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Use the standard infinite-plane boundary condition: the sheet is an equipotential, with no imposed uniform background electric field and the electric potential tending to zero far from the localized charges. The images of $+Q$ at $z=d$ and $-2Q$ at $z=2d$ are $-Q$ at $z=-d$ and $+2Q$ at $z=-2d$.

At the point $z=0^{+}$ directly below the real charges, horizontal components vanish. The first real-image pair contributes $-2k_{e}Q/d^2$ along $z$, while the second contributes $+2k_e(2Q)/(2d{)}^2=+k_{e}Q/d^2$. ThereforeFor $Q>0$, the electric field points downwards into the sheet. The expression is the one-sided vacuum limit; the electric field inside the conducting material is zero. A prescribed uniform background field would be an additional boundary condition and would change this result.

Let $x\ge 0$ denote the radial distance from the axis in the plane, and write $r_h=(x^2+h^2{)}^{1/2}$. A charge $q$ at height $h$ and its image $-q$ give a combined normal electric field $-2k_{e}qh/r_h^3$ at $z=0^{+}$; the tangential components cancel. Superposing the two pairs yieldsThe conductor boundary condition gives $\sigma ={ϵ}_0{E}_z$, soTo integrate the induced surface charge, use concentric annuli of area $2\pi xdx$. Since ${\int }_0^{\infty }x(x^2+h^2{)}^{-3/2}dx=1/h$,The local surface charge density is negative near the origin but positive farther out; integrating only the negative patch would miss the total. The total real charge is $-Q$, so the induced $+Q$ also agrees with the sum of the two image charges. An isolated finite plate constrained to zero net charge would require a different global boundary-value problem; the formula is the infinite-plane solution just specified.

Set the normal electric field to zero and cancel its nonzero overall factor:Taking the power $2/3$ gives $4^{2/3}(x^2+d^2)=x^2+4d^2$, henceThe numerical value is independent of the nonzero magnitude of $Q$. Beyond this radius the signed normal electric field is positive. Thus its magnitude first falls to zero, rises again and eventually decays; it is not globally monotone.

For $x\gg d$, expanding each denominator givesThis is a dipole rather than a monopole tail. The real-plus-image charge distribution has zero total charge and dipole moment $p_z=-6Qd$; in its equatorial plane a dipole produces $E_z=-k_e{p}_z/x^3$, confirming the coefficient and sign.

Spherical symmetry and the vacuum Gauss's law imply $\mathbf{E}=A\hat{\mathbf{r}}/r^2$ in the gap, where $A$ is constant. Integrating the electric field between the two equipotentials givesConsequentlyThe inner charge is $Q=4\pi {ϵ}_{0}A$, which reproduces the spherical capacitor capacitance $4\pi {ϵ}_{0}ab/(b-a)$. The radial direction is outward for positive $V_1$ and inward for negative $V_1$; the formula represents the signed field, not just its magnitude.

The electric field magnitude falls as $1/r^2$ throughout the vacuum gap, so its maximum occurs just outside the inner sphere:ThusInside the metal the electric field is zero. The one-sided gap value is the relevant value for dielectric breakdown; the ideal zero-thickness conducting boundary itself should not be assigned a single field value across its surface-charge discontinuity.

Inserting an independently biased conducting sphere makes two vacuum gaps, each with its own constant radial electric flux. Let $A_1$ and $A_2$ be the coefficients of $1/r^2$. The two potential drops areThereforeThe middle conductor can carry net electric charge, so $A_1$ and $A_2$ need not be equal. Its charge is $4\pi {ϵ}_0(A_2-A_1)$, consistent with the jump in the normal electric field. For $0<V_2<V_1$ both fields point outwards; outside that interval one of the gap fields reverses direction.

The integral Gauss law in vacuum is ${\oint }_{\partial V}\mathbf{E}\cdot d\mathbf{S}=Q_{\mathrm{enc}}/{ϵ}_0$. A single isolated point charge has spherical symmetry, so its electric field is radial and constant in magnitude on a centred sphere. The flux is $4\pi r^2{E}_r$, giving Coulomb's lawIn electrostatic equilibrium, a conductor is an equipotential: a tangential electric field would move its mobile charges. The electric field inside the conducting material is zero. A thin pillbox straddling its surface therefore gives $E_{n}A=\sigma A/{ϵ}_0$, henceHere $\sigma$ is the charge density on the particular face and $\hat{\mathbf{n}}$ points into vacuum. For an isolated thin plate with total charge divided equally between two faces, each face instead has half that total areal density. The local boundary result must not be confused with the $\sigma /(2{ϵ}_0)$ field of an isolated nonconducting infinite charged sheet.

