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The method of images replaces electrical conductors by fictitious image charges outside the physical region. Choose the image charges so that the physical electric charges and their images produce the prescribed constant electric potential on every conducting boundary. The uniqueness theorem for the electrostatic potential then identifies this constructed electric potential with the physical solution in the charge-containing region. Several conducting planes can require several, or infinitely many, image charges; a single reflected image is sufficient for one infinite plane.

Put the plane at $z=0$ and the point charge at $(0,0,d)$. For $z>0$, use an image charge $-q$ at $(0,0,-d)$:This vanishes on the plane. An arbitrary constant can be added if a different reference electric potential is chosen. Excluding the point charge's own electric field, Coulomb's law givesThe separation from the image charge is $2d$, and the force is attractive for either sign of $q$. No imposed external electric field is included.

At electrostatic equilibrium the electric field in the metal is zero. The empty cavity also has zero electric field, by the uniqueness theorem for the electrostatic potential. The electric charge lies on the outer surface. For the isolated conducting sphere, spherical symmetry and Gauss's law giveTaking the electric potential to vanish at infinity and using $E_r=-d\varphi /dr$ givesThe electric potential is continuous at the surface, while the normal electric field jumps by $\sigma /{ϵ}_0$, with surface charge density $\sigma =Q/(4\pi R^2)$.

Electrical contact equalizes electric potential and preserves total electric charge. The usual isolated-sphere approximation assigns the two conducting spheres the capacitances $4\pi {ϵ}_{0}R$ and $8\pi {ϵ}_{0}R$. It then givesThese are accurate when the conducting spheres are far apart and joined by a thin wire. Literal touching changes the answer: each conducting sphere polarizes the other, so its electric potential is not its isolated value. The source does not specify an approximation which would remove this effect.

For completeness, the exact result for ideal externally tangent conducting spheres can be obtained by the method of images. Let their radii be $a,b$, their centre separation $D=a+b$, and their common electric potential $V$. Begin with central image charges $4\pi {ϵ}_{0}Va$ and $4\pi {ϵ}_{0}Vb$. Reflect each external image charge $q^{\prime }$ at centre-distance $s$ into the other conducting sphere of radius $a$: the new image charge is $-a{q}^{\prime }/s$ at centre-distance $a^2/s$ toward the original. Repeat, alternately, in the two conducting spheres. Pairing successive opposite-sign terms gives the convergent net electric chargesEach reflection cancels a nonconstant surface electric potential; the remaining central term fixes $V$. The total internal image charge equals the exterior electric flux divided by $1/{ϵ}_0$, so these sums give the physical net electric charges.

For $a=R,b=2R$, multiply both dimensionless sums by the common factor $4\pi {ϵ}_{0}V(2R/3)$. Their values follow by writing reciprocals as integrals and setting $t=u^3$:Dividing by $S_a+S_b$ enforces conservation of electric charge and yieldsThis concrete touching-sphere result disproves applying the isolated $1:2$ sharing rule to literal contact.

For the negative test electric charge, use the same change in the fixed point charge's electric potential:The external agent extracts energy, since the final Coulomb potential energy is lower. Again the path must avoid the fixed point charge, and the work done by the electric field is $-W_{-}$.

The fixed point charge produces electric potential $\varphi (r)=k_{e}q/r$, where $k_e=1/(4\pi {ϵ}_0)$. The external quasistatic work is the increase in Coulomb potential energy:The endpoints have radii $2d$ and $d$, irrespective of their opposite coordinate signs. One must use a path avoiding the singular point charge at the origin. The electric field is conservative on such paths, so the endpoint calculation suffices; a literal straight path through the origin is inadmissible. The work done by the electric field has the opposite sign.

