $Q$ in medium 1, the potential in medium 2 can be represented by the source $Q_t=2{ϵ}_{2}Q/({ϵ}_1+{ϵ}_2)$ using medium 2's Coulomb prefactor $1/(4\pi {ϵ}_2)$. The same potential uses $2{ϵ}_{1}Q/({ϵ}_1+{ϵ}_2)$ if medium 1's prefactor is retained instead. These different effective source labels do not change the physical field. A decomposition $Q_t=Q+Q^{\prime }$ must state which prefactor it uses.

The image of a charge $q$ at height $z$ is $-q$ at height $-z$. Their separation is $2z$, giving force $-q^2/(16\pi {\epsilon }_0{z}^2)$. Starting at rest at height $d$, energy conservation gives ${\dot{z}}^2=q^2(1/z-1/d)/(8\pi {\epsilon }_{0}m)$. Integrating the reciprocal speed to the plane yields the stated finite time in the electrostatic Newtonian model. The induced-conductor potential energy is $-q^2/(16\pi {\epsilon }_{0}z)$; using the full image-pair energy would double the force incorrectly.

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 charge $q$ at $\mathbf{p}$, $R=|\mathbf{p}|>a$, outside a grounded sphere of radius $a$, the displayed image gives zero potential on the sphere and decay at infinity. On $|\mathbf{x}|=a$, one has $|\mathbf{x}-{\mathbf{p}}^{\prime }|=(a/R)|\mathbf{x}-\mathbf{p}|$, proving boundary cancellation. The induced force is inward, of magnitude $q^{2}aR/[4\pi {\epsilon }_0(R^2-a^2{)}^2]$, from Coulomb's law applied to the image field. The self-field is excluded.

Place equal positive point charges at distances $d$ on opposite sides of a grounded conducting sphere of radius $a$, and set $x=d/a>1$. Each physical charge has an image $-qa/d$ at distance $a^2/d$ from the centre on its own side. The repulsion from the other physical charge balances attraction to both images precisely when the displayed equation holds. At large $x$, the two sides behave as $1/(4x^2)$ and $2/x^3$, explaining the approximate root $x\simeq 8$. A constant nonzero sphere potential adds a central image and changes the force balance.

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.