Introduce cylinder length $\ell$ and line charge density $\lambda =Q/\ell$; only charge per length matters in the long-cylinder approximation. A coaxial cylindrical Gaussian surface gives $E_{r}2\pi r\ell =Q/{ϵ}_0$. Choosing the outer cylinder's electric potential to be zero,The capacitance per unit length follows from $C=Q/[V(a)-V(b)]$:If $Q$ is intended as a line charge rather than a finite total charge, replace $\lambda$ by that $Q$. Neglecting end effects, equal opposite cylinder charges make the exterior electric field vanish; the potential difference between the conductors remains the quantity defining their capacitance.

Define $\mathbf{d}$ to point from the negative charge to the positive charge. The electric dipole moment is $\mathbf{p}=q\mathbf{d}$. In a uniform electric field, the two forces cancel, so the net force is zero. They nevertheless form a couple with torqueFor a positive angular displacement increasing $\theta$, the restoring torque component is $-pE\sin \theta$. Thus $dU/d\theta =pE\sin \theta$ andwith the additive constant chosen to make $U=0$ at $\theta =\pi /2$. Alignment is a stable minimum and antialignment an unstable maximum. The zero-force result uses a uniform field; a field gradient can exert a net force on a dipole.

The Lorentz force is $\mathbf{F}=q(\mathbf{E}+\mathbf{v}\times \mathbf{B})$. With no electric field, the exact nonrelativistic component equations for the given quadrupole areThe magnetic force does no work, so the total speed stays constant. In the paraxial approximation, $\dot{z}\simeq v$ and its small change is second order in the transverse motion. For $q>0$, put $\kappa =\sqrt{qA/(mv)}$. Initial travel parallel to the $z$ axis means zero entrance slopes, yieldingThus $x$ is focused while $y$ is defocused. The first focus in the focusing plane occurs atFor negative $q$, interchange the focusing and defocusing planes and use $|q|$ in $\kappa$. A single quadrupole magnet gives a line focus of a general beam, not simultaneous focusing of both transverse coordinates onto one point. The magnet must extend at least to the computed focus, and the slopes must remain small. For a relativistic beam, the corresponding paraxial factor is ${\kappa }^2=|q|A/p$, with $p$ the longitudinal relativistic momentum.

Use a cross-section normal to the sheets' line of intersection: the positive sheet is $x=0$, and the negative sheet is $y=0$. A Gaussian pillbox gives a single infinite sheet's electric field magnitude $\sigma /(2{ϵ}_0)$ on either side. Superposition consequently gives, away from the sheets,The electric field points away from the positively charged sheet and towards the negatively charged sheet, bisecting each right angle. The electric field lines are straight within each quadrant: in the first quadrant, for example, $dy/dx=-1$, so $x+y$ is constant along a line. They start on $x=0$ and finish on $y=0$. At a sheet, the normal component has the usual surface-charge jump, so the quadrant formula specifies the appropriate one-sided limits rather than a unique electric field on the charged surface.

Place eight identical cubes around the charged corner to make one larger cube, with the point charge at its centre. Each large face has electric flux $q/(6{ϵ}_0)$ and consists of four equivalent small faces. Each of the original cube's three faces opposite the charged corner therefore hasOn each of its three faces containing the corner, the electric field of the point charge obeying Coulomb's law is tangent to the face at every regular point; its outward-normal component is zero. Defining the face integral by omitting the singular corner givesA boundary point charge does not satisfy the ordinary nonsingular hypotheses for simply counting it as an interior charge in Gauss's law. The $1/8$ result expresses the octant solid angle and the specified improper surface integrals. Moving a charge towards the corner from inside a cube before integrating is a different limiting operation; its total closed-surface electric flux stays $q/{ϵ}_0$ until it reaches the boundary.

