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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.

The electric polarization $P$ is electric dipole moment per unit volume. A polarized body acts as the bound volume and surface charge densityIndeed, summing the scalar potential of its dipoles and integrating by parts givesThis is exactly the potential of ${\rho }_{\mathrm{b}}$ and ${\sigma }_{\mathrm{b}}$.

For the central free charge, Gauss's law for the electric displacement field givesThus on the sphere,

Use $\mathbf{E}=-\nabla \varphi$ and Gauss's law $\nabla \cdot \mathbf{E}=\rho /{ϵ}_0$. The divergence theorem and the decay at infinity giveMultiplying by $1/2$ proves that the two expressions for the electrostatic energy agree.

The uniform volume charge density in the thick shell isSpherical symmetry and the integral form of Gauss's law giveThus $|\mathbf{E}|$ is zero inside, rises continuously through the charged region, reaches $Q/(4\pi {ϵ}_0{b}^2)$ at $b$, and then decays as $r^{-2}$. Taking $\varphi (\infty )=0$ and integrating $E=-d\varphi /dr$ givesThe electric potential is continuous, constant inside, decreases smoothly across the charge, and continues as $1/r$ outside.

As $b\to a$, the charge becomes a surface shell. ThenThe potential remains continuous, but the normal electric field jumps by $Q/(4\pi {ϵ}_0{a}^2)$ across the surface charge density.

The field-energy expression givesThe charge-potential expression gives the same result directly because the entire charge lies where the potential equals $\varphi (a)$:This is the electrostatic energy of a uniformly charged spherical shell.

Since $\varphi (a)=Q/(4\pi {ϵ}_{0}a)$ is proportional to $Q$,to first order. This is the work required to bring the additional charge $\delta Q$ from infinity to the potential created by the charge already present.

The method of images replaces the earthed plane by an image charge $-q$ at $(-a,0,0)$. The two Coulomb potentials cancel at $x=0$, so uniqueness for the Dirichlet problem makes the resulting field the physical field in $x>0$. With $\mathbf{r}=(x,y,z)$ and ${\mathbf{e}}_x=(1,0,0)$,and

The electric multipole expansion from part (a) gives, for $r\gg a$,Thus the leading field is that of an electric dipole.

On the plane,Taking the plane normal to be $+{\mathbf{e}}_x$ and using polar coordinates,With the outward normal of the region $x>0$, the sign is reversed. This is consistent with Gauss's law: all electric flux from the real charge terminates on the grounded conductor.

The electrostatic boundary condition gives the induced surface charge densityand its integral isIn the plane $z=0$, the field lines leave the positive charge, meet the conductor normally, and are the right-half-plane portions of the field lines joining the real charge to its negative image.

For the thin spherical charge layer limit, smoothness of $h$ givesThus the shell's total electric charge tends to $4\pi {\rho }_0{a}^3{h}^{\prime }(1)$ while its thickness tends to zero. At each fixed $r>a$, the observation point eventually lies outside the shell, whereas at each fixed $r<a$ the interior electric field is unchanged. The limiting electric field away from the sphere is thereforeIts outward-minus-inward jump isPhysically, the limit is a uniformly charged ball of charge density ${\rho }_0$, together with an infinitesimally thin spherical shell of uniform surface charge densityIndeed its electric charge is $4\pi a^2\sigma$, and the standard normal electric field jump is $\sigma /{ϵ}_0$. The surface charge density can have either sign, or vanish, according to $h^{\prime }(1)$.

The order of limits matters. For every finite $\delta >0$, taking the two one-sided limits at $r=a$ first gives zero jump. The displayed nonzero jump is obtained by first taking $\delta \to 0$ at fixed points on either side, then approaching $a$. The value at the surface itself need not coincide with either trace of the limiting electric field.