The electric dipole moment is $\mathbf{p}=qa\hat{\mathbf{y}}$. Write $\theta$ for the angle measured from $+y$, with $\hat{\mathbf{\theta }}$ pointing toward increasing $\theta$. The exact electric potential isFor $r\gg a$, expand the reciprocals to first order in $a/r$. Their common $1/r$ terms cancel, giving the electric dipole approximationThe radial electric field follows from $E_r=-\partial \varphi /\partial r$, and the tangential electric field from $E_{\theta }=-(1/r)\partial \varphi /\partial \theta$. The omitted terms are of relative order $a^2/r^2$ away from zeros of a component. Multiplying by the test electric charge gives the force componentsThe electric field lines emerge from the positive electric charge and terminate on the negative electric charge. They are mirror-symmetric about the electric dipole axis and bow out across its equatorial plane. In the point-electric dipole limit their meridional curves obey $dr/(rd\theta )=E_r/E_{\theta }=2\cot \theta$, hence $r=C{\sin }^2\theta$.

Interpret the coordinates as a plane in ordinary three-dimensional electrostatics. Set $k_e=1/(4\pi {ϵ}_0)$, where ${ϵ}_0$ is the vacuum permittivity. The work required to bring the second point charge in quasistatically is its electric charge times the first charge's electric potential:The individual point-charge self-energies are excluded; this is the electrostatic potential energy of point charges relative to infinite separation.

By the superposition principle, for $\mathbf{r}=(x,y)$ and $\mathbf{d}=(d,0)$ the electric field isAt the first position both contributions point toward positive $x$, whereas at the second they oppose one another. ThusAt $(d/2,d)$ the two distances are $\sqrt{5}d/2$. The vertical electric field components cancel and the horizontal components add:Dimensional convention. The source's description of space as two-dimensional cannot literally mean two-dimensional electrostatics with the same inverse-square formula. In a genuinely two-dimensional Gauss's law, the circumference grows as $r$, giving an electric field proportional to $1/r$ and a logarithmic electric potential. For example, integrating $1/r$ from $d$ to infinity diverges, so the same finite reference potential energy at infinite separation does not exist. The displayed results use the conventional three-dimensional Coulomb's law restricted to the stated plane.

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 $k_e{Q}^2/(2r)$.

For this space-charge-filled parallel-plate capacitor, place the first plate at $x=0$ and the second at $x=d=x_o$, ignoring fringing field of a capacitor. With vacuum permittivity between the plates and fixed charge density $\rho$, Poisson's equation and the prescribed boundary electric potentials giveIntegrating twice and imposing both values yieldsThe electric field is minus the gradient of the electric potential, henceThus the electric potential is a parabola, concave downward for $\rho >0$ and upward for $\rho <0$. The signed electric field is a straight line. Its magnitude has a V-shaped corner if its zerolies inside the gap; otherwise its magnitude is a single straight segment. An interior zero requires $|V_1-V_2|<|\rho |d^2/(2{ϵ}_0)$. At that point the electric potential has its interior extremum. For $\rho =0$, these reduce to a linear electric potential and a uniform electric field.

In SI units, Maxwell's equations in vacuum, allowing charge density $\rho$ and electric current density $\mathbf{J}$, areThe source-free specialization sets $\rho =0$ and $\mathbf{J}=0$; it cannot describe the charged objects at their locations. Applying the divergence theorem to the first equation gives Gauss's law:For an isolated point charge $q_1$, spherical symmetry makes the electric field radial and constant in magnitude on a concentric sphere. Therefore $4\pi r^2{E}_r=q_1/{ϵ}_0$. Multiplying by a second point charge gives Coulomb's law,The force points away from the first charge for like signs and toward it for opposite signs.