In source-free vacuum, Maxwell's equations readTake the curl of the third equation and use the curl of the curl identity $\nabla \times (\nabla \times \mathbf{E})=\nabla (\nabla \cdot \mathbf{E})-{\nabla }^2\mathbf{E}$. ThenThe prescribed plane wave travels in the $+z$ direction. Its polarization is along $x$, and Faraday's law requires $B_y=(k/\omega )E_x$ with $\omega =ck$. The wave's magnetic flux density is thereforeThe direction is also fixed by $\mathbf{E}\times \mathbf{B}$ pointing along the propagation direction. A superposed static uniform magnetic field is not fixed by the oscillatory electric field; the expression above is the magnetic field belonging to the wave itself.

Define $V=\varphi (r_2)-\varphi (r_1)$ and $\eta =\ln (r_2/r_1)>0$. Cylindrical symmetry and Gauss's law imply a radial electric field proportional to $1/r$. Integrating $\mathbf{E}=-\nabla \varphi$ fixes the sign:If the inner cylinder has linear charge density ${\lambda }_q$, a coaxial Gaussian surface gives $2\pi r{E}_r={\lambda }_q/{ϵ}_0$, so ${\lambda }_q=-2\pi {ϵ}_{0}V/\eta$. Capacitance is positive regardless of the chosen terminal polarity:End effects are neglected for the long cylinders, as required by the cylindrical approximation.

The electromagnetic energy density in vacuum is $u_E={ϵ}_0{E}^2/2$. For a length ${\ell }_z$ of the coaxial capacitor, integrate over annular shells:Since $C=C^{\prime }{\ell }_z=2\pi {ϵ}_0{\ell }_z/\eta$, this becomesThe integral is over the inter-cylinder electric field, and gives the stored energy, independently of which electrode is assigned zero electric potential.

With $V=+20\mathrm{V}$ in the outer-minus-inner convention, $\eta =\ln 1.5=0.405465$ and the radial electric field at the orbit is $E_r=-986.52\mathrm{V}{\mathrm{m}}^{-1}$. The signed roots areUsing the proton mass and charge givesThe speeds are therefore $48.8$ and $96.7\mathrm{km}{\mathrm{s}}^{-1}$, in opposite senses of rotation. Both are nonrelativistic. The slower counterclockwise orbit has an outward magnetic force; the faster clockwise orbit has an inward magnetic force. Reversing $\mathbf{B}$ interchanges the senses, while reversing the electrode polarity changes the quadratic and may eliminate real orbits; its sign must therefore be specified.

The central point charge and spherical boundaries give a spherically symmetric charge distribution. A conductor in electrostatic equilibrium has zero electric field within its material. Consequently a spherical Gaussian surface inside the metal encloses zero net electric charge, so the inner surface carries $-q$. The shell's prescribed total electric charge then leaves $-2q$ on its outer surface. Gauss's law givesThe electric field is singular at the ideal central point charge. At each charged surface the normal component has distinct one-sided limits, as required by the electrostatic boundary conditions at a conductor. Spherical symmetry makes the outer surface charge density uniform:

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.

The current density is the moving charge density times its velocity:For a cylindrical Gaussian surface of radius $r<R$ and length $\ell$, cylindrical symmetry makes the electric field radial. Gauss's law gives $E(2\pi r\ell )=\rho \pi r^2\ell /{ϵ}_0$. A circular Ampère's circuital law contour encloses electric current $I(r)=\rho v_0\pi r^2$, giving $B(2\pi r)={\mu }_{0}I(r)$. HenceThe right-hand rule fixes the positive azimuthal direction of the magnetic field for positive electric charge moving along $+z$. Both the electric field and the magnetic field vanish on the axis. These are the steady long-cylinder electric field and magnetic field; end effects and changes of the prescribed beam distribution are excluded.

The integral and differential forms of Gauss's law in electrostatics areHere $\partial \mathcal{V}$ is a closed surface bounding a volume $\mathcal{V}$, $d\mathbf{A}=\hat{\mathbf{n}}dA$ is its outward area vector, $\mathbf{E}$ is the electric field, $\rho$ is total volume charge density, and $Q_{\mathrm{enc}}={\int }_{\mathcal{V}}\rho d^{3}r$ is enclosed electric charge. The divergence theorem converts one form to the other. For ideal sheets or point charges, interpret the differential equation with their distributional charge densities.