Resolve the angular surface charge density into Legendre polynomials:For a single degree $\ell$, write the common surface potential as $V_{\ell }{P}_{\ell }$. Regularity and decay require radial factors $(r/R{)}^{\ell }$ inside and $(R/r{)}^{\ell +1}$ outside. Their exterior-minus-interior radial derivative at $R$ is $-(2\ell +1)V_{\ell }{P}_{\ell }/R$. Thus spherical harmonic matching with a derivative jump gives $V_{\ell }=R{\sigma }_{\ell }/(2\ell +1)$.

Only degrees zero and two occur, yieldingBoth expressions have surface value $2AR/3-2AR{P}_2/15$, and their derivative difference is $-2A/3+(2A/3)P_2=-A{\sin }^2\theta$, checking both interface conditions. The interior quadratic is regular at the origin, and the exterior potential tends to zero.

The leading far-field term is the monopole term. Indeed the total charge is $Q=R^2\int \sigma d\Omega =8\pi A{R}^2/3$, and the normalization of the specified jump gives $Q/(4\pi r)=2A{R}^2/(3r)$. No extra permittivity factor is inserted into the supplied normalization. Regularity here means finite piecewise smooth potential; its normal derivative necessarily jumps at the charged surface.

Boundedness at the origin allows only $r^n$ for nonconstant interior modes; decay at infinity allows only $r^{-n}$ outside. Write the common surface potential in its Fourier series as ${\sum }_{n\ge 1}{v}_n(\theta )$, where $v_n=A_n\cos n\theta +B_n\sin n\theta$. Continuity then makes the corresponding interior and exterior factors $(r/R{)}^n$ and $(R/r{)}^n$.

Their outward radial derivatives at the surface are $n{v}_n/R$ inside and $-n{v}_n/R$ outside. Thus the jump is $-2n{v}_n/R$. Equating it to minus the prescribed surface-charge mode givesTherefore the potential of a charged cylindrical shell with angular Fourier data isEither branch gives the same value at $R$. The mean angular mode is zero: outside, neither a constant nor a logarithm can survive the condition at infinity; continuity then forces the interior constant to zero. Consequently $\Phi (0)=0$.

The absence of a mean term in the surface charge density is essential. A nonzero mean would require a logarithmic exterior potential and would be incompatible with decay at infinity in this two-dimensional reduction. The normalization follows the stated derivative jump, with no additional permittivity factor introduced.

Treat each angular Fourier mode independently. Boundedness at the axis excludes the negative radial powers and the logarithm inside; decay at infinity excludes the positive powers, constant and logarithm outside. For the cosine mode writeContinuity at $R$ gives $B_n=A_n{R}^{2n}$. The jump of the normal derivative is thenThus $A_n=a_n{R}^{1-n}/(2n)$ and $B_n=a_n{R}^{n+1}/(2n)$. The sine mode has the same relation with $b_n$. Therefore the potential of a charged cylindrical shell with angular Fourier data isThere is no angular mean term: the given surface charge density has zero average. The interior constant is zero because its boundary value must match the vanishing exterior mean. A nonzero mean charge per unit length would instead produce a logarithmic electrostatic potential, incompatible with the specified decay at infinity.

The plane separates the exterior into two half-domains. Use an odd Legendre extension for an equatorial Dirichlet plane in each. On the upper hemisphere, extend $u+2u^2$ to the odd function $u+2u|u|$ on the whole sphere. On the lower hemisphere, the appropriate odd extension is $u-2u|u|$. Define, for odd $\ell$,The orthogonality and parity of the Legendre polynomials give the solution in both half-domains asSet its value on $u=0$ to zero. The first coefficients are $b_1=3/2$ and $b_3=7/12$. All odd Legendre polynomials vanish at zero, so the plane condition holds. On the sphere the sum is $2u|u|$, and multiplication by $\mathrm{sgn}(u)$ recovers $2u^2$ on either side. Thus the original sphere data and decay at infinity both hold. A jump of the normal derivative at the sheet is permitted; the sheet can support induced surface charge density.