For the spherical capacitor, a Gaussian sphere between the shells encloses $q$, so $E_r=q/(4\pi {ϵ}_0{r}^2)$ for $a<r<b$. Integrating the electric field inward from the outer shell yieldsAn arbitrary constant $V(b)$ remains unless a reference is specified. The capacitance is determined by the potential difference:For the subsequent charging operation, take the initial outer shell to be earthed, or interpret the stated $V$ as the inner-to-outer voltage with the outer reference zero. Isolating the inner shell fixes its electric charge at $q=CV$. Removing the outer shell leaves an isolated conducting sphere. Its electric field and electric potential, now with zero at infinity, areConsequentlyThese changes of electric potential are compatible with constant electric charge. Without the initial reference convention the absolute stated $V$ does not determine $q=CV$: for example, if the outer shell were initially isolated and neutral, the absolute inner electric potential would instead be $q/(4\pi {ϵ}_{0}a)$.

Finally take the connected spheres to be sufficiently far apart that the long wire's capacitance and their mutual influence can be neglected. Their common electric potential requiresso $q_a=q/3$ and $q_{2a}=2q/3$. The magnitudes of the surface electric fields areThis is the usual widely separated-sphere approximation. A wire enforces an equipotential, but alone does not make this ratio exact for arbitrary arrangements. As a concrete counterexample, if the second shell enclosed the first concentrically and they were connected, all net electric charge would reside on the exterior and the inner sphere's surface electric field would be zero.

Neglect the entrance and exit capacitor fringing fields and use the coaxial electric field between the cylindrical plates. Let $U=V(R_1)-V(R_2)$, with $R_2>R_1$. The radial electric field isFor an electron, of electric charge $-e$, to follow the circular path $R=(R_1+R_2)/2$, its inward centripetal force must be $m_e{v}_0^2/R$. ThereforeThe inner plate must be at higher electric potential: the electric field points outward and the electron's electric force inward. The electric force is perpendicular to the velocity, so no work is done and the speed remains $v_0$. The required voltage is independent of the opening angle $\varphi$, which only sets the transit time $R\varphi /v_0$. For a narrow gap, $\ln (R_2/R_1)\simeq (R_2-R_1)/R$. This is a nonrelativistic, negligible-capacitor fringing field answer; finite open cylindrical sections do not have an exactly radial electric field at their ends.

Take the grounded plane as $z=0$. The method of images replaces it, in $z>0$, by an image charge $-q$ at $z=-d$ and an image charge $+q$ at $z=-2d$. These make the electric potential zero on the plane; the uniqueness theorem for the electrostatic potential establishes the solution in the physical half-space.

At $z=3d/2$, both real point charges contribute upward electric fields, each of magnitude $kq/(d/2{)}^2$. The nearer image charge contributes downward and the farther one upward. HenceThis is upward for $q>0$; retaining signed $q$ automatically handles the opposite sign. No self-electric field is involved because the evaluation point is distinct from every point charge.

For image charges in a sixty-degree conducting wedge, use polar coordinates in a cross-section normal to the planes' intersection, putting the physical wedge at $0<\theta <\pi /3$ and the real point charge at $(r,\theta )=(d,\pi /6)$. Repeated reflection in the two grounded faces produces five image charges, all at radius $d$ and the same coordinate along the intersection:Equivalently, the positive copies are at $\theta =\pi /6+2k\pi /3$ and the negative copies at $\theta =-\pi /6+2k\pi /3$, for $k=0,1,2$. Only the original point charge lies inside the physical wedge.

For any observation point on either face, reflection pairs every positive point charge with an equidistant negative one. Their Coulomb potentials cancel, so both grounded faces have zero electric potential. The electric potential also tends to zero at infinity. The uniqueness theorem for the electrostatic potential then identifies this method of images construction with the physical solution.

Let $U>0$ be the drop in electric potential along the positive electric charge's path. By conservation of energy, $qU=m{v}^2/2$, so the nonrelativistic exit speed is $v=\sqrt{2qU/m}$. Apply a uniform magnetic field perpendicular to the incident velocity and to the desired orbital plane. The magnetic Lorentz force is perpendicular to the velocity, does no work, and supplies the centripetal force:For example, $\mathbf{v}=v\hat{\mathbf{x}}$ and $\mathbf{B}=B\hat{\mathbf{z}}$ give initial deflection along $-y$ for positive $q$. Reversing the magnetic field reverses the bending. A uniform electric field alone would give constant acceleration and a parabolic path, not uniform circular motion. The displayed result assumes $qU\ll m{c}^2$; the exact relativistic momentum relation would give $R=\sqrt{q^2{U}^2/c^2+2mqU}/(qB)$.