Differentiate the Green-function representation, valid away from the charged sheet:For $0<x<L$, the two inverse tangent limits at $y\downarrow 0$ are $\pi /2$ and $-\pi /2$; approaching from below reverses both signs. ThusThis agrees with the normal electric field jump obtained by applying Gauss's law to a thin pillbox enclosing unit surface charge density.

First use the usual idealization of a grounded infinite conducting plane at $z=0$, with the real point charge $+q$ at $(0,0,d)$. The method of images replaces the metal, for purposes of calculating the exterior electric field, by an image charge $-q$ at $(0,0,-d)$. The exact electric potential in $z>0$ isIt vanishes on the plane, and the uniqueness theorem for the electrostatic potential identifies it with the physical exterior solution. The tangential electric field cancels on the plane, while the normal components add:The electric field inside the conductor is zero, so the electrostatic boundary condition gives $\sigma ={ϵ}_0{E}_z(0^{+})$. Integrating the surface charge density over the facing plane yieldsThe exact image expression is essential near the charge: using the far-field electric dipole approximation at $\rho$ comparable with $d$ would be invalid.

Electrical boundary condition. The source does not specify grounding or the initial total electric charge of the slab. The boxed total is the grounded infinite-plane answer and the large-plane approximation to the charge on its nearby face. An initially neutral, finite, electrically isolated slab retains zero total electric charge; an approximately $-q$ accumulation near the real charge is then balanced by approximately $+q$ on its remote face and edges. This gives a concrete counterexample to interpreting $-q$ as the unconditional total charge of every large metal slab. The local image formula assumes the infinite-plane limit with no applied background electric field.

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.

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:

Write the uniform surface charge density on the annulus as $\sigma =Q/[\pi (b^2-a^2)]$, and let $x$ be signed distance along its axis. A ring of radius $r$ and width $dr$ contains electric charge $dq=2\pi \sigma rdr$. Its transverse electric field components cancel in pairs, and its electric potential is $dV=dq/[4\pi {ϵ}_0\sqrt{x^2+r^2}]$. Setting $V(\infty )=0$ and integrating gives the axial field of a uniformly charged annulus:At $x=0$, each element's electric field cancels that of the diametrically opposite element, so $\mathbf{E}=0$. There is no surface charge at the centre because $a>0$; the axis electric field is continuous there.

For positive $Q$ and $x>0$, differentiate the signed axial electric field:Its stationary condition is $a^{4/3}(x^2+b^2)=b^{4/3}(x^2+a^2)$. Factoring with $A=a^{2/3}$ and $B=b^{2/3}$ givesThe derivative starts positive, has exactly one positive zero and is negative at large $x$, so this is the positive-side maximum. For either sign of $Q$, the electric field magnitude has equal maxima at $x=\pm x_{\max }$. As $b\to a$, the thin-ring limit is $x_{\max }\to a/\sqrt{2}$.

Use the electric dipole above a grounded conducting plane approximation, valid for $d\ll a$. Let $\hat{\mathbf{z}}$ point upward, and choose $\mathbf{p}=p(\sin \theta \hat{\mathbf{x}}+\cos \theta \hat{\mathbf{z}})$. Reflecting the two constituent point charges and reversing their signs produces an image electric dipole momentat height $-a$. Thus the parallel electric dipole moment reverses and the normal one is unchanged. At the real electric dipole, the separation from its image is $2a\hat{\mathbf{z}}$, giving the image electric fieldThe physical torque is $\mathbf{\tau }=\mathbf{p}\times {\mathbf{E}}_{\mathrm{im}}$, soFor $0<\theta <\pi /2$, its direction reduces $\theta$, tending to align the electric dipole with the upward normal. In general it favours either normal orientation; a perfectly parallel orientation has zero torque but is unstable to tilting.

The conductor-response potential energy includes one half of the real-image interaction:This factor accounts for the induced surface charge changing as the physical source is assembled. Differentiation with respect to height at fixed orientation givesThere is no horizontal force, since translating the entire arrangement parallel to the plane leaves $U$ unchanged. For example, the normal and parallel orientations give respective downward force magnitudes $3p^2/(32\pi {ϵ}_0{a}^4)$ and $3p^2/(64\pi {ϵ}_0{a}^4)$.

A single physical point charge induces total plane electric charge $-q$. One can verify this by integrating its induced surface charge density, $-qh/[2\pi ({\varrho }^2+h^2{)}^{3/2}]$, over the plane. The two physical electric charges therefore induce opposite totals, givingThe local surface charge density is generally nonzero. The zero total holds for the finite neutral pair as well as its ideal electric dipole limit.

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 $\sigma =Q/(4\pi R^2)$.

A large electrical conductor between the point charges provides electrostatic shielding: one must not retain their original mutual Coulomb force through an ideal infinite plate. In the usual broad-plate approximation, let $\sigma$ denote the unperturbed surface charge density on each face. The electrostatic boundary condition gives an outward electric field $E_0=\sigma /{ϵ}_0$. Neglecting local electrostatic induction, the rod's tension then satisfies $mg\tan \varphi =Q{E}_0$, givingThere are two physical qualifications. First, if $\sigma$ means the total charge per projected plate area, shared equally between the two faces, then $E_0=\sigma /(2{ϵ}_0)$ and the same approximation gives $\varphi \simeq 0.989^{\circ }$. The factor of two is a convention about the specified surface charge density, not a different Gauss's law.

Second, a point charge near an electrical conductor makes its surface charge density nonuniform. For an infinite plane with prescribed background electric field $E_0$, the method of images adds an attractive force $k{Q}^2/(4x^2)$, where $x=l\sin \varphi$. The more accurate equilibrium equation isIts larger, stable root is approximately $1.96^{\circ }$ for the per-face convention, or $0.877^{\circ }$ for the total-area convention. This ideal-plane refinement is not an exact answer for an unspecified finite plate: its dimensions and the meaning of the quoted surface charge density would be needed. In particular, adding a uniform-sheet electric field to the original mutual repulsion would describe a different, nonshielding model.

In vacuum, Gauss's law states that the outward electric flux through a closed surface is the enclosed electric charge divided by the vacuum permittivity:Take opposite electric charges $\pm Q$ on the two electrical conductors, with surface charge densities $\pm \sigma =\pm Q/A$. A Gaussian pillbox crossing one isolated large sheet gives $2ES=\sigma S/{ϵ}_0$, so each sheet contributes $\sigma /(2{ϵ}_0)$ on either side. By the superposition principle, the electric fields add between the sheets and cancel outside. Ignoring the fringing field of a capacitor, the gap electric field is $E=\sigma /{ϵ}_0$. The electric potential difference and capacitance of this parallel-plate capacitor are consequentlyThe approximation requires $d$ much smaller than both transverse sheet dimensions. A uniform filling with permittivity $ϵ$ replaces ${ϵ}_0$ by $ϵ$.

Let the original cylinder axis be $\mathbf{r}=0$, and let the hole axis be at transverse vector $\mathbf{a}$, with $|\mathbf{a}|=a$. The charge density is reproduced by a full cylinder of radius $R_1$ and charge density $+\rho$, superposed with a cylinder of radius $R_2$ centred at $\mathbf{a}$ and charge density $-\rho$. This uses the superposition principle for fixed insulating electric charges; the hole is not a hole in an electrical conductor.

For a complete uniformly charged infinite cylinder of radius $R$, a coaxial Gaussian surface of length $\ell$ gives, by Gauss's law,Consequently defineThe complete electric field, including the exterior if desired, isIn the insulating material, $|\mathbf{r}|<R_1$ and $s=|\mathbf{r}-\mathbf{a}|>R_2$, soInside the hole both contributions use their interior formulas, and all dependence on $\mathbf{r}$ cancels:The electric field in the hole is uniform, points from the original axis towards the hole axis for $\rho >0$, and vanishes for a concentric hole. The two expressions agree at the hole wall, as expected with no imposed surface charge density there. Its uniformity is the useful uniform electric field in an off-axis cylindrical cavity result.

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.