Take the grounded electrical conductor to occupy $z\le 0$ and approximate the negligible-radius charged sphere by a point charge $Q$ at ${\mathbf{r}}_{+}=D\hat{\mathbf{z}}$. Put a fictitious image charge $-Q$ at ${\mathbf{r}}_{-}=-D\hat{\mathbf{z}}$. The constructed electric potential is zero on $z=0$, has the correct point charge singularity, and vanishes at infinity. The uniqueness theorem for the electrostatic potential makes it the physical electric potential for $z>0$:Taking its negative gradient gives the requested electric field outside the small sphere:On the plane at cylindrical radius $r$, the tangential components cancel and the normal component isThe electrostatic boundary conditions at a conductor, with outward unit normal $+\hat{\mathbf{z}}$ and zero interior electric field, give the induced surface charge densityFinally integrate over annuli of area $2\pi rdr$:The image charge is a computational device, not a real charge below the plane. For a finite conducting sphere, its induced polarization would change the point-source approximation; the stated negligible radius is essential.

The ideal voltage-feedback Operational amplifier has zero input currents and, while stable negative feedback holds it in its linear operating range, equal input voltages: $i_{+}=i_{-}=0$ and $v_{+}=v_{-}$. The first rule follows from infinite input impedance; the second is the large-open-loop-gain limit of $v_o=A(v_{+}-v_{-})$. Equal input electric potential is a virtual short, not an electrically conducting connection.

For an inverting amplifier with the noninverting input grounded, $v_{-}=0$ and node-current balance gives $v_i/R_i=-v_o/R_f$, hence $v_o=-R_f{v}_i/R_i$. For a noninverting amplifier, the feedback divider gives $v_{-}=v_o{R}_g/(R_f+R_g)=v_i$, hence $v_o=(1+R_f/R_g)v_i$.

One should check output swing, input common-mode range, slew rate, Gain-bandwidth product and stability before using these limits in a real circuit. Saturation, positive feedback or operation beyond bandwidth invalidates the equal-voltage rule; finite input bias currents and offset voltage can also matter.^[ref]

Use an inverting operational amplifier. Ground the non-inverting input, connect the source to the inverting input through $R_{\mathrm{in}}$, and connect $R_f$ from the output back to that same input.

In linear operation the ideal operational amplifier draws no input electric current, and feedback holds the inverting input at the same electric potential as the grounded non-inverting input. This node is a virtual ground. Kirchhoff's current law there givesThe source supplies current $I_i=V_i/R_{\mathrm{in}}$, so its input impedance is $R_{\mathrm{in}}$. The required design is thereforeOnly these two resistors and the single ideal operational amplifier are needed in the signal circuit; its power supplies are implicit. The result assumes an input small enough to keep the output within its linear operating range.

The method of images replaces conductors by fictitious charges chosen to reproduce their boundary potentials in the physical domain. The uniqueness theorem for the electrostatic potential then identifies the resulting physical exterior field; the fictitious interior field is not the field inside the conductor.

For a source at $d\hat{\mathbf{z}}$, let $r^{\prime }=a^2/d$ and $q^{\prime }=-qa/d$. At a point on $r=a$, the distances $R$ and $R^{\prime }$ to source and image obey $R^{\prime }=(a/d)R$. Hence $q/R+q^{\prime }/R^{\prime }=0$ everywhere on the sphere. This proves the image charge for a grounded conducting sphere construction.

With real charges at $\pm d\hat{\mathbf{z}}$, superpose two images of charge $-qa/d$ at $\pm a^2\hat{\mathbf{z}}/d$. For the right-hand charge, take positive force away from the centre. Its electrostatic force isPutting $x=d/a$, the equal-charge balance around a grounded sphere requiresFor large $x$, the right side is approximately $2/x^3$, so $x\simeq 8$. This is an approximate equilibrium, not an exact cancellation at $x=8$.

To hold the sphere at electric potential $V$, add a central image $Q_0=4\pi {ϵ}_{0}aV$. The grounded pair already contributes zero potential on its surface. ThereforeAt $d=8a$ this is $F_0(8a)+qV/(64a)$, or approximately $qV/(64a)$ if the small grounded residual is neglected as in the preceding approximation. The charge on the sphere is $Q_0-2qa/d$; fixing its potential is different from independently fixing its net charge.

In a charge-free region of homogeneous permittivity, Gauss's law and $\mathbf{E}=-\nabla V$ give Laplace's equation, ${\nabla }^{2}V=0$. Prescribed surface electric potentials specify a Dirichlet problem. Its Uniqueness of the Dirichlet problem permits replacing conductors by fictitious charges outside the physical solution region: a harmonic candidate with the same boundary conditions is the physical solution. For an exterior problem its behaviour at infinity must also be prescribed. For a proof, subtract two candidate potentials to obtain $u$ with zero boundary values. Green's first identity gives $\int |\nabla u{|}^2{d}^{3}x=0$, so $u$ is constant and its boundary value fixes it to zero. Conditions on a normal electric field instead determine a potential only up to an additive constant when the usual compatibility condition holds.

A cylindrical Gaussian surface around an infinite line charge gives $E_r=\lambda /(2\pi {ϵ}_{0}r)$. The electric potential difference is thereforeAn isolated infinite line charge has no finite potential zero at infinity. For the opposite pair, however, the logarithmic growth terms cancel and one can choose $V\to 0$ there. Define $r_b^2=(x-b{)}^2+y^2$ and $r_c^2=(x-c{)}^2+y^2$. superposition givesOn $x^2+y^2=bc$,Thus the circle, and its cylindrical extension, is an equipotential. This is an Apollonius circle: a constant ratio of distances gives a circle rather than a constant distance from either charge.

For the cylinder above the plane, call its centre height $H>a$ and set $s=\sqrt{H^2-a^2}$. The cylinder-plane image charges are $+\lambda$ at height $s$ and $-\lambda$ at height $-s$. Their distances are equal on the grounded plane, while their distance ratio is $(H+s)/a$ on the circular surface. The positive fictitious line lies inside the metal, so the exterior is charge-free and Uniqueness of the Dirichlet problem applies. A Gaussian surface enclosing the physical cylinder confirms that its charge per unit length is indeed $\lambda$. Its voltage isThe physical field energy per unit length at fixed cylinder charge is $U^{\prime }=\lambda V_c/2$. Differentiate with respect to height, keeping $a$ and $\lambda$ fixed:The electrostatic attraction per unit length of a cylinder above a plane is downward for either sign of charge. Placing the fictitious line at the cylinder centre would give the wrong answer at this separation. The long-cylinder approximation neglects the ends.

For a smooth isolated conducting sphere, Gauss's law gives the surface electric field $E_s=Q/(4\pi {ϵ}_0{a}^2)$ and electric potential $V=Q/(4\pi {ϵ}_{0}a)=a{E}_s$. At the specified dielectric strength, $V_{\max }=3.0\times 10^3\mathrm{V}$. The capacitance of an isolated conducting sphere is $C=4\pi {ϵ}_{0}a$, so its capacitor energy isThis is the ideal breakdown-limited energy of a conducting sphere, taking air's permittivity approximately equal to vacuum permittivity and neglecting nearby objects and surface irregularities. The vacuum permittivity used numerically is $8.8542\times 10^{-12}\mathrm{F}{\mathrm{m}}^{-1}$.^[